Chapter 4 Quadratic Equations (Additional Questions)

Explanation:

A quadratic equation is a polynomial equation of the second degree. It can be written in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers, and the coefficient $a$ is not equal to zero ($a \neq 0$). The key characteristic is that the highest power of the variable ($x$) in the equation must be exactly 2 after all terms are simplified and combined.

Let's examine each given option:

(A) $x^3 - x^2 + 1 = 0$

In this equation, the highest power of the variable $x$ is 3 (from the term $x^3$). Therefore, this is a cubic equation, not a quadratic equation.

(B) $2x^2 - 3\sqrt{x} + 5 = 0$

This equation contains the term $-3\sqrt{x}$. The term $\sqrt{x}$ can be written as $x^{1/2}$. For an expression to be a polynomial (and thus for an equation to be a polynomial equation), the powers of the variable must be non-negative integers. Since the power $1/2$ is not an integer, this equation is not a polynomial equation. Therefore, it cannot be a quadratic equation.

(C) $(x+1)^2 = x^2 - 5$

Let's expand and simplify this equation:

First, expand the left side using the formula $(a+b)^2 = a^2 + 2ab + b^2$:

$(x+1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1$

Now, substitute this expansion back into the original equation:

$x^2 + 2x + 1 = x^2 - 5$

To determine the degree of the equation, move all terms to one side and combine like terms:

$x^2 + 2x + 1 - (x^2 - 5) = 0$

$x^2 + 2x + 1 - x^2 + 5 = 0$

Combine the $x^2$ terms, the $x$ terms, and the constant terms:

$(x^2 - x^2) + 2x + (1 + 5) = 0$

$0x^2 + 2x + 6 = 0$

This simplifies to:

$2x + 6 = 0$

This equation is of the form $ax + b = 0$, where $a=2$ and $b=6$. The highest power of $x$ is 1. This is a linear equation, not a quadratic equation (as the coefficient of $x^2$ is $0$).

(D) $(x-2)(x+3) = x^2 + 5x$

Let's expand and simplify this equation:

First, expand the left side using the distributive property (or FOIL method):

$(x-2)(x+3) = x(x+3) - 2(x+3)$

$= x^2 + 3x - 2x - 6$

Combine the $x$ terms:

$= x^2 + x - 6$

Now, substitute this expansion back into the original equation:

$x^2 + x - 6 = x^2 + 5x$

Move all terms to one side and combine like terms:

$x^2 + x - 6 - (x^2 + 5x) = 0$

$x^2 + x - 6 - x^2 - 5x = 0$

Combine the $x^2$ terms, the $x$ terms, and the constant terms:

$(x^2 - x^2) + (x - 5x) - 6 = 0$

$0x^2 - 4x - 6 = 0$

This simplifies to:

$-4x - 6 = 0$

This equation is of the form $ax + b = 0$, where $a=-4$ and $b=-6$. The highest power of $x$ is 1. This is a linear equation, not a quadratic equation (as the coefficient of $x^2$ is $0$).

Based on the definition of a quadratic equation, which requires the highest power of the variable to be 2 with a non-zero coefficient after simplification, none of the provided options (A), (B), (C), or (D) are quadratic equations as written.

Option (A) is cubic.

Option (B) is not a polynomial equation.

Options (C) and (D) simplify to linear equations.

There appears to be a typographical error in the question or the options as presented, as typically a multiple-choice question of this nature would include exactly one correct answer among the choices.

Explanation:

The standard form of a quadratic equation is given as $ax^2 + bx + c = 0$. In this equation, $x$ is the variable, and $a, b,$ and $c$ are coefficients which are real numbers.

A polynomial equation is classified by its degree, which is the highest power of the variable in the equation after it has been simplified. A quadratic equation is specifically a polynomial equation of the second degree.

For the equation $ax^2 + bx + c = 0$ to be of the second degree, the term with the highest power, which is $ax^2$, must be present. This means the coefficient of the $x^2$ term, which is $a$, cannot be zero.

If $a$ were equal to zero ($a=0$), the equation would become:

$(0)x^2 + bx + c = 0$

which simplifies to:

$bx + c = 0$

This simplified equation is a linear equation (assuming $b \neq 0$), not a quadratic equation, because the highest power of $x$ is 1.

Therefore, for $ax^2 + bx + c = 0$ to be a quadratic equation, the essential condition on the coefficients is that the coefficient of the $x^2$ term must not be zero.

Thus, $a \neq 0$.

There are no restrictions on the values of $b$ or $c$ for the equation to be quadratic. For example, $x^2 - 4 = 0$ (where $a=1, b=0, c=-4$) and $2x^2 + 3x = 0$ (where $a=2, b=3, c=0$) are both valid quadratic equations.

Looking at the options:

(A) $a \neq 0$: This is the correct condition for the equation to be quadratic.

(B) $b \neq 0$: The equation can be quadratic even if $b=0$ (e.g., $x^2 + 5 = 0$).

(C) $c \neq 0$: The equation can be quadratic even if $c=0$ (e.g., $3x^2 - 2x = 0$).

(D) $a=0$: If $a=0$, the equation is linear, not quadratic.

The correct answer is (A).

Explanation:

The degree of a polynomial equation is the highest power of the variable present in the equation after it has been simplified and all like terms have been combined.

A quadratic equation is defined as a polynomial equation of the second degree. Its standard form is given by $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and the coefficient $a$ is not equal to zero ($a \neq 0$).

In the standard form $ax^2 + bx + c = 0$, the powers of the variable $x$ are 2 (in $ax^2$), 1 (in $bx = bx^1$), and 0 (in $c = cx^0$). The highest power among 2, 1, and 0 is 2.

Since the coefficient $a$ is required to be non-zero ($a \neq 0$) for the equation to be quadratic, the term $ax^2$ with the highest power of 2 is always present. Thus, the highest power of $x$ is 2.

Therefore, the degree of a quadratic equation is always 2.

Comparing this with the given options:

(A) 0: A degree of 0 corresponds to a constant equation (e.g., $5=0$, which has no variable, or $5=5$).

(B) 1: A degree of 1 corresponds to a linear equation (e.g., $2x+3=0$).

(C) 2: A degree of 2 corresponds to a quadratic equation.

(D) 3: A degree of 3 corresponds to a cubic equation (e.g., $x^3 - x^2 + 1 = 0$).

The correct answer is (C).

Explanation:

A root of a quadratic equation is a value of the variable that satisfies the equation. In other words, if a number is a root of an equation, substituting that number for the variable will make the equation true.

Given the quadratic equation $x^2 - kx + 6 = 0$ and that 2 is a root.

This means that when we substitute $x=2$ into the equation, the left-hand side will be equal to the right-hand side (which is 0).

Substitute $x=2$ into the equation:

$(2)^2 - k(2) + 6 = 0$

Simplify and solve for $k$:

$4 - 2k + 6 = 0$

Combine the constant terms:

$10 - 2k = 0$

Add $2k$ to both sides of the equation:

Divide both sides by 2 to find the value of $k$:

Thus, the value of $k$ is 5.

Let's check if $x=2$ is indeed a root for the equation $x^2 - 5x + 6 = 0$:

Substitute $x=2$:

$(2)^2 - 5(2) + 6 = 4 - 10 + 6 = 10 - 10 = 0$

Since the left side equals the right side (0), $x=2$ is indeed a root when $k=5$.

Comparing the calculated value of $k=5$ with the given options:

(A) 5

(B) -5

(C) 4

(D) -4

The correct answer is (A).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the sum of its roots (let's call them $\alpha$ and $\beta$) is given by the formula:

Sum of roots $= \alpha + \beta = -\frac{b}{a}$

The given quadratic equation is $3x^2 - 5x + 2 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we can identify the coefficients:

$a = 3$

$b = -5$

$c = 2$

Now, we can calculate the sum of the roots using the formula $-\frac{b}{a}$:

Sum of roots $= -\frac{b}{a} = -\frac{(-5)}{3}$

Sum of roots $= \frac{5}{3}$

Thus, the sum of the roots of the equation $3x^2 - 5x + 2 = 0$ is $\frac{5}{3}$.

Comparing this result with the given options:

(A) $\frac{5}{3}$

(B) $-\frac{5}{3}$

(C) $\frac{2}{3}$

(D) $-\frac{2}{3}$

The correct answer is (A).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the product of its roots (let's call them $\alpha$ and $\beta$) is given by the formula:

Product of roots $= \alpha \cdot \beta = \frac{c}{a}$

The given quadratic equation is $2x^2 + 7x - 4 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we can identify the coefficients:

$a = 2$

$b = 7$

$c = -4$

Now, we can calculate the product of the roots using the formula $\frac{c}{a}$:

Product of roots $= \frac{c}{a} = \frac{-4}{2}$

Product of roots $= -2$

Thus, the product of the roots of the equation $2x^2 + 7x - 4 = 0$ is $-2$.

Comparing this result with the given options:

(A) $\frac{7}{2}$

(B) $-\frac{7}{2}$

(C) $\frac{4}{2} = 2$

(D) $-\frac{4}{2} = -2$

The correct answer is (D).

Explanation:

To find the roots of the quadratic equation $x^2 - 7x + 12 = 0$ by factorization, we need to express the quadratic polynomial as a product of two linear factors.

The given equation is in the standard form $ax^2 + bx + c = 0$, where $a=1$, $b=-7$, and $c=12$.

In the factorization method for a quadratic $x^2 + bx + c = 0$, we look for two numbers whose sum is $b$ and whose product is $c$.

Here, we need two numbers whose sum is $-7$ and whose product is $12$.

Let the two numbers be $p$ and $q$. We need:

$p + q = -7$

$p \cdot q = 12$

Let's list the pairs of factors of 12 and check their sums:

Factors of 12: (1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)

Sum of factors:

$1 + 12 = 13$

$2 + 6 = 8$

$3 + 4 = 7$

$-1 + (-12) = -13$

$-2 + (-6) = -8$

$-3 + (-4) = -7$

The pair of numbers that satisfy both conditions (sum is -7 and product is 12) is $-3$ and $-4$.

Now, we rewrite the middle term $-7x$ as the sum of two terms using these numbers, i.e., $-3x$ and $-4x$.

$x^2 - 7x + 12 = 0$

$x^2 - 3x - 4x + 12 = 0$

Next, we group the terms and factor out the common factors from each group:

$(x^2 - 3x) + (-4x + 12) = 0$

Factor $x$ from the first group: $x(x - 3)$

Factor $-4$ from the second group: $-4(x - 3)$

So, the equation becomes:

$x(x - 3) - 4(x - 3) = 0$

Now, factor out the common binomial factor $(x - 3)$:

$(x - 3)(x - 4) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

So, we set each factor equal to zero to find the roots:

$x - 3 = 0 \quad$ or $\quad x - 4 = 0$

Solving for $x$ in each case:

$x - 3 = 0 \implies x = 3$

$x - 4 = 0 \implies x = 4$

The roots of the quadratic equation $x^2 - 7x + 12 = 0$ are 3 and 4.

Comparing the roots with the given options:

(A) 3, 4

(B) -3, -4

(C) 3, -4

(D) -3, 4

The correct answer is (A).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the discriminant is a value that determines the nature of the roots (real or complex, distinct or equal) without actually calculating the roots.

The quadratic formula for finding the roots of $ax^2 + bx + c = 0$ is given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The expression under the square root in the quadratic formula, $b^2 - 4ac$, is called the discriminant.

It is often denoted by the Greek letter delta ($\Delta$).

Discriminant $= \Delta = b^2 - 4ac$

The value of the discriminant helps determine the nature of the roots:

• If $b^2 - 4ac > 0$, the equation has two distinct real roots.
• If $b^2 - 4ac = 0$, the equation has two equal real roots (a repeated root).
• If $b^2 - 4ac < 0$, the equation has no real roots (it has two distinct complex conjugate roots).

Comparing the formula $b^2 - 4ac$ with the given options:

(A) $b^2 + 4ac$: This is not the correct formula for the discriminant.

(B) $b^2 - 4ac$: This is the correct formula for the discriminant.

(C) $\sqrt{b^2 - 4ac}$: This is the square root of the discriminant, which appears in the quadratic formula, but it is not the discriminant itself.

(D) $-b \pm \sqrt{b^2 - 4ac}$: This is the numerator part of the quadratic formula for finding the roots, not the discriminant.

The correct answer is (B).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$ (where $a, b, c$ are real numbers and $a \neq 0$), the discriminant is given by the formula $D = b^2 - 4ac$.

The value of the discriminant helps us determine the nature of the roots of the quadratic equation without explicitly calculating them using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

There are three cases for the discriminant, each corresponding to a different nature of the roots:

1. If $D = b^2 - 4ac > 0$:

The expression under the square root in the quadratic formula ($\sqrt{D}$) is a positive real number. This results in two different values for $\frac{-b \pm \sqrt{D}}{2a}$, one with $+$ and one with $-$. Therefore, the equation has two distinct real roots.

2. If $D = b^2 - 4ac = 0$:

The expression under the square root is zero ($\sqrt{0} = 0$). The quadratic formula becomes $x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$. This means there is only one value for the root, or rather, two equal real roots.

3. If $D = b^2 - 4ac < 0$:

The expression under the square root is a negative number. The square root of a negative number is not a real number; it's an imaginary number. This leads to two distinct complex conjugate roots. In this case, there are no real roots.

The question states that the discriminant is greater than zero ($D > 0$). Based on the analysis above, when $D > 0$, the quadratic equation has two distinct real roots.

Comparing this conclusion with the given options:

(A) Real and distinct: This matches our conclusion when $D > 0$.

(B) Real and equal: This occurs when $D = 0$.

(C) No real roots: This occurs when $D < 0$.

(D) Complex: This is a general term for roots that are not purely real. When $D < 0$, the roots are complex and distinct. When $D > 0$, the roots are real, which are a subset of complex numbers, but the term "complex" in options usually implies non-real roots. Option (A) is more specific and accurate for $D > 0$ in the context of distinguishing from real and equal or no real roots.

The correct answer is (A).

Explanation:

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ (where $a, b, c$ are real numbers and $a \neq 0$) is given by $D = b^2 - 4ac$. It is a crucial part of the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ and determines the nature of the roots.

We analyze the nature of the roots based on the value of the discriminant:

1. If $D > 0$: The term $\sqrt{D}$ is a positive real number. The quadratic formula becomes $x = \frac{-b \pm \sqrt{D}}{2a}$, which gives two different real values for $x$: $\frac{-b + \sqrt{D}}{2a}$ and $\frac{-b - \sqrt{D}}{2a}$. The roots are real and distinct.

2. If $D = 0$: The term $\sqrt{D}$ is $\sqrt{0} = 0$. The quadratic formula becomes $x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$. In this case, both possible values for $x$ are the same. The equation has two real and equal roots (sometimes called a repeated root or a double root).

3. If $D < 0$: The term $\sqrt{D}$ is the square root of a negative number, which is an imaginary number. The roots are given by $x = \frac{-b \pm \sqrt{|D|}i}{2a}$ (where $i = \sqrt{-1}$). These roots are complex conjugates and are not real. The equation has no real roots.

The question asks for the nature of the roots when the discriminant is equal to zero ($D = 0$). According to our analysis, when $D = 0$, the roots are real and equal.

Comparing this conclusion with the given options:

(A) Real and distinct: This happens when $D > 0$.

(B) Real and equal: This happens when $D = 0$.

(C) No real roots: This happens when $D < 0$.

(D) Irrational: Real roots are irrational if the discriminant is positive and not a perfect square. Equal roots (when $D=0$) are always rational if the coefficients $a$ and $b$ are rational, since the root is $\frac{-b}{2a}$. This option is not the primary or defining characteristic when $D=0$.

The correct answer is (B).

Explanation:

To determine the nature of the roots of a quadratic equation, we examine the value of its discriminant. For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the discriminant, denoted by $D$ or $\Delta$, is calculated using the formula:

$D = b^2 - 4ac$

The given quadratic equation is $x^2 + x + 1 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we can identify the coefficients:

$a = 1$ (coefficient of $x^2$)

$b = 1$ (coefficient of $x$)

$c = 1$ (constant term)

Now, we calculate the discriminant $D$ by substituting the values of $a$, $b$, and $c$ into the formula:

$D = (1)^2 - 4(1)(1)$

$D = 1 - 4$

$D = -3$

The value of the discriminant is $D = -3$.

The nature of the roots is determined as follows:

• If $D > 0$, the quadratic equation has two distinct real roots.
• If $D = 0$, the quadratic equation has two equal real roots (a repeated root).
• If $D < 0$, the quadratic equation has no real roots (it has two distinct complex conjugate roots).

Since the calculated discriminant $D = -3$ is less than zero ($D < 0$), the quadratic equation $x^2 + x + 1 = 0$ has no real roots.

Comparing our conclusion with the given options:

(A) Real and distinct: This is true only when $D > 0$.

(B) Real and equal: This is true only when $D = 0$.

(C) No real roots: This is true when $D < 0$. Our discriminant is $-3$, which is less than 0.

(D) Rational: This relates to the discriminant being a perfect square (and non-negative), which determines if real roots are rational. Since $D < 0$, the roots are not real, so this option is not applicable here in the context of describing the primary nature.

The correct answer is (C).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$ (where $a, b, c$ are real numbers and $a \neq 0$), the nature of its roots is determined by the value of the discriminant, $D$, given by the formula:

$D = b^2 - 4ac$

The conditions for the nature of roots based on the discriminant are:

• If $D > 0$, the roots are real and distinct.
• If $D = 0$, the roots are real and equal.
• If $D < 0$, there are no real roots (the roots are complex and distinct).

We need to find the equation that has real and equal roots. This occurs when the discriminant $D = 0$. We will calculate the discriminant for each given option:

(A) $x^2 - 4x + 4 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=-4$, $c=4$.

Discriminant $D = b^2 - 4ac = (-4)^2 - 4(1)(4) = 16 - 16 = 0$.

Since $D=0$, the roots are real and equal.

(B) $x^2 - 5x + 6 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=-5$, $c=6$.

Discriminant $D = b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1$.

Since $D=1 > 0$, the roots are real and distinct.

(C) $x^2 + 2x + 3 = 0$

Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=2$, $c=3$.

Discriminant $D = b^2 - 4ac = (2)^2 - 4(1)(3) = 4 - 12 = -8$.

Since $D=-8 < 0$, there are no real roots.

(D) $x^2 - 9 = 0$

This can be written as $x^2 + 0x - 9 = 0$. Comparing with $ax^2 + bx + c = 0$, we have $a=1$, $b=0$, $c=-9$.

Discriminant $D = b^2 - 4ac = (0)^2 - 4(1)(-9) = 0 - (-36) = 36$.

Since $D=36 > 0$, the roots are real and distinct ($x^2=9 \implies x = \pm 3$).

The only quadratic equation among the given options that has a discriminant equal to zero is $x^2 - 4x + 4 = 0$.

The correct answer is (A).

Explanation:

We are asked to find the roots of the quadratic equation $x^2 - 2x + 1 = 0$. We can solve this equation using the factorization method, the completing the square method, or the quadratic formula.

Method 1: Factorization

The given equation is $x^2 - 2x + 1 = 0$. This is a quadratic trinomial of the form $ax^2 + bx + c = 0$ with $a=1$, $b=-2$, and $c=1$.

We look for two numbers whose product is $ac = (1)(1) = 1$ and whose sum is $b = -2$. The numbers are $-1$ and $-1$, since $(-1) \times (-1) = 1$ and $(-1) + (-1) = -2$.

We rewrite the middle term, $-2x$, as the sum of $-x$ and $-x$:

$x^2 - x - x + 1 = 0$

Now, we group the terms and factor by grouping:

$(x^2 - x) + (-x + 1) = 0$

Factor out common terms from each group:

$x(x - 1) - 1(x - 1) = 0$

Notice that $(x - 1)$ is a common factor in both terms. Factor it out:

$(x - 1)(x - 1) = 0$

This can be written as:

$(x - 1)^2 = 0$

For the equation $(x - 1)^2 = 0$ to be true, the factor $(x - 1)$ must be equal to zero:

$x - 1 = 0$

Solving for $x$:

$x = 1$

Since the equation is $(x-1)(x-1)=0$, both factors give the same root. Thus, the roots are 1 and 1.

Method 2: Using the Discriminant and Quadratic Formula

The equation is $x^2 - 2x + 1 = 0$. Here, $a=1$, $b=-2$, $c=1$.

First, calculate the discriminant $D = b^2 - 4ac$:

$D = (-2)^2 - 4(1)(1) = 4 - 4 = 0$

Since the discriminant $D=0$, the quadratic equation has real and equal roots.

Now, use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-2) \pm \sqrt{0}}{2(1)}$

$x = \frac{2 \pm 0}{2}$

This gives the roots:

$x_1 = \frac{2 + 0}{2} = \frac{2}{2} = 1$

$x_2 = \frac{2 - 0}{2} = \frac{2}{2} = 1$

So the roots are 1 and 1.

Comparing the roots with the given options:

(A) 1, -1

(B) 1, 1

(C) 2, 1

(D) -1, -1

The correct answer is (B).

Explanation:

We are given a problem involving two consecutive positive integers and their product.

Let the first positive integer be $x$.

Since the integers are consecutive, the next positive integer will be one more than the first one. So, the second consecutive positive integer is $x+1$.

We are given that the product of these two integers is 306.

The product of $x$ and $x+1$ is written as $x \times (x+1)$ or simply $x(x+1)$.

So, the relationship described in the problem can be written as an equation:

To represent this situation as a quadratic equation in standard form ($ax^2 + bx + c = 0$), we need to expand the left side and move all terms to one side of the equation, setting the other side to zero.

Expand the left side of the equation $x(x+1) = 306$:

$x \times x + x \times 1 = 306$

$x^2 + x = 306$

Now, move the constant term (306) from the right side to the left side. When we move a term across the equality sign, we change its sign.

$x^2 + x - 306 = 0$

This equation, $x^2 + x - 306 = 0$, is a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a=1$, $b=1$, and $c=-306$. This equation represents the given situation.

Now let's compare our derived equation with the given options:

(A) $x(x+1) + 306 = 0 \implies x^2 + x + 306 = 0$ (Incorrect, the constant term has the wrong sign)

(B) $x(x+1) - 306 = 0 \implies x^2 + x - 306 = 0$ (Correct, this matches our derived equation)

(C) $x + (x+1) = 306 \implies 2x + 1 = 306 \implies 2x - 305 = 0$ (Incorrect, this is a linear equation representing the *sum* being 306, not the product)

(D) $x(x+1) = -306 \implies x^2 + x = -306 \implies x^2 + x + 306 = 0$ (Incorrect, this equation implies the product is -306, not 306, and it simplifies to the same incorrect equation as option A)

The quadratic equation that represents the situation is $x(x+1) - 306 = 0$, which is equivalent to $x^2 + x - 306 = 0$.

The correct answer is (B).

Explanation:

Let's analyze the Assertion (A) and the Reason (R) separately.

Evaluation of Assertion (A):

The assertion states that the equation $(x-2)^2 + 1 = 2x - 3$ is a quadratic equation.

To verify this, we need to simplify the given equation and see if it can be written in the standard form of a quadratic equation, $ax^2 + bx + c = 0$, with $a \neq 0$.

The given equation is:

$(x-2)^2 + 1 = 2x - 3$

First, expand the term $(x-2)^2$ using the formula $(a-b)^2 = a^2 - 2ab + b^2$:

$(x-2)^2 = x^2 - 2(x)(2) + (2)^2 = x^2 - 4x + 4$

Substitute this expansion back into the equation:

$x^2 - 4x + 4 + 1 = 2x - 3$

Combine the constant terms on the left side:

$x^2 - 4x + 5 = 2x - 3$

Now, move all terms to one side to set the equation to zero. Subtract $2x$ from both sides and add 3 to both sides:

$x^2 - 4x - 2x + 5 + 3 = 0$

Combine the like terms:

$x^2 + (-4x - 2x) + (5 + 3) = 0$

$x^2 - 6x + 8 = 0$

The simplified equation is $x^2 - 6x + 8 = 0$.

This equation is in the standard form $ax^2 + bx + c = 0$, with $a=1$, $b=-6$, and $c=8$. Since the coefficient of the $x^2$ term, $a=1$, is not equal to zero, this is indeed a quadratic equation.

Therefore, Assertion (A) is true.

Evaluation of Reason (R):

The reason states that a quadratic equation is any equation that can be put in the form $ax^2 + bx + c = 0$, where $a \neq 0$.

This statement is the formal and correct definition of a quadratic equation in one variable. The condition $a \neq 0$ is essential because if $a=0$, the $x^2$ term vanishes, and the equation becomes a linear equation ($bx+c=0$, assuming $b \neq 0$).

Therefore, Reason (R) is true.

Relationship between Assertion (A) and Reason (R):

Assertion (A) makes a claim about a specific equation being quadratic. Reason (R) provides the definition of what constitutes a quadratic equation.

Our method for verifying Assertion (A) was to simplify the given equation and check if it fits the definition provided in Reason (R). Since the simplified equation ($x^2 - 6x + 8 = 0$) matches the form $ax^2 + bx + c = 0$ with $a \neq 0$ (as stated in R), Reason (R) provides the fundamental criterion that explains why Assertion (A) is true.

Thus, Reason (R) is the correct explanation of Assertion (A).

Based on the analysis, both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

The correct answer is (A).

Explanation:

Let's analyze the Assertion (A) and the Reason (R) separately to determine their truth values.

Evaluation of Assertion (A):

The assertion claims that the roots of the quadratic equation $x^2 + 4x + 5 = 0$ are real.

To determine the nature of the roots of a quadratic equation $ax^2 + bx + c = 0$, we examine its discriminant, $D = b^2 - 4ac$.

• If $D > 0$, the roots are real and distinct.
• If $D = 0$, the roots are real and equal.
• If $D < 0$, the roots are not real (they are complex and distinct).

For the given equation $x^2 + 4x + 5 = 0$, the coefficients are $a=1$, $b=4$, and $c=5$.

Let's calculate the discriminant:

$D = b^2 - 4ac = (4)^2 - 4(1)(5)$

$D = 16 - 20$

$D = -4$

Since the discriminant $D = -4$ is less than zero ($D < 0$), the roots of the equation $x^2 + 4x + 5 = 0$ are not real. They are complex conjugate roots.

Therefore, Assertion (A) is false.

Evaluation of Reason (R):

The reason states that the discriminant of $x^2 + 4x + 5 = 0$ is $4^2 - 4(1)(5) = 16 - 20 = -4$, which is less than 0.

As calculated above, the discriminant of $x^2 + 4x + 5 = 0$ with $a=1, b=4, c=5$ is indeed $D = (4)^2 - 4(1)(5) = 16 - 20 = -4$.

The statement that $-4$ is less than 0 is also true.

Therefore, Reason (R) is true.

Conclusion:

We have determined that Assertion (A) is false and Reason (R) is true.

Let's look at the options based on these findings:

(A) Both A and R are true and R is the correct explanation of A. (Incorrect, A is false)

(B) Both A and R are true but R is not the correct explanation of A. (Incorrect, A is false)

(C) A is true but R is false. (Incorrect, A is false and R is true)

(D) A is false but R is true. (Correct, this matches our findings)

The correct answer is (D).

Explanation:

For a quadratic equation $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the discriminant, $D = b^2 - 4ac$, determines the nature of the roots. The relationship between the discriminant and the nature of the roots is as follows:

(i) $D > 0$: When the discriminant is positive, the term $\sqrt{D}$ in the quadratic formula ($\frac{-b \pm \sqrt{D}}{2a}$) is a positive real number. This leads to two different values for the roots, $\frac{-b + \sqrt{D}}{2a}$ and $\frac{-b - \sqrt{D}}{2a}$. Both roots are real and distinct.

This matches with (c) Real and distinct roots.

(ii) $D = 0$: When the discriminant is zero, the term $\sqrt{D}$ is $\sqrt{0} = 0$. The quadratic formula becomes $x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$. Both values of the roots are the same. The equation has two real and equal roots.

This matches with (d) Real and equal roots.

(iii) $D < 0$: When the discriminant is negative, the term $\sqrt{D}$ involves the square root of a negative number, which is an imaginary number. The roots are $\frac{-b \pm \sqrt{|D|}i}{2a}$, which are complex conjugate roots. These roots are not real.

This matches with (a) No real roots.

(iv) $D \ge 0$: This condition means the discriminant is either greater than zero ($D > 0$) or equal to zero ($D = 0$).

• If $D > 0$, the roots are real and distinct.
• If $D = 0$, the roots are real and equal.

In both cases ($D > 0$ or $D = 0$), the roots are real. So, $D \ge 0$ means the equation has real roots (either distinct or equal).

This matches with (b) Real roots.

Summarizing the matches:

• (i) $D > 0$ matches (c) Real and distinct roots.
• (ii) $D = 0$ matches (d) Real and equal roots.
• (iii) $D < 0$ matches (a) No real roots.
• (iv) $D \ge 0$ matches (b) Real roots.

The correct matching sequence is (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b).

Comparing this sequence with the given options:

(A) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

(B) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)

(C) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b)

(D) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)

Option (A) matches the correct sequence.

The correct answer is (A).

Solution:

Let the length of the rectangular park be $l$ metres and the breadth be $b$ metres, as given in the problem.

According to the first condition, the breadth of the rectangular park is 3 m less than its length.

The area of the rectangular park is given by the product of its length and breadth.

Substitute the expression for $b$ from the relationship $b=l-3$:

Now, consider the existing triangular park. It is an isosceles triangle with base 12 m and altitude 5 m.

The area of a triangle is given by the formula $\frac{1}{2} \times \text{base} \times \text{altitude}$.

Calculate the area of the triangular park:

According to the second condition, the area of the rectangular park is 4 metres more than the area of the triangular park.

Substitute the expressions for the areas into this equation:

Simplify the equation:

This equation represents the area of the rectangular park according to the given conditions. To see if it matches any of the options, let's look at the options provided:

(A) $l(l-3) = \frac{1}{2}(12)(5) + 4$

(B) $l(l+3) = \frac{1}{2}(12)(5) + 4$

(C) $l(3-l) = 60 + 4$

(D) $l(l-3) + 4 = 30$

Let's simplify the right-hand side of options (A) and (B):

$\frac{1}{2}(12)(5) + 4 = 30 + 4 = 34$

So, option (A) is $l(l-3) = 34$. This matches our derived equation.

Option (B) is $l(l+3) = 34$. This is incorrect because the breadth is $l-3$, not $l+3$.

Option (C) is $l(3-l) = 60 + 4 = 64$. The left side is $3l - l^2$. So, $3l - l^2 = 64$, or $l^2 - 3l + 64 = 0$. This is incorrect, and the breadth is $l-3$, not $3-l$ (assuming $l>3$). Also, the area of the triangle is 30, not related to 60 here.

Option (D) is $l(l-3) + 4 = 30$. Rearranging this gives $l(l-3) = 30 - 4 = 26$. This is incorrect because the area of the rectangle is the area of the triangle *plus* 4, not minus 4 from 30.

The equation representing the area of the rectangular park according to the given conditions is $l(l-3) = \frac{1}{2}(12)(5) + 4$.

The correct answer is (A).

Solution:

From the Case Study described in Question 18, we have the following information:

• Length of the rectangular park = $l$ metres
• Breadth of the rectangular park = $b$ metres
• Breadth is 3 m less than the length: $b = l - 3$
• Area of the rectangular park = $l \times b = l(l-3)$
• Area of the existing isosceles triangular park = $\frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times 12 \times 5 = 30$ square metres
• Area of the rectangular park is 4 metres more than the area of the triangular park.

Based on the last condition, we can write the equation for the area of the rectangular park:

To find the length $l$, we need to solve this quadratic equation. Expand the left side and move the constant term to the left side to get the standard form $al^2 + bl + c = 0$:

$l^2 - 3l = 34$

$l^2 - 3l - 34 = 0$

This is a quadratic equation with coefficients $a=1$, $b=-3$, and $c=-34$. We can find the roots using the quadratic formula $l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

First, calculate the discriminant, $D = b^2 - 4ac$:

$D = (-3)^2 - 4(1)(-34)$

$D = 9 - (-136)$

$D = 9 + 136$

Now, substitute the values of $a$, $b$, and $D$ into the quadratic formula:

$l = \frac{-(-3) \pm \sqrt{145}}{2(1)}$

$l = \frac{3 \pm \sqrt{145}}{2}$

The two roots are $l_1 = \frac{3 + \sqrt{145}}{2}$ and $l_2 = \frac{3 - \sqrt{145}}{2}$.

Since $l$ represents the length of a park, it must be a positive value. $\sqrt{145}$ is a positive real number (approximately 12.04). The value $\frac{3 - \sqrt{145}}{2}$ will be negative (since $3 < \sqrt{145}$). Therefore, the length of the park is $l = \frac{3 + \sqrt{145}}{2}$.

Note that $\sqrt{145}$ is not an integer, so the length $l = \frac{3 + \sqrt{145}}{2}$ is an irrational number. Let's approximate its value:

Since $12^2 = 144$ and $13^2 = 169$, $\sqrt{145}$ is slightly greater than 12.

$l \approx \frac{3 + 12.04}{2} \approx \frac{15.04}{2} \approx 7.52$ metres.

Comparing this value with the given options (A) 7, (B) 9, (C) 11, (D) 13, we see that none of the options match the calculated length based on the problem statement as written. This indicates a potential inconsistency or typo in the problem statement's numerical values or the provided options.

It is common in such multiple-choice questions that the intended problem has integer solutions matching one of the options. Let's consider the possibility of a typo in the numbers provided in the case study, assuming that one of the options for length (7, 9, 11, or 13) is the correct integer length and the breadth is still $l-3$. The area of the rectangle would then be $l(l-3)$. The area of the triangle is 30. The condition is Area_rectangle = Area_triangle + 4.

• If $l=7$, Area_rectangle $= 7(7-3) = 7(4) = 28$. Difference from triangle area = $28 - 30 = -2$. (Requires difference to be -2, not 4)
• If $l=9$, Area_rectangle $= 9(9-3) = 9(6) = 54$. Difference from triangle area = $54 - 30 = 24$. (Requires difference to be 24, not 4)
• If $l=11$, Area_rectangle $= 11(11-3) = 11(8) = 88$. Difference from triangle area = $88 - 30 = 58$.
• If $l=13$, Area_rectangle $= 13(13-3) = 13(10) = 130$. Difference from triangle area = $130 - 30 = 100$.

None of the integer options for $l$ satisfy the condition that the rectangular area is exactly 4 more than the triangular area (30). However, if the intended area of the rectangular park was 28 m$^2$, then the length would be 7m ($l(l-3)=28 \implies l^2-3l-28=0 \implies (l-7)(l+4)=0 \implies l=7$). This would mean the area of the rectangle is 2m *less* than the area of the triangle, not 4m more. Alternatively, if the altitude of the triangle was intended to be 4m instead of 5m, the triangle area would be $\frac{1}{2} \times 12 \times 4 = 24$ m$^2$. Then Area_rectangle $= 24 + 4 = 28$ m$^2$, which leads to $l=7$m. This seems the most likely intended scenario given option (A).

Assuming there was a typo in the triangle's altitude and it was meant to be 4m instead of 5m, the area of the triangle would be 24 m$^2$. The area of the rectangular park would then be $24 + 4 = 28$ m$^2$. The equation for the length $l$ would be $l(l-3) = 28$, which is $l^2 - 3l - 28 = 0$. Factoring this equation, we look for two numbers that multiply to -28 and add to -3. These numbers are -7 and 4. So, the equation can be factored as $(l-7)(l+4) = 0$. The roots are $l=7$ and $l=-4$. Since length must be positive, $l=7$.

Based on the high probability of a typo in the question designed to yield an integer answer from the options, and the plausibility of a small error in one of the given numbers (like the triangle's altitude), we infer that the intended length is 7m, corresponding to the scenario where the rectangular area is 28 m$^2$.

The correct answer, based on the likely intended question with a minor typo, is (A).

Explanation:

The formula for finding the roots of a quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$) is known as the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This formula is a general solution that can be used to find the roots of any quadratic equation, regardless of whether it is factorable or not. The derivation of this formula involves transforming the standard quadratic equation into a form where the variable $x$ is isolated.

The standard method used to derive the quadratic formula from the general form $ax^2 + bx + c = 0$ is the method of completing the square.

The steps involved in deriving the formula by completing the square are:

1. Divide the entire equation by $a$ (since $a \neq 0$).

2. Move the constant term $\frac{c}{a}$ to the right side of the equation.

3. Add the square of half the coefficient of $x$ (which is $(\frac{b}{2a})^2 = \frac{b^2}{4a^2}$) to both sides of the equation to make the left side a perfect square trinomial.

4. Rewrite the left side as the square of a binomial, $(x + \frac{b}{2a})^2$, and simplify the right side.

5. Take the square root of both sides of the equation.

6. Solve for $x$ by isolating the $x$ term.

This process leads directly to the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Comparing this method with the given options:

(A) Factorization: Factorization is a method for solving quadratic equations, but the quadratic formula is not derived *by* factorization. Instead, factorization can sometimes be applied to quadratic equations *after* they are solved using the formula (or other methods), or the formula can be used when factorization is difficult or impossible.

(B) Completing the square: This is the standard algebraic method used to derive the quadratic formula.

(C) Substitution: Substitution is a technique used in solving equations (often systems of equations or simplifying complex expressions), but it is not the primary method for deriving the general quadratic formula from $ax^2+bx+c=0$.

(D) Graphical method: The graphical method helps visualize the roots (x-intercepts) of a quadratic equation, but it does not provide an algebraic derivation of the quadratic formula itself.

Therefore, the quadratic formula is derived using the method of completing the square.

The correct answer is (B).

Solution:

We are asked to find the roots of the quadratic equation $x^2 - x - 2 = 0$. A root of an equation is a value of the variable that makes the equation true when substituted.

We can find the roots by solving the equation. We will use the factorization method here, as it is suitable for this equation.

The given equation is $x^2 - x - 2 = 0$.

We need to find two numbers that multiply to the constant term $c = -2$ and add up to the coefficient of the $x$ term, $b = -1$.

Let the two numbers be $p$ and $q$. We need $p \cdot q = -2$ and $p + q = -1$.

Let's consider pairs of factors of -2:

• $1$ and $-2$: Product is $1 \times (-2) = -2$. Sum is $1 + (-2) = -1$. This pair works.
• $-1$ and $2$: Product is $(-1) \times 2 = -2$. Sum is $-1 + 2 = 1$. This pair does not work.

The two numbers are $1$ and $-2$. We can rewrite the middle term $-x$ as $+1x - 2x$:

$x^2 + x - 2x - 2 = 0$

Now, group the terms and factor out the common factor from each group:

$(x^2 + x) + (-2x - 2) = 0$

$x(x + 1) - 2(x + 1) = 0$

Factor out the common binomial term $(x + 1)$:

$(x + 1)(x - 2) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Set each factor equal to zero and solve for $x$:

$x + 1 = 0 \implies x = -1$

$x - 2 = 0 \implies x = 2$

The roots of the equation $x^2 - x - 2 = 0$ are -1 and 2.

Now, let's check which of the given options are the roots we found:

• (A) -1: Yes, this is one of the roots.
• (B) 1: No, this is not one of the roots.
• (C) 2: Yes, this is the other root.
• (D) -2: No, this is not one of the roots.

The options that are roots of the equation are (A) and (C).

Alternatively, we could test each option by substituting the value into the equation $x^2 - x - 2 = 0$ and checking if the result is 0.

Check option (A) $x=-1$: $(-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0$. True.

Check option (B) $x=1$: $(1)^2 - (1) - 2 = 1 - 1 - 2 = -2$. False.

Check option (C) $x=2$: $(2)^2 - (2) - 2 = 4 - 2 - 2 = 0$. True.

Check option (D) $x=-2$: $(-2)^2 - (-2) - 2 = 4 + 2 - 2 = 4$. False.

Both methods confirm that the roots are -1 and 2.

The options that are roots of the equation are (A) and (C).

Explanation:

A quadratic equation in the standard form $ax^2 + bx + c = 0$ (where $a, b,$ and $c$ are real numbers and $a \neq 0$) has real and equal roots if and only if its discriminant ($D$) is equal to zero.

The discriminant is given by the formula:

$D = b^2 - 4ac$

For the equation $4x^2 - 12x + k = 0$ to have real and equal roots, the discriminant must be 0:

$D = 0$

First, identify the coefficients of the given quadratic equation by comparing it with the standard form $ax^2 + bx + c = 0$:

$a = 4$

$b = -12$

$c = k$

Now, substitute these coefficients into the discriminant formula and set it equal to 0:

$D = (-12)^2 - 4(4)(k) = 0$

Calculate the square of $b$ and the product $4ac$:

$(-12)^2 = 144$

$4(4)(k) = 16k$

Substitute these values back into the discriminant equation:

$144 - 16k = 0$

Now, solve this linear equation for $k$. Add $16k$ to both sides of the equation:

Divide both sides by 16:

To perform the division $\frac{144}{16}$, you can recognize that $16 \times 9 = 144$ or perform the division:

Thus, the value of $k$ for which the quadratic equation $4x^2 - 12x + k = 0$ has real and equal roots is 9.

Comparing the calculated value of $k=9$ with the given options:

(A) 9

(B) -9

(C) 36

(D) 0

The correct answer is (A).

Solution:

Let the uniform speed of the train be $x$ km/hr.

The distance covered is 480 km.

The formula relating distance, speed, and time is: Time = Distance / Speed.

In the first case, the time taken to cover 480 km at a uniform speed of $x$ km/hr is:

In the second case, the speed is 8 km/hr less than the uniform speed. So, the new speed is $(x-8)$ km/hr.

The time taken to cover the same distance (480 km) at this reduced speed is:

According to the problem statement, if the speed had been 8 km/hr less, it would have taken 3 hours more to cover the same distance. This means the time taken in the second case ($T_2$) is 3 hours more than the time taken in the first case ($T_1$).

Substitute the expressions for $T_1$ and $T_2$ into this equation:

To find the equation that represents this situation among the given options, we can rearrange this equation:

Subtract $\frac{480}{x}$ from both sides of the equation:

This equation represents the situation described in the problem.

Let's compare this equation with the given options:

(A) $\frac{480}{x} - \frac{480}{x-8} = 3$ (This implies $T_1 - T_2 = 3$, which is incorrect as $T_2$ should be greater than $T_1$)

(B) $\frac{480}{x-8} - \frac{480}{x} = 3$ (This implies $T_2 - T_1 = 3$, which is correct)

(C) $\frac{480}{x} - 3 = \frac{480}{x-8}$ (Rearranging gives $\frac{480}{x} - \frac{480}{x-8} = 3$, same as option A - incorrect)

(D) $480x = (x-8)(3)$ (This equation does not relate distance, speed, and time correctly for the given scenario - incorrect)

The equation $\frac{480}{x-8} - \frac{480}{x} = 3$ correctly represents the situation.

If we were to convert this into the standard quadratic form $ax^2 + bx + c = 0$, we would find the common denominator on the left side:

$\frac{480x - 480(x-8)}{x(x-8)} = 3$

$\frac{480x - 480x + 3840}{x^2 - 8x} = 3$

$\frac{3840}{x^2 - 8x} = 3$

Multiply both sides by $x^2 - 8x$ (assuming $x \neq 0$ and $x \neq 8$):

$3840 = 3(x^2 - 8x)$

$3840 = 3x^2 - 24x$

Move all terms to one side:

$0 = 3x^2 - 24x - 3840$

Divide by 3:

$x^2 - 8x - 1280 = 0$

This is the quadratic equation that would be solved to find the value of $x$. However, the question asks for the equation representing the situation among the given options, which is in fractional form.

The correct equation from the options is $\frac{480}{x-8} - \frac{480}{x} = 3$.

The correct answer is (B).

Solution:

We are given a problem involving two consecutive even integers.

Let the smaller even integer be $x$, as specified in the problem.

Consecutive even integers differ by 2. For example, if one even integer is 6, the next is $6+2=8$. If one is -10, the next is $-10+2=-8$.

So, if the smaller even integer is $x$, the next consecutive even integer is $x+2$.

The problem states that the sum of the squares of these two consecutive even integers is 100.

Square of the first integer ($x$): $x^2$

Square of the second integer ($x+2$): $(x+2)^2$

The sum of these squares is 100:

This equation represents the situation described in the problem. We can expand it to see the standard quadratic form, but the question asks for the equation as is.

Expanding the equation: $x^2 + (x^2 + 4x + 4) = 100$

$2x^2 + 4x + 4 = 100$

Move the constant term to the left side:

$2x^2 + 4x + 4 - 100 = 0$

$2x^2 + 4x - 96 = 0$

Divide by 2:

$x^2 + 2x - 48 = 0$

This is the standard quadratic form, but we need to choose the equation from the given options.

Let's compare our derived equation $x^2 + (x+2)^2 = 100$ with the given options:

(A) $x^2 + (x+2)^2 = 100$: This matches the equation we derived directly from the problem statement.

(B) $x^2 + (x+1)^2 = 100$: This equation represents the sum of squares of two consecutive *integers* ($x$ and $x+1$), not necessarily even integers.

(C) $x^2 + (x+2)^2 + 100 = 0$: This equation is equivalent to $x^2 + (x+2)^2 = -100$, which means the sum of two squares is -100. The square of any real number is non-negative, so the sum of two squares cannot be a negative number like -100 for real $x$. This does not represent the given situation.

(D) $x + x+2 = 100$: This equation simplifies to $2x + 2 = 100$. This represents the *sum* of the two consecutive even integers being 100, not the sum of their *squares*. This is a linear equation.

The equation that correctly represents the situation where the sum of the squares of two consecutive even integers ($x$ and $x+2$) is 100 is $x^2 + (x+2)^2 = 100$.

The correct answer is (A).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the nature of the roots (whether they are real or not, and if real, whether they are distinct or equal) is determined by the value of its discriminant, $D$.

The discriminant is given by the formula:

$D = b^2 - 4ac$

The relationship between the discriminant and the nature of the roots is as follows:

• If $D > 0$: The equation has two real and distinct roots.
• If $D = 0$: The equation has two real and equal roots (a repeated real root).
• If $D < 0$: The equation has no real roots (it has two distinct complex conjugate roots).

The question asks for the condition under which a quadratic equation has real roots. This condition includes both the case where the roots are distinct and real ($D > 0$) and the case where the roots are equal and real ($D = 0$).

Combining these two conditions, the roots are real if the discriminant $D$ is either greater than zero or equal to zero. This is expressed by the inequality:

Let's compare this condition with the given options:

(A) $D < 0$: This corresponds to no real roots.

(B) $D = 0$: This corresponds to real and equal roots (which are real roots).

(C) $D > 0$: This corresponds to real and distinct roots (which are real roots).

(D) $D \ge 0$: This corresponds to both cases where the roots are real ($D > 0$ or $D = 0$).

Therefore, a quadratic equation has real roots if the discriminant $D$ is greater than or equal to zero.

The correct answer is (D).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the sum of its roots (let's call them $\alpha$ and $\beta$) is given by the formula:

The given quadratic equation is $x^2 + 5x + 6 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we can identify the coefficients:

$a = 1$

$b = 5$

$c = 6$

The problem states that the roots are -2 and -3, and their sum is $(-2) + (-3) = -5$.

Let's calculate the sum of the roots using the formula $-\frac{b}{a}$ with the identified coefficients:

The calculated sum of the roots using the formula $(-b/a)$ is $-5$. The problem also states that the sum of the given roots $(-2 \text{ and } -3)$ is $-5$.

We need to find the option that correctly relates the sum of the roots $(-5)$ to the coefficients using the formula.

Let's evaluate the options:

(A) $-(-5)/1 = 5$, matches $-(b/a)$. The value $-(-5)/1 = 5$. The value $-(b/a) = -5/1 = -5$. $5$ does not match $-5$. Incorrect.

(B) $-(-5)/1 = 5$, matches $b/a$. The value $-(-5)/1 = 5$. The value $b/a = 5/1 = 5$. $5$ matches $5$. However, the sum of the roots is $-5$, not $5$. This option misrepresents the calculated sum.

(C) $-5/1 = -5$, matches $-(b/a)$. The value $-5/1 = -5$. The value $-(b/a) = -5/1 = -5$. $-5$ matches $-5$. This correctly states the sum is $-5/1$ and that this matches the formula $-b/a$. Correct.

(D) $-5/1 = -5$, matches $b/a$. The value $-5/1 = -5$. The value $b/a = 5/1 = 5$. $-5$ does not match $5$. Incorrect.

The sum of the roots is indeed $-5$, and this value matches the formula $-(b/a)$ for the given equation.

The correct answer is (C).

Explanation:

Let $P(x)$ be a quadratic polynomial. A quadratic polynomial in one variable $x$ is an expression of the form $ax^2 + bx + c$, where $a, b,$ and $c$ are real numbers and $a \neq 0$.

A quadratic equation is formed when a quadratic polynomial is set equal to zero. For example, setting the polynomial $P(x) = ax^2 + bx + c$ to zero gives the quadratic equation $ax^2 + bx + c = 0$.

The roots of this quadratic equation are the values of the variable $x$ that satisfy the equation, i.e., the values of $x$ for which $ax^2 + bx + c = 0$ is true.

The zeros of a polynomial $P(x)$ are defined as the values of the variable $x$ for which the value of the polynomial is zero, i.e., the values of $x$ for which $P(x) = 0$.

Comparing the definitions, we see that the roots of the equation $P(x) = 0$ are precisely the same values of $x$ as the zeros of the polynomial $P(x)$. Both terms refer to the values of the variable that make the expression $ax^2 + bx + c$ equal to zero.

Let's consider the given options:

• (A) Coefficients are the numerical values $a, b,$ and $c$ in the polynomial/equation. They are not the roots or zeros.
• (B) The Degree is the highest power of the variable in the polynomial (which is 2 for a quadratic). It is a property of the polynomial, not the roots or zeros.
• (C) Zeros of the polynomial are the values of $x$ for which the polynomial equals zero. These are the same as the roots of the equation formed by setting the polynomial to zero.
• (D) The Discriminant ($b^2 - 4ac$) is a value derived from the coefficients that tells us about the nature of the roots (real, distinct, equal, complex), but it is not the roots themselves.

Therefore, the roots of the quadratic equation are the zeros of the corresponding polynomial.

The correct answer is (C).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the roots are real and equal if and only if its discriminant ($D$) is equal to zero.

The discriminant is given by the formula:

$D = b^2 - 4ac$

For the given quadratic equation $2x^2 - 6x + k = 0$ to have real and equal roots, the discriminant must be equal to 0:

$D = 0$

Identify the coefficients of the given equation by comparing it with the standard form $ax^2 + bx + c = 0$:

$a = 2$

$b = -6$

$c = k$

Substitute these values into the discriminant formula $D = b^2 - 4ac$ and set it equal to 0:

$(-6)^2 - 4(2)(k) = 0$

Calculate the terms:

$36 - 8k = 0$

Now, solve this linear equation for $k$. Add $8k$ to both sides of the equation:

Divide both sides by 8:

Simplify the fraction:

$k = \frac{\cancel{36}^{9}}{\cancel{8}_{2}}$

Express the result as a decimal:

Thus, the value of $k$ for which the quadratic equation $2x^2 - 6x + k = 0$ has real and equal roots is 4.5.

Comparing the calculated value of $k=4.5$ with the given options:

(A) 3

(B) 4.5

(C) 6

(D) 9

The correct answer is (B).

Solution:

Let the son's age be $x$ years, as specified in the problem.

According to the first condition, the age of the father is two years more than three times the age of his son.

• Three times the age of his son is $3 \times x = 3x$ years.
• Two years more than three times the age of his son is $3x + 2$ years.

So, the father's age is $(3x + 2)$ years.

According to the second condition, the product of their ages is 320.

The product of the son's age ($x$) and the father's age ($3x+2$) is $x \times (3x+2)$, which can be written as $x(3x+2)$.

Setting this product equal to 320:

This equation represents the given information. Let's expand it to see the quadratic form:

$3x^2 + 2x = 320$

Move the constant term to the left side to get the standard form $ax^2 + bx + c = 0$:

$3x^2 + 2x - 320 = 0$

Now, let's compare the equation we derived, $x(3x+2) = 320$, with the given options:

(A) $x(3x+2) = 320$: This exactly matches the equation we derived.

(B) $x(2x+3) = 320$: This represents the father's age as $2x+3$, which is incorrect according to the problem statement.

(C) $x(3x-2) = 320$: This represents the father's age as $3x-2$, which is incorrect according to the problem statement.

(D) $x+(3x+2) = 320$: This represents the sum of their ages being 320, not the product of their ages.

The equation that represents the given information is $x(3x+2) = 320$.

The correct answer is (A).

Solution:

From the Case Study in Question 29, we established that if the son's age is $x$ years, then the father's age is $(3x+2)$ years, and the equation representing the product of their ages being 320 is:

To find the age of the son ($x$), we need to solve this quadratic equation. First, convert it to the standard form $ax^2 + bx + c = 0$:

Expand the left side:

$3x^2 + 2x = 320$

Move the constant term to the left side:

$3x^2 + 2x - 320 = 0$

This is a quadratic equation with $a=3$, $b=2$, and $c=-320$. We can solve this using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

First, calculate the discriminant $D = b^2 - 4ac$:

$D = (2)^2 - 4(3)(-320)$

$D = 4 - (-3840)$

$D = 4 + 3840$

Now, substitute the values of $a$, $b$, and $D$ into the quadratic formula:

$x = \frac{-2 \pm \sqrt{3844}}{2(3)}$

$x = \frac{-2 \pm \sqrt{3844}}{6}$

We need to find the square root of 3844. Let's calculate $\sqrt{3844}$:

By estimation or calculation, $\sqrt{3844} = 62$.

Substitute the value of the square root into the formula to find the roots:

$x = \frac{-2 \pm 62}{6}$

This gives two possible values for $x$:

$x_1 = \frac{-2 + 62}{6} = \frac{60}{6} = 10$

$x_2 = \frac{-2 - 62}{6} = \frac{-64}{6} = -\frac{32}{3}$

Since $x$ represents the age of the son, it must be a positive value. The root $-\frac{32}{3}$ is negative, so it is not a valid age in this context.

The valid age for the son is $x = 10$ years.

Let's check this solution:

Son's age = 10 years.

Father's age = $3(10) + 2 = 30 + 2 = 32$ years.

Product of their ages = $10 \times 32 = 320$. This matches the condition given in the problem.

Comparing the calculated age of the son with the given options:

(A) 8 years

(B) 10 years

(C) 12 years

(D) 15 years

The age of the son is 10 years.

The correct answer is (B).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the nature of the roots is determined by the value of its discriminant, which is calculated as $D = b^2 - 4ac$.

The discriminant tells us whether the roots are real or complex, and whether they are distinct or equal. The conditions are as follows:

• If $D = b^2 - 4ac > 0$: The quadratic equation has two distinct real roots.
• If $D = b^2 - 4ac = 0$: The quadratic equation has two equal real roots (a repeated root).
• If $D = b^2 - 4ac < 0$: The quadratic equation has no real roots. Instead, it has two distinct complex conjugate roots.

The question asks for the condition under which the quadratic equation has no real roots.

Based on the analysis of the discriminant, this occurs when the discriminant $D$ is less than zero.

So, the condition is $b^2 - 4ac < 0$.

Comparing this condition with the given options:

(A) $b^2 - 4ac > 0$: Real and distinct roots.

(B) $b^2 - 4ac = 0$: Real and equal roots.

(C) $b^2 - 4ac < 0$: No real roots.

(D) $b^2 - 4ac \ge 0$: Real roots (either distinct or equal).

The condition for a quadratic equation to have no real roots is $b^2 - 4ac < 0$.

The correct answer is (C).

Explanation:

For a quadratic equation with roots $\alpha$ and $\beta$, the relationship between the roots and the coefficients is given by:

• Sum of roots: $\alpha + \beta = -\frac{b}{a}$
• Product of roots: $\alpha \cdot \beta = \frac{c}{a}$

Conversely, if the sum and product of the roots of a quadratic equation are known, the equation can be formed directly using the formula:

In this problem, we are given:

Sum of the roots $= 7$

Product of the roots $= 12$

Substitute these values into the formula for forming the quadratic equation:

$x^2 - (7)x + (12) = 0$

This is the quadratic equation whose roots have a sum of 7 and a product of 12.

Comparing this equation with the given options:

(A) $x^2 + 7x + 12 = 0$ (The coefficient of $x$ is +7, but it should be -7)

(B) $x^2 - 7x + 12 = 0$ (This matches the derived equation)

(C) $x^2 + 7x - 12 = 0$ (The coefficient of $x$ is +7, and the constant term is -12)

(D) $x^2 - 7x - 12 = 0$ (The constant term is -12, but it should be +12)

The correct equation is $x^2 - 7x + 12 = 0$.

The correct answer is (B).

Explanation:

To find the roots of the quadratic equation $x^2 - 5x = 0$, we need to find the values of $x$ that satisfy the equation.

The given equation is $x^2 - 5x = 0$. This is an incomplete quadratic equation because the constant term is zero ($c=0$). It is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=-5$, and $c=0$.

We can solve this equation efficiently by factorization.

Notice that both terms on the left side, $x^2$ and $-5x$, have a common factor of $x$. We can factor out $x$ from the expression $x^2 - 5x$:

$x(x) - x(5) = x(x - 5)$

So, the equation becomes:

$x(x - 5) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for $x$:

Factor 1: $x = 0$

Factor 2: $x - 5 = 0$

Solve the second equation for $x$:

$x - 5 = 0$

Add 5 to both sides:

So, the roots of the equation $x^2 - 5x = 0$ are $x = 0$ and $x = 5$.

Let's check if these values satisfy the original equation:

For $x=0$: $(0)^2 - 5(0) = 0 - 0 = 0$. The equation is satisfied.

For $x=5$: $(5)^2 - 5(5) = 25 - 25 = 0$. The equation is satisfied.

Comparing the roots 0 and 5 with the given options:

(A) 0, 5

(B) 0, -5

(C) 5, -5

(D) 0, 0

The correct answer is (A).

Solution:

Let the uniform speed of the train be $x$ km/hr.

The distance covered is 300 km.

The relationship between distance, speed, and time is given by: Time = Distance / Speed.

Time taken at the uniform speed ($T_1$) is:

If the speed had been 5 km/hr more, the new speed would be $(x+5)$ km/hr.

Time taken at this increased speed ($T_2$) is:

According to the problem, the journey with the increased speed ($T_2$) takes 1 hour less than the journey with the uniform speed ($T_1$).

This can be expressed as:

Rearranging this equation, we get:

Substitute the expressions for $T_1$ and $T_2$ into this equation:

This equation represents the situation described in the problem. We can expand it to obtain a standard quadratic equation if needed for solving, but the question asks for the equation representing the situation as given in the options.

Let's compare our derived equation with the given options:

(A) $\frac{300}{x} - \frac{300}{x+5} = 1$: This matches the equation we derived.

(B) $\frac{300}{x+5} - \frac{300}{x} = 1$: This equation implies $T_2 - T_1 = 1$, meaning the second journey took 1 hour *more* than the first, which contradicts the problem statement.

(C) $\frac{300}{x} + \frac{300}{x+5} = 1$: This equation implies the *sum* of the two travel times is 1 hour, which is incorrect.

(D) $300x = (x+5)(1)$: This equation does not correctly relate distance, speed, and time; it seems to mix different quantities incorrectly.

The equation that correctly represents the situation is $\frac{300}{x} - \frac{300}{x+5} = 1$.

The correct answer is (A).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the roots of the equation are the values of $x$ that make the equation true. These values of $x$ are also known as the zeros of the corresponding quadratic polynomial $y = ax^2 + bx + c$.

Graphically, the zeros of the polynomial $y = ax^2 + bx + c$ are the x-coordinates of the points where the graph of the polynomial (which is a parabola) intersects the x-axis. These intersection points are the x-intercepts of the graph.

The number and nature of the roots of a quadratic equation are determined by the discriminant, $D = b^2 - 4ac$:

• If $D > 0$: The equation has two distinct real roots. This means there are two different real values of $x$ for which $y=0$. The graph of the parabola intersects the x-axis at exactly two distinct points.
• If $D = 0$: The equation has two equal real roots (a single distinct real root with multiplicity 2). This means there is exactly one real value of $x$ for which $y=0$. The graph of the parabola touches the x-axis at exactly one point (the vertex lies on the x-axis).
• If $D < 0$: The equation has no real roots (it has two distinct complex conjugate roots). This means there are no real values of $x$ for which $y=0$. The graph of the parabola does not intersect the x-axis at all (it lies entirely above or below the x-axis).

The question states that the quadratic equation $ax^2 + bx + c = 0$ has real and distinct roots. According to our analysis, this condition corresponds to the case where the discriminant $D > 0$.

When the discriminant is greater than zero ($D > 0$), the graph of the corresponding polynomial $y = ax^2 + bx + c$ intersects the x-axis at exactly two distinct points.

Comparing this conclusion with the given options:

(A) Exactly one: This occurs when $D=0$ (real and equal roots).

(B) Exactly two: This occurs when $D>0$ (real and distinct roots).

(C) No: This occurs when $D<0$ (no real roots).

(D) At least one: This occurs when $D \ge 0$ (real roots, which can be distinct or equal). While "exactly two" is a subset of "at least one", option (B) is a more precise description for the specific condition "real and distinct roots" given in the question.

The correct answer is (B).

Explanation:

We are asked to identify which of the given options is NOT a method used to solve a quadratic equation.

Let's examine each option:

(A) Factorization: This is a common method used to solve quadratic equations, especially when the quadratic polynomial can be easily factored into a product of two linear expressions. If $ax^2 + bx + c = 0$ can be written as $(px+q)(rx+s)=0$, then the roots are found by setting each factor to zero: $px+q=0$ and $rx+s=0$. This is a valid method.

(B) Completing the square: This is an algebraic technique used to solve quadratic equations by transforming the equation into the form $(x+h)^2 = k$. Taking the square root of both sides then allows solving for $x$. This method is also fundamental in deriving the quadratic formula. This is a valid method.

(C) Division algorithm: The division algorithm for polynomials is used to divide one polynomial by another, resulting in a quotient and a remainder. For example, it's used to find factors or simplify rational expressions. While polynomial division can be part of the process of finding roots *if* a factor (and thus a root) is already known (e.g., dividing $P(x)$ by $(x-r)$ if $r$ is a root), the division algorithm itself is not a method used *directly* to find the roots of a general quadratic equation $ax^2+bx+c=0$.

(D) Quadratic formula: This is a direct formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, that provides the roots of any quadratic equation $ax^2 + bx + c = 0$ based on its coefficients $a$, $b$, and $c$. It is a comprehensive method that works for all quadratic equations. This is a valid method.

Based on the analysis, Factorization, Completing the square, and the Quadratic formula are standard methods for solving quadratic equations. The Division algorithm, while a tool in polynomial algebra, is not typically classified as a primary method for solving a quadratic equation directly from its general form.

The correct answer is (C).

Explanation:

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, where $a, b,$ and $c$ are real numbers and $a \neq 0$, the nature of its roots is determined by the value of its discriminant, $D = b^2 - 4ac$.

A quadratic equation has real roots if and only if its discriminant is greater than or equal to zero ($D \ge 0$). Real roots include both the case of distinct real roots ($D > 0$) and equal real roots ($D = 0$).

The given quadratic equation is $x^2 - 2mx + 1 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we identify the coefficients:

$a = 1$

$b = -2m$

$c = 1$

Now, we calculate the discriminant $D$ using the formula $D = b^2 - 4ac$:

$D = (-2m)^2 - 4(1)(1)$

$D = 4m^2 - 4$

For the quadratic equation to have real roots, the discriminant must be greater than or equal to zero:

To solve this inequality for $m$, we can divide both sides by 4:

We can factor the left side as a difference of squares, $m^2 - 1^2 = (m-1)(m+1)$:

The expression $(m-1)(m+1)$ is zero when $m=1$ or $m=-1$. These values divide the number line into intervals. The product $(m-1)(m+1)$ is non-negative when both factors have the same sign (both positive or both negative) or when one or both factors are zero.

• If $m \ge 1$, then $m-1 \ge 0$ and $m+1 \ge 2 > 0$. The product $(m-1)(m+1) \ge 0 \times 2 = 0$.
• If $m \le -1$, then $m-1 \le -2 < 0$ and $m+1 \le 0$. The product $(m-1)(m+1) \ge (-2) \times 0 = 0$.
• If $-1 < m < 1$, then $m-1 < 0$ and $m+1 > 0$. The product $(m-1)(m+1) < 0$.

So, the inequality $m^2 - 1 \ge 0$ holds when $m \le -1$ or $m \ge 1$.

The condition for the quadratic equation $x^2 - 2mx + 1 = 0$ to have real roots is $m \le -1$ or $m \ge 1$.

Comparing this result with the given options:

(A) $m > 1$ or $m < -1$ (Excludes $m=1$ and $m=-1$, which give real and equal roots)

(B) $m \ge 1$ or $m \le -1$ (Includes $m=1$ and $m=-1$, correctly representing real roots)

(C) $-1 < m < 1$ (Represents no real roots)

(D) $m = 1$ or $m = -1$ (Represents real and equal roots specifically, not all real roots)

The correct answer is (B).

Explanation:

The given quadratic equation is $kx^2 - 4x + 2 = 0$.

For this equation to be quadratic, the coefficient of $x^2$ must be non-zero, so $k \neq 0$.

Comparing the given equation with the standard form $ax^2 + bx + c = 0$, the coefficients are:

$a = k$

$b = -4$

$c = 2$

The sum of the roots of a quadratic equation is given by the formula $-\frac{b}{a}$.

The product of the roots of a quadratic equation is given by the formula $\frac{c}{a}$.

According to the problem statement, the sum of the roots is equal to the product of the roots.

To solve for $k$, we can subtract $\frac{2}{k}$ from both sides:

A fraction is equal to zero only if its numerator is zero. In this case, the numerator is 2, which is not zero. Therefore, there is no value of $k$ that satisfies the equation $\frac{2}{k} = 0$.

This means that, as stated, the quadratic equation $kx^2 - 4x + 2 = 0$ cannot have its sum of roots equal to its product of roots for any valid quadratic coefficient $k \neq 0$.

However, since this is a multiple choice question with options, it suggests there might be a typographical error in the problem statement. A common type of problem leads to $k=4$. This occurs if the constant term was intended to be $k$ instead of 2, i.e., the equation was $kx^2 - 4x + k = 0$.

Let's assume the intended equation was $kx^2 - 4x + k = 0$ (where $a=k, b=-4, c=k, k \neq 0$).

Sum of roots $= -\frac{b}{a} = -\frac{-4}{k} = \frac{4}{k}$

Product of roots $= \frac{c}{a} = \frac{k}{k} = 1$ (since $k \neq 0$)

Setting sum equal to product: $\frac{4}{k} = 1$

Solving for $k$: $4 = k$

The value $k=4$ is among the options. If $k=4$, the equation is $4x^2 - 4x + 4 = 0$, which is a valid quadratic equation. Its sum of roots is $4/4=1$ and its product of roots is $4/4=1$. The condition is satisfied.

Assuming the likely intended question corresponds to option (C), the value of $k$ is 4.

The correct answer is (C).

Solution:

The given quadratic equation is $x^2 - 8x + k = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we can identify the coefficients:

$a = 1$

$b = -8$

$c = k$

We are given that one root of the equation is 5.

For a quadratic equation $ax^2 + bx + c = 0$, let the two roots be $\alpha$ and $\beta$.

The sum of the roots is given by the formula $\alpha + \beta = -\frac{b}{a}$.

Let the given root be $\alpha = 5$. Let the other root be $\beta$.

Using the formula for the sum of roots:

Substitute the known values ($\alpha=5$, $a=1$, $b=-8$):

Now, solve for $\beta$:

Subtract 5 from both sides of the equation:

The other root of the equation is 3.

We can also find the value of $k$ using the product of the roots. The product of the roots is given by $\alpha \cdot \beta = \frac{c}{a}$.

Substitute the roots $\alpha=5$ and $\beta=3$, and the coefficients $a=1$ and $c=k$:

So the equation is $x^2 - 8x + 15 = 0$. The roots of $x^2 - 8x + 15 = 0$ are indeed 5 and 3 (since $(x-5)(x-3) = x^2 - 3x - 5x + 15 = x^2 - 8x + 15$).

Comparing the other root $\beta=3$ with the given options:

(A) 3

(B) -3

(C) 5

(D) -5

The correct answer is (A).

Solution:

We are given a rectangular garden with an area of $150 \text{ m}^2$.

Let the breadth of the rectangular garden be $x$ metres, as stated in the problem.

According to the condition given, the length of the garden is 5 m more than its breadth.

So, the length of the rectangular garden is $(x+5)$ metres.

The formula for the area of a rectangle is Length $\times$ Breadth.

We are given that the area is $150 \text{ m}^2$.

Area $= \text{Length} \times \text{Breadth}$

Rewrite the equation with the terms rearranged and the variable on the left side:

This equation represents the situation described in the problem. We can expand it to get the standard quadratic form $ax^2 + bx + c = 0$ if needed for solving:

Expand the left side:

$x^2 + 5x = 150$

Move the constant term to the left side:

$x^2 + 5x - 150 = 0$

Now, let's compare the equation we derived, $x(x+5) = 150$, with the given options:

(A) $x(x+5) = 150$: This exactly matches the equation we derived directly from the problem statement.

(B) $x(x-5) = 150$: This equation implies the length is $x-5$, which would mean the breadth ($x$) is 5m more than the length. This contradicts the problem statement that the length is 5m more than the breadth.

(C) $x + (x+5) = 150$: This equation simplifies to $2x + 5 = 150$. This represents the *sum* of the breadth and length (or half the perimeter if multiplied by 2) being 150, not the area.

(D) $2(x + x+5) = 150$: This equation simplifies to $2(2x+5) = 150$, which is $4x+10=150$. This represents the *perimeter* of the rectangle being 150, not the area.

The equation that can be used to find the dimensions (specifically the breadth $x$) is $x(x+5) = 150$.

The correct answer is (A).

A quadratic equation is a polynomial equation of the second degree in a single variable.

The standard form of a quadratic equation is given by:

$ax^2 + bx + c = 0$

where $x$ represents a variable, and $a$, $b$, and $c$ are coefficients.

It is essential that the coefficient $a$ is not equal to zero ($a \neq 0$), otherwise, the equation would be linear, not quadratic.

In the standard form $ax^2 + bx + c = 0$:

The coefficients are:

a: The coefficient of $x^2$. It is also called the leading coefficient.

b: The coefficient of $x$.

c: The constant term.

To check if the given equation is a quadratic equation, we need to simplify it and see if it can be written in the standard form $ax^2 + bx + c = 0$, where $a \neq 0$.

The given equation is:

$(x - 2)^2 + 1 = 2x - 3$

Expand the term $(x - 2)^2$ using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$(x - 2)^2 = x^2 - 2(x)(2) + (-2)^2 = x^2 - 4x + 4$

Substitute this back into the original equation:

$(x^2 - 4x + 4) + 1 = 2x - 3$

Simplify the left side:

$x^2 - 4x + 5 = 2x - 3$

Move all terms to one side to set the equation to zero:

$x^2 - 4x + 5 - 2x + 3 = 0$

Combine like terms:

$x^2 + (-4x - 2x) + (5 + 3) = 0$

$x^2 - 6x + 8 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a=1$, $b=-6$, and $c=8$.

Since the coefficient of $x^2$, which is $a=1$, is not zero, the equation is a quadratic equation.

Conclusion: Yes, the equation $(x - 2)^2 + 1 = 2x - 3$ is a quadratic equation.

A root of a quadratic equation is a value of the variable that satisfies the equation, i.e., makes the equation true. To check if a value is a root, we substitute the value into the equation and verify if the left-hand side equals the right-hand side (which is 0 in this case).

The given quadratic equation is $x^2 + x - 2 = 0$.

Check for $x = 1$:

Substitute $x = 1$ into the equation:

$(1)^2 + (1) - 2$

Evaluate the expression:

$1 + 1 - 2$

$2 - 2$

$0$

Since the left-hand side equals $0$, which is the right-hand side of the equation, $x = 1$ satisfies the equation.

Conclusion for $x = 1$: $x = 1$ is a root of the quadratic equation $x^2 + x - 2 = 0$.

Check for $x = -2$:

Substitute $x = -2$ into the equation:

$(-2)^2 + (-2) - 2$

Evaluate the expression:

$4 - 2 - 2$

$2 - 2$

$0$

Since the left-hand side equals $0$, which is the right-hand side of the equation, $x = -2$ satisfies the equation.

Conclusion for $x = -2$: $x = -2$ is a root of the quadratic equation $x^2 + x - 2 = 0$.

Final Answer: Both $x = 1$ and $x = -2$ are roots of the given quadratic equation.

We need to find the roots of the quadratic equation $x^2 - 3x - 10 = 0$ using the factorisation method. This involves splitting the middle term ($bx$) into two terms such that their sum is $bx$ and their product is equal to the product of the coefficient of $x^2$ ($a$) and the constant term ($c$).

The given equation is $x^2 - 3x - 10 = 0$.

Here, $a=1$, $b=-3$, and $c=-10$.

We need to find two numbers whose sum is $-3$ and whose product is $a \times c = 1 \times (-10) = -10$.

Let the two numbers be $p$ and $q$. We are looking for $p+q = -3$ and $pq = -10$.

Consider the factors of $-10$. Possible pairs are $(1, -10)$, $(-1, 10)$, $(2, -5)$, $(-2, 5)$.

Let's check the sum of these pairs:

• $1 + (-10) = -9$ (Not -3)
• $-1 + 10 = 9$ (Not -3)
• $2 + (-5) = -3$ (This is -3)
• $-2 + 5 = 3$ (Not -3)

The pair of numbers we are looking for is $2$ and $-5$.

Now, split the middle term $-3x$ as $+2x - 5x$:

$x^2 + 2x - 5x - 10 = 0$

Group the terms and factor out the common factors from each group:

$(x^2 + 2x) + (-5x - 10) = 0$

$x(x + 2) - 5(x + 2) = 0$

Factor out the common binomial term $(x + 2)$:

$(x + 2)(x - 5) = 0$

For the product of two terms to be zero, at least one of the terms must be zero. So, we set each factor equal to zero:

$x + 2 = 0 \quad$ or $\quad x - 5 = 0$

Solve each linear equation for $x$:

From $x + 2 = 0$, we get $x = -2$.

From $x - 5 = 0$, we get $x = 5$.

The roots of the quadratic equation $x^2 - 3x - 10 = 0$ are $x = -2$ and $x = 5$.

The roots are -2 and 5.

We need to find the roots of the quadratic equation $2x^2 + x - 6 = 0$ using the factorisation method, which involves splitting the middle term.

The given equation is $2x^2 + x - 6 = 0$.

This is in the standard form $ax^2 + bx + c = 0$, with $a=2$, $b=1$, and $c=-6$.

We need to find two numbers whose sum is $b=1$ and whose product is $a \times c = 2 \times (-6) = -12$.

Let the two numbers be $p$ and $q$. We are looking for $p+q = 1$ and $pq = -12$.

Consider the factors of $-12$. The pairs of factors are $(1, -12), (-1, 12), (2, -6), (-2, 6), (3, -4), (-3, 4)$.

Let's check the sum of these pairs:

• $1 + (-12) = -11$
• $-1 + 12 = 11$
• $2 + (-6) = -4$
• $-2 + 6 = 4$
• $3 + (-4) = -1$
• $-3 + 4 = 1$

The pair of numbers whose sum is 1 and product is -12 is $-3$ and $4$.

Now, we split the middle term $x$ as $-3x + 4x$:

$2x^2 - 3x + 4x - 6 = 0$

Group the first two terms and the last two terms:

$(2x^2 - 3x) + (4x - 6) = 0$

Factor out the common factor from each group:

$x(2x - 3) + 2(2x - 3) = 0$

Now, factor out the common binomial term $(2x - 3)$:

$(2x - 3)(x + 2) = 0$

For the product of two terms to be zero, at least one of the terms must be zero. So, we set each factor equal to zero:

$2x - 3 = 0 \quad$ or $\quad x + 2 = 0$

Solve each linear equation for $x$:

From $2x - 3 = 0$:

$2x = 3$

$x = \frac{3}{2}$

From $x + 2 = 0$:

$x = -2$

The roots of the quadratic equation $2x^2 + x - 6 = 0$ are $x = \frac{3}{2}$ and $x = -2$.

The roots are $\frac{3}{2}$ and -2.

We are asked to find the roots of the quadratic equation $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ using the factorisation method.

The given equation is in the standard form $ax^2 + bx + c = 0$, where $a = \sqrt{2}$, $b = 7$, and $c = 5\sqrt{2}$.

We need to find two numbers whose sum is $b=7$ and whose product is $a \times c$.

Calculate the product $ac$:

$ac = (\sqrt{2}) \times (5\sqrt{2}) = 5 \times (\sqrt{2})^2 = 5 \times 2 = 10$

So, we need two numbers whose sum is $7$ and whose product is $10$.

Let's list pairs of factors of 10 and their sums:

• (1, 10): Sum = $1 + 10 = 11$ (Not 7)
• (2, 5): Sum = $2 + 5 = 7$ (This is 7)

The two numbers are 2 and 5.

Now, split the middle term $7x$ as $2x + 5x$ in the original equation:

$\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$

Group the terms and factor out common factors from each group:

$(\sqrt{2}x^2 + 2x) + (5x + 5\sqrt{2}) = 0$

Factor out $\sqrt{2}x$ from the first group (note that $2 = \sqrt{2} \times \sqrt{2}$):

$\sqrt{2}x(x + \frac{2}{\sqrt{2}}) + (5x + 5\sqrt{2}) = 0$

Simplify $\frac{2}{\sqrt{2}}$: $\frac{2}{\sqrt{2}} = \frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}} = \sqrt{2}$.

Factor out 5 from the second group:

$\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$

Now, factor out the common binomial term $(x + \sqrt{2})$:

$(x + \sqrt{2})(\sqrt{2}x + 5) = 0$

For the product of two factors to be zero, at least one of the factors must be zero. Set each factor equal to zero:

$x + \sqrt{2} = 0 \quad$ or $\quad \sqrt{2}x + 5 = 0$

Solve the first equation for $x$:

$x + \sqrt{2} = 0 \implies x = -\sqrt{2}$

Solve the second equation for $x$:

$\sqrt{2}x + 5 = 0 \implies \sqrt{2}x = -5 \implies x = -\frac{5}{\sqrt{2}}$

The roots of the quadratic equation $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ are $x = -\sqrt{2}$ and $x = -\frac{5}{\sqrt{2}}$.

The roots are $-\sqrt{2}$ and $-\frac{5}{\sqrt{2}}$.

The quadratic formula is a formula that provides the solutions (or roots) of a quadratic equation in standard form $ax^2 + bx + c = 0$, where $a \neq 0$.

The formula for finding the roots of $ax^2 + bx + c = 0$ is given by:

$\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

In this formula:

• $x$ represents the variable.
• $a$, $b$, and $c$ are the coefficients of the quadratic equation, with $a \neq 0$.
• The term $b^2 - 4ac$ is called the discriminant, which determines the nature of the roots.

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula $\Delta = b^2 - 4ac$. The value of the discriminant tells us about the nature of the roots.

The given quadratic equation is $2x^2 - 4x + 3 = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we identify the coefficients:

$a = 2$

$b = -4$

$c = 3$

Now, we calculate the discriminant using the formula $\Delta = b^2 - 4ac$:

$\Delta = (-4)^2 - 4(2)(3)$

$\Delta = 16 - 24$

$\Delta = -8$

The discriminant is $\Delta = -8$.

The nature of the roots is determined by the sign of the discriminant:

• If $\Delta > 0$, the equation has two distinct real roots.
• If $\Delta = 0$, the equation has two equal real roots.
• If $\Delta < 0$, the equation has no real roots (the roots are complex).

In this case, the discriminant is $\Delta = -8$, which is less than 0 ($\Delta < 0$).

Conclusion: Since the discriminant is negative, the quadratic equation $2x^2 - 4x + 3 = 0$ has no real roots. The roots are complex.

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula $\Delta = b^2 - 4ac$. The value of the discriminant helps us determine the nature of the roots without actually solving for them.

The given quadratic equation is $3x^2 - 2x + \frac{1}{3} = 0$.

Comparing this equation with the standard form $ax^2 + bx + c = 0$, we identify the coefficients:

$a = 3$

$b = -2$

$c = \frac{1}{3}$

Now, we calculate the discriminant using the formula $\Delta = b^2 - 4ac$:

$\Delta = (-2)^2 - 4(3)\left(\frac{1}{3}\right)$

$\Delta = 4 - 4\left(\frac{\cancel{3}}{\cancel{3}}\right)$

$\Delta = 4 - 4(1)$

$\Delta = 4 - 4$

$\Delta = 0$

The discriminant is $\Delta = 0$.

The nature of the roots depends on the value of the discriminant:

• If $\Delta > 0$, there are two distinct real roots.
• If $\Delta = 0$, there are two equal real roots.
• If $\Delta < 0$, there are no real roots (complex roots).

In this case, the discriminant is $\Delta = 0$.

Conclusion: Since the discriminant is equal to zero, the quadratic equation $3x^2 - 2x + \frac{1}{3} = 0$ has two equal real roots.

Given: The quadratic equation $2x^2 + kx + 3 = 0$.

To Find: The value(s) of $k$ for which the equation has two equal roots.

Solution:

A quadratic equation $ax^2 + bx + c = 0$ has two equal real roots if and only if its discriminant ($\Delta$) is equal to zero.

The discriminant is given by the formula $\Delta = b^2 - 4ac$.

Comparing the given equation $2x^2 + kx + 3 = 0$ with the standard form $ax^2 + bx + c = 0$, we identify the coefficients:

$a = 2$

$b = k$

$c = 3$

For the equation to have two equal roots, the discriminant must be zero:

$\Delta = 0$

$b^2 - 4ac = 0$

Substitute the values of $a$, $b$, and $c$ into the discriminant formula:

$(k)^2 - 4(2)(3) = 0$

$k^2 - 8(3) = 0$

$k^2 - 24 = 0$

Now, solve for $k$:

$k^2 = 24$

Take the square root of both sides:

$k = \pm \sqrt{24}$

Simplify the square root of 24 by finding its prime factors:

So, $24 = 2 \times 2 \times 2 \times 3 = 2^2 \times 6$.

Therefore, $\sqrt{24} = \sqrt{2^2 \times 6} = \sqrt{2^2} \times \sqrt{6} = 2\sqrt{6}$.

Substitute the simplified square root back into the equation for $k$:

$k = \pm 2\sqrt{6}$

The possible values of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two equal roots are $2\sqrt{6}$ and $-2\sqrt{6}$.

The values of $k$ are $2\sqrt{6}$ and $-2\sqrt{6}$.

Given: The equation $kx(x - 2) + 6 = 0$ has two equal roots.

To Find: The value(s) of $k$ satisfying this condition.

Solution:

First, we need to rewrite the given equation in the standard quadratic form $ax^2 + bx + c = 0$.

The given equation is $kx(x - 2) + 6 = 0$.

Expand the expression:

$kx^2 - 2kx + 6 = 0$

This equation is now in the standard form $ax^2 + bx + c = 0$, where:

$a = k$

$b = -2k$

$c = 6$

For the equation to be a quadratic equation, the coefficient of $x^2$ must not be zero, i.e., $a \neq 0$. Thus, $k \neq 0$.

A quadratic equation has two equal real roots if and only if its discriminant is equal to zero. The discriminant ($\Delta$) is given by the formula $\Delta = b^2 - 4ac$.

Set the discriminant equal to zero:

$\Delta = 0$

$b^2 - 4ac = 0$

Substitute the identified coefficients $a=k$, $b=-2k$, and $c=6$ into the discriminant equation:

$(-2k)^2 - 4(k)(6) = 0$

$4k^2 - 24k = 0$

Now, solve this equation for $k$. Factor out the common term $4k$:

$4k(k - 6) = 0$

This equation is satisfied if either $4k = 0$ or $k - 6 = 0$.

Case 1: $4k = 0$

$k = \frac{0}{4}$

$k = 0$

Case 2: $k - 6 = 0$

$k = 6$

We found two possible values for $k$: $0$ and $6$. However, for the original equation to be a quadratic equation, the coefficient $a$ must be non-zero. In this case, $a=k$.

If $k=0$, the original equation becomes $0 \times x(x-2) + 6 = 0$, which simplifies to $6 = 0$. This is a false statement, meaning there are no solutions (roots) for this equation. It is not a quadratic equation and does not have two equal roots.

If $k=6$, the equation becomes $6x(x-2) + 6 = 0$, or $6x^2 - 12x + 6 = 0$. Dividing by 6, we get $x^2 - 2x + 1 = 0$, which is $(x-1)^2 = 0$. This is a quadratic equation and has two equal roots ($x=1$).

Therefore, the value of $k$ for which the given equation is a quadratic equation and has two equal roots is $k=6$.

The value of $k$ is 6.

Consider a standard quadratic equation:

$ax^2 + bx + c = 0$

where $a$, $b$, and $c$ are coefficients with $a \neq 0$.

Let $\alpha$ and $\beta$ be the roots of this quadratic equation.

The relationship between the sum of the roots ($\alpha + \beta$) and the coefficients is given by:

$\alpha + \beta = -\frac{b}{a}$

The relationship between the product of the roots ($\alpha \beta$) and the coefficients is given by:

$\alpha \beta = \frac{c}{a}$

In summary, for a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:

Sum of roots: $\mathbf{\alpha + \beta = -\frac{b}{a}}$

Product of roots: $\mathbf{\alpha \beta = \frac{c}{a}}$

Given: The quadratic equation is $3x^2 + (2k + 1)x - 9 = 0$. The sum of the roots is $-1$.

To Find: The value of $k$.

Solution:

The standard form of a quadratic equation is $ax^2 + bx + c = 0$.

Comparing the given equation $3x^2 + (2k + 1)x - 9 = 0$ with the standard form, we identify the coefficients:

$a = 3$

$b = 2k + 1$

$c = -9$

The relationship between the sum of the roots ($\alpha + \beta$) and the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula:

$\alpha + \beta = -\frac{b}{a}$

We are given that the sum of the roots is $-1$. So, we have:

$\alpha + \beta = -1$

Equating the two expressions for the sum of the roots:

$-\frac{b}{a} = -1$

Substitute the values of $a$ and $b$ from the given equation:

$-\frac{(2k + 1)}{3} = -1$

Now, solve for $k$:

Multiply both sides by 3:

Given: The quadratic equation is $2x^2 - 6x + k = 0$. The product of its roots is $4$.

To Find: The value of $k$.

Solution:

The standard form of a quadratic equation is $ax^2 + bx + c = 0$.

Comparing the given equation $2x^2 - 6x + k = 0$ with the standard form, we identify the coefficients:

$a = 2$

$b = -6$

$c = k$

The relationship between the product of the roots ($\alpha \beta$) and the coefficients of a quadratic equation $ax^2 + bx + c = 0$ is given by the formula:

$\alpha \beta = \frac{c}{a}$

We are given that the product of the roots is $4$. So, we have:

$\alpha \beta = 4$

Equating the two expressions for the product of the roots:

$\frac{c}{a} = 4$

Substitute the values of $a$ and $c$ from the given equation into this relationship:

$\frac{k}{2} = 4$

To solve for $k$, multiply both sides of the equation by 2:

$k = 4 \times 2$

$k = 8$

The value of $k$ for which the product of the roots of the given quadratic equation is $4$ is $8$.

The value of $k$ is 8.

Given:

Sum of roots, $S = \frac{1}{4}$

Product of roots, $P = -1$

To Form: A quadratic equation.

Solution:

A quadratic equation with roots $\alpha$ and $\beta$ can be written in the form:

$x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0$

$x^2 - (\alpha + \beta)x + (\alpha \beta) = 0$

Substitute the given sum of roots ($S = \frac{1}{4}$) and product of roots ($P = -1$) into this form:

$x^2 - \left(\frac{1}{4}\right)x + (-1) = 0$

$x^2 - \frac{1}{4}x - 1 = 0$

To clear the fraction and obtain an equation with integer coefficients, multiply the entire equation by the least common multiple of the denominators, which is 4:

$4 \times \left(x^2 - \frac{1}{4}x - 1\right) = 4 \times 0$

$4x^2 - 4 \times \frac{1}{4}x - 4 \times 1 = 0$

$4x^2 - x - 4 = 0$

This is a quadratic equation with the specified sum and product of roots.

The quadratic equation is $4x^2 - x - 4 = 0$.

Given: The quadratic equation $3x^2 - 4kx + 4 = 0$.

To Find: The values of $k$ for which the equation has real roots.

Solution:

A quadratic equation $ax^2 + bx + c = 0$ has real roots if and only if its discriminant ($\Delta$) is greater than or equal to zero.

The discriminant is given by the formula $\Delta = b^2 - 4ac$.

So, the condition for real roots is $\Delta \geq 0$, which means $b^2 - 4ac \geq 0$.

Compare the given equation $3x^2 - 4kx + 4 = 0$ with the standard form $ax^2 + bx + c = 0$ to identify the coefficients:

$a = 3$

$b = -4k$

$c = 4$

Substitute these coefficients into the discriminant inequality $b^2 - 4ac \geq 0$:

$(-4k)^2 - 4(3)(4) \geq 0$

$16k^2 - 12(4) \geq 0$

$16k^2 - 48 \geq 0$

Now, solve this inequality for $k$:

Add 48 to both sides:

$16k^2 \geq 48$

Divide both sides by 16:

$k^2 \geq \frac{48}{16}$

$k^2 \geq 3$

To solve the inequality $k^2 \geq 3$, we take the square root of both sides. Remember that taking the square root introduces both positive and negative solutions, and for an inequality, this splits into two cases:

$k \geq \sqrt{3}$ or $k \leq -\sqrt{3}$

Thus, the quadratic equation $3x^2 - 4kx + 4 = 0$ has real roots when the value of $k$ is less than or equal to $-\sqrt{3}$ or greater than or equal to $\sqrt{3}$.

The values of $k$ are $k \leq -\sqrt{3}$ or $k \geq \sqrt{3}$.

Given: The quadratic equation $2x^2 + ax - 6 = 0$. One root of the equation is $x = 2$.

To Find: The value of $a$ and the other root.

Solution:

Since $x = 2$ is a root of the quadratic equation $2x^2 + ax - 6 = 0$, it must satisfy the equation when substituted for $x$.

Substitute $x = 2$ into the equation:

$2(2)^2 + a(2) - 6 = 0$

$2(4) + 2a - 6 = 0$

$8 + 2a - 6 = 0$

$2 + 2a = 0$

Solve for $a$:

$2a = -2$

$a = \frac{-2}{2}$

$a = -1$

So, the value of $a$ is $-1$.

Now that we have the value of $a$, the quadratic equation is $2x^2 + (-1)x - 6 = 0$, which is:

$2x^2 - x - 6 = 0$

To find the other root, let the roots of the equation be $\alpha$ and $\beta$. We are given one root, say $\alpha = 2$.

For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $\alpha + \beta = -\frac{B}{A}$ and the product of the roots is $\alpha \beta = \frac{C}{A}$.

Comparing $2x^2 - x - 6 = 0$ with $Ax^2 + Bx + C = 0$, we have $A=2$, $B=-1$, and $C=-6$.

Using the product of roots relationship:

$\alpha \beta = \frac{C}{A}$

Substitute the known root $\alpha = 2$ and the coefficients $A=2$, $C=-6$:

$2 \times \beta = \frac{-6}{2}$

$2 \beta = -3$

Solve for $\beta$:

$\beta = \frac{-3}{2}$

Alternatively, using the sum of roots relationship:

$\alpha + \beta = -\frac{B}{A}$

Substitute $\alpha = 2$, $A=2$, $B=-1$:

$2 + \beta = -\frac{(-1)}{2}$

$2 + \beta = \frac{1}{2}$

$\beta = \frac{1}{2} - 2$

$\beta = \frac{1}{2} - \frac{4}{2}$

$\beta = \frac{1 - 4}{2}$

$\beta = -\frac{3}{2}$

Both methods give the same other root.

The value of $a$ is -1 and the other root is $-\frac{3}{2}$.

We are asked to solve the quadratic equation $x^2 - 2ax + a^2 - b^2 = 0$ by factorisation.

The given equation is:

$x^2 - 2ax + (a^2 - b^2) = 0$

This equation is in the standard form $Ax^2 + Bx + C = 0$, where $A=1$, $B=-2a$, and $C=a^2 - b^2$.

We notice that the constant term $C = a^2 - b^2$ is a difference of squares, which can be factored as $(a-b)(a+b)$.

So, the equation can be written as:

$x^2 - 2ax + (a-b)(a+b) = 0$

For factorisation by splitting the middle term, we need to find two numbers whose sum is the coefficient of $x$, which is $-2a$, and whose product is the constant term, which is $(a-b)(a+b)$.

Consider the terms $-(a-b)$ and $-(a+b)$.

Their product is: $[-(a-b)][-(a+b)] = (a-b)(a+b) = a^2 - b^2$.

Their sum is: $-(a-b) + -(a+b) = -a + b - a - b = -2a$.

Thus, the two numbers are $-(a-b)$ and $-(a+b)$.

Now, we split the middle term $-2ax$ using these two numbers:

$x^2 - (a-b)x - (a+b)x + (a-b)(a+b) = 0$

Group the terms and factor out common factors from each group:

$(x^2 - (a-b)x) - ((a+b)x - (a-b)(a+b)) = 0$

Factor $x$ from the first group and $(a+b)$ from the second group:

$x(x - (a-b)) - (a+b)(x - (a-b)) = 0$

Now, factor out the common binomial term $(x - (a-b))$:

$(x - (a-b))(x - (a+b)) = 0$

$(x - a + b)(x - a - b) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Set each factor equal to zero:

$x - a + b = 0 \quad$ or $\quad x - a - b = 0$

Solve the first equation for $x$:

$x - a + b = 0 \implies x = a - b$

Solve the second equation for $x$:

$x - a - b = 0 \implies x = a + b$

The roots of the quadratic equation $x^2 - 2ax + a^2 - b^2 = 0$ are $a-b$ and $a+b$.

The roots are $a-b$ and $a+b$.

Given: Total number of trees to be planted = 17956.

The number of rows is equal to the number of columns.

To Find: The number of rows.

Solution:

Let the number of rows be $n$.

Since the number of rows and columns are the same, the number of columns is also $n$.

The total number of trees is the product of the number of rows and the number of columns.

Total Trees = (Number of rows) $\times$ (Number of columns)

$17956 = n \times n$

$17956 = n^2$

To find the number of rows $n$, we need to find the square root of 17956.

$n = \sqrt{17956}$

We will find the square root of 17956 using the long division method.

Pair the digits from right to left: $\overline{1} \; \overline{79} \; \overline{56}$.

The first period is 1. The largest square less than or equal to 1 is $1^2 = 1$. Write 1 in the quotient and as the divisor. Subtract 1 from 1.

Bring down the next period, 79. The new dividend is 79.

Double the current quotient (1), which is $1 \times 2 = 2$. Write 2 and a blank for the next digit of the divisor (2_).

Find the largest digit $d$ such that $2d \times d \leq 79$. If $d=3$, $23 \times 3 = 69$. If $d=4$, $24 \times 4 = 96$ (too large). So the digit is 3.

Write 3 in the quotient and as the next digit of the divisor (23). Multiply $23 \times 3 = 69$. Subtract 69 from 79.

Bring down the next period, 56. The new dividend is 1056.

Double the current quotient (13), which is $13 \times 2 = 26$. Write 26 and a blank for the next digit of the divisor (26_).

Find the largest digit $d'$ such that $26d' \times d' \leq 1056$. The last digit of 1056 is 6. A number ending in 4 or 6 gives a square ending in 6. Let's try $d'=4$. $264 \times 4 = 1056$. This matches exactly.

Write 4 in the quotient and as the next digit of the divisor (264). Multiply $264 \times 4 = 1056$. Subtract 1056 from 1056.

The remainder is 0. The square root is 134.

The long division is shown below:

Thus, $\sqrt{17956} = 134$.

So, $n = 134$.

The number of rows is 134.

The number of rows is 134.

Given: The product of two consecutive positive integers is 306.

To Frame: A quadratic equation to find the integers.

Solution:

Let the first positive integer be represented by the variable $x$.

Since the two integers are consecutive and positive, the next consecutive positive integer will be one more than the first one.

So, the second consecutive positive integer is $x + 1$.

The problem states that the product of these two integers is 306.

Therefore, we can write the equation:

$(x) \times (x + 1) = 306$

Now, we need to express this equation in the standard quadratic form $ax^2 + bx + c = 0$.

Expand the left side of the equation:

$x \times x + x \times 1 = 306$

$x^2 + x = 306$

To get the standard form, move the constant term from the right side to the left side by subtracting 306 from both sides:

$x^2 + x - 306 = 306 - 306$

$x^2 + x - 306 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a=1$, $b=1$, and $c=-306$. Since $a \neq 0$, this is a quadratic equation.

Solving this equation for $x$ would give the first positive integer, and then $x+1$ would give the second integer.

The required quadratic equation is $x^2 + x - 306 = 0$.

Given:

Area of the rectangular park = 120 sq meters.

Length is 2 meters more than its breadth.

Let the breadth of the park be $x$ meters.

To Frame: A quadratic equation based on the given information.

Solution:

Let the breadth of the rectangular park be $x$ meters.

According to the problem statement, the length is 2 meters more than the breadth.

So, the length of the park is $(x + 2)$ meters.

The formula for the area of a rectangle is given by:

Area = Length $\times$ Breadth

We are given that the area is 120 sq meters. Substitute the expressions for length and breadth into the area formula:

$(x + 2) \times x = 120$

Now, we need to expand and rearrange this equation into the standard quadratic form $ax^2 + bx + c = 0$.

Distribute $x$ on the left side:

$x \times x + 2 \times x = 120$

$x^2 + 2x = 120$

Move the constant term from the right side to the left side by subtracting 120 from both sides:

$x^2 + 2x - 120 = 120 - 120$

$x^2 + 2x - 120 = 0$

This equation is in the form $ax^2 + bx + c = 0$, with $a=1$, $b=2$, and $c=-120$. Since $a \neq 0$, it is a quadratic equation.

Solving this quadratic equation for $x$ would give the breadth of the park, and $(x+2)$ would give the length.

The required quadratic equation is $x^2 + 2x - 120 = 0$.

Given: $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 5x + 6 = 0$.

To Find: The value of $\alpha^2 + \beta^2$.

Solution:

The given quadratic equation is $x^2 - 5x + 6 = 0$.

This is in the standard form $ax^2 + bx + c = 0$.

Comparing the coefficients, we have:

$a = 1$

$b = -5$

$c = 6$

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, the sum and product of the roots are given by:

Sum of roots: $\alpha + \beta = -\frac{b}{a}$

Product of roots: $\alpha \beta = \frac{c}{a}$

Calculate the sum of the roots for the given equation:

$\alpha + \beta = -\frac{(-5)}{1} = 5$

Calculate the product of the roots for the given equation:

$\alpha \beta = \frac{6}{1} = 6$

We need to find the value of $\alpha^2 + \beta^2$. We know the algebraic identity:

$(\alpha + \beta)^2 = \alpha^2 + 2\alpha \beta + \beta^2$

Rearranging this identity to solve for $\alpha^2 + \beta^2$:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$

Substitute the values of $(\alpha + \beta)$ and $(\alpha \beta)$ that we calculated:

$\alpha^2 + \beta^2 = (5)^2 - 2(6)$

$\alpha^2 + \beta^2 = 25 - 12$

$\alpha^2 + \beta^2 = 13$

The value of $\alpha^2 + \beta^2$ is 13.

Given: The quadratic equation is $ax^2 + bx + c = 0$, where $a \neq 0$. One root of the equation is the reciprocal of the other.

To Show: $a = c$.

Proof:

Let the roots of the quadratic equation $ax^2 + bx + c = 0$ be $\alpha$ and $\beta$.

According to the given condition, one root is the reciprocal of the other. Let's assume that $\beta$ is the reciprocal of $\alpha$.

So, we have:

$\beta = \frac{1}{\alpha}$

For a quadratic equation in the standard form $ax^2 + bx + c = 0$, the product of the roots is given by the formula:

Product of roots ($\alpha \beta$) $= \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{c}{a}$

Now, substitute the condition $\beta = \frac{1}{\alpha}$ into the product of roots equation (i):

$\alpha \left(\frac{1}{\alpha}\right) = \frac{c}{a}$

Simplify the left side of the equation:

The term $\alpha$ in the numerator and denominator cancels out (assuming $\alpha \neq 0$, which must be true for its reciprocal $\frac{1}{\alpha}$ to exist as a root):

$1 = \frac{c}{a}$

Multiply both sides of the equation by $a$:

$1 \times a = \frac{c}{a} \times a$

$a = c$

This shows that if one root of the quadratic equation $ax^2 + bx + c = 0$ is the reciprocal of the other, then the coefficient of $x^2$ is equal to the constant term.

Hence, $a = c$.

Given: The equation $x^2 + k(2x + k - 1) + 2 = 0$ has real and equal roots.

To Find: The value(s) of $k$ satisfying this condition.

Solution:

First, we need to rewrite the given equation in the standard quadratic form $Ax^2 + Bx + C = 0$.

The given equation is $x^2 + k(2x + k - 1) + 2 = 0$.

Expand the term with $k$:

$x^2 + 2kx + k(k - 1) + 2 = 0$

Further simplify the constant term:

$x^2 + 2kx + k^2 - k + 2 = 0$

This equation is now in the standard form $Ax^2 + Bx + C = 0$, where:

$A = 1$

$B = 2k$

$C = k^2 - k + 2$

A quadratic equation has real and equal roots if and only if its discriminant ($\Delta$) is equal to zero.

The discriminant is given by the formula $\Delta = B^2 - 4AC$.

Set the discriminant equal to zero:

$\Delta = 0$

$B^2 - 4AC = 0$

Substitute the identified coefficients $A=1$, $B=2k$, and $C=k^2 - k + 2$ into the discriminant equation:

$(2k)^2 - 4(1)(k^2 - k + 2) = 0$

$4k^2 - 4(k^2 - k + 2) = 0$

Distribute the $-4$ on the left side:

$4k^2 - 4k^2 + 4k - 8 = 0$

Combine like terms:

$(4k^2 - 4k^2) + (4k - 8) = 0$

$0 + 4k - 8 = 0$

$4k - 8 = 0$

Solve the linear equation for $k$:

Add 8 to both sides:

$4k = 8$

Divide both sides by 4:

$k = \frac{8}{4}$

$k = 2$

The coefficient of $x^2$ in the standard form is $A=1$, which is non-zero for any value of $k$. Thus, the equation is always quadratic. The condition for real and equal roots is met when $k=2$.

The value of $k$ is 2.

Given:

The altitude of a right triangle is 7 cm less than its base.

The hypotenuse of the right triangle is 13 cm.

To Frame: The quadratic equation to find the base.

Solution:

Let the base of the right triangle be $x$ cm.

According to the problem, the altitude is 7 cm less than the base.

So, the altitude of the right triangle is $(x - 7)$ cm.

Note that for the altitude to be a positive length, we must have $x - 7 > 0$, which means $x > 7$.

In a right triangle, the relationship between the base, altitude (perpendicular side), and hypotenuse is given by the Pythagorean theorem:

$(\text{Base})^2 + (\text{Altitude})^2 = (\text{Hypotenuse})^2$

Substitute the expressions for the base, altitude, and the value of the hypotenuse into the Pythagorean theorem:

$(x)^2 + (x - 7)^2 = (13)^2$

Expand and simplify the equation:

$x^2 + (x^2 - 2(x)(7) + (-7)^2) = 169$

$x^2 + (x^2 - 14x + 49) = 169$

$x^2 + x^2 - 14x + 49 = 169$

Combine the $x^2$ terms and move the constant term from the right side to the left side:

$2x^2 - 14x + 49 - 169 = 0$

$2x^2 - 14x - 120 = 0$

To simplify the equation, divide the entire equation by the common factor 2:

$\frac{2x^2}{2} - \frac{14x}{2} - \frac{120}{2} = \frac{0}{2}$

$x^2 - 7x - 60 = 0$

This equation is in the standard quadratic form $ax^2 + bx + c = 0$, with $a=1$, $b=-7$, and $c=-60$. Since $a \neq 0$, it is a quadratic equation. Solving this equation would give the possible values for the base $x$. We would then need to check if the solution for $x$ is positive and greater than 7 to be physically valid in the context of the problem.

The required quadratic equation is $x^2 - 7x - 60 = 0$.

The method of completing the square is a technique used to solve quadratic equations by transforming one side of the equation into a perfect square trinomial. This allows us to easily take the square root of both sides and solve for the variable.

Here are the general steps to solve a quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$) by completing the square:

1. Divide by the coefficient of $x^2$: If $a \neq 1$, divide every term in the equation by $a$ to make the coefficient of $x^2$ equal to 1. The equation becomes $x^2 + \frac{b}{a}x + \frac{c}{a} = 0$.
2. Move the constant term: Transpose the constant term $\frac{c}{a}$ to the right side of the equation: $x^2 + \frac{b}{a}x = -\frac{c}{a}$.
3. Complete the square on the left side: Identify the coefficient of $x$, which is $\frac{b}{a}$. Take half of this coefficient ($\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}$) and square it $\left(\left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}\right)$. Add this value to both sides of the equation: $x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$.
4. Factor the left side: The left side is now a perfect square trinomial and can be factored as $\left(x + \frac{b}{2a}\right)^2$. So, $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}$.
5. Simplify the right side: Combine the terms on the right side by finding a common denominator: $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$.
6. Take the square root of both sides: Take the square root of both sides of the equation. Remember to include both the positive and negative roots on the right side: $x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$.
7. Solve for $x$: Isolate $x$ by transposing $\frac{b}{2a}$ to the right side: $x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$. Combine the terms on the right side to get the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. While the quadratic formula is derived from completing the square, the final step in this method is usually to calculate the values directly from $x + \frac{b}{2a} = \pm \sqrt{\text{Right Side}}$.

Now, let's solve the equation $2x^2 - 7x + 3 = 0$ using the method of completing the square.

Step 1: Divide by the coefficient of $x^2$.

The coefficient of $x^2$ is 2. Divide the entire equation by 2:

$\frac{2x^2}{2} - \frac{7x}{2} + \frac{3}{2} = \frac{0}{2}$

$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$

Step 2: Move the constant term to the right side.

Transpose $\frac{3}{2}$ to the right side:

$x^2 - \frac{7}{2}x = -\frac{3}{2}$

Step 3: Complete the square on the left side.

The coefficient of $x$ is $-\frac{7}{2}$. Half of this coefficient is $\frac{1}{2} \times \left(-\frac{7}{2}\right) = -\frac{7}{4}$.

The square of half the coefficient is $\left(-\frac{7}{4}\right)^2 = \frac{(-7)^2}{4^2} = \frac{49}{16}$.

Add $\frac{49}{16}$ to both sides of the equation:

$x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}$

Step 4: Factor the left side.

The left side is now a perfect square of the form $(x + p)^2$, where $p$ is half the coefficient of $x$. So, $p = -\frac{7}{4}$.

The left side factors as $\left(x - \frac{7}{4}\right)^2$.

$\left(x - \frac{7}{4}\right)^2 = -\frac{3}{2} + \frac{49}{16}$

Step 5: Simplify the right side.

Find a common denominator for the terms on the right side. The common denominator for 2 and 16 is 16.

$-\frac{3}{2} = -\frac{3 \times 8}{2 \times 8} = -\frac{24}{16}$

So, the right side becomes:

$-\frac{24}{16} + \frac{49}{16} = \frac{49 - 24}{16} = \frac{25}{16}$

The equation is now:

$\left(x - \frac{7}{4}\right)^2 = \frac{25}{16}$

Step 6: Take the square root of both sides.

Take the square root of both sides, including both positive and negative roots on the right:

$x - \frac{7}{4} = \pm \sqrt{\frac{25}{16}}$

$x - \frac{7}{4} = \pm \frac{\sqrt{25}}{\sqrt{16}}$

$x - \frac{7}{4} = \pm \frac{5}{4}$

Step 7: Solve for $x$.

Isolate $x$ by adding $\frac{7}{4}$ to both sides:

$x = \frac{7}{4} \pm \frac{5}{4}$

This gives two possible solutions for $x$:

Case 1: Using the positive sign

$x_1 = \frac{7}{4} + \frac{5}{4} = \frac{7 + 5}{4} = \frac{12}{4} = 3$

Case 2: Using the negative sign

$x_2 = \frac{7}{4} - \frac{5}{4} = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$

The roots of the quadratic equation $2x^2 - 7x + 3 = 0$ are $x = 3$ and $x = \frac{1}{2}$.

The roots are 3 and $\frac{1}{2}$.

We will derive the quadratic formula by applying the method of completing the square to the general quadratic equation $ax^2 + bx + c = 0$, where $a \neq 0$.

Given Equation:

$ax^2 + bx + c = 0 \quad$ (where $a \neq 0$)

Step 1: Divide the equation by $a$.

This step ensures that the coefficient of $x^2$ is 1.

$\frac{ax^2}{a} + \frac{bx}{a} + \frac{c}{a} = \frac{0}{a}$

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

Step 2: Move the constant term to the right side of the equation.

Subtract $\frac{c}{a}$ from both sides:

$x^2 + \frac{b}{a}x = -\frac{c}{a}$

Step 3: Complete the square on the left side.

To complete the square, we need to add a term to the left side that makes it a perfect square trinomial. This term is the square of half the coefficient of $x$.

The coefficient of $x$ is $\frac{b}{a}$.

Half of the coefficient of $x$ is $\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}$.

The square of half the coefficient of $x$ is $\left(\frac{b}{2a}\right)^2 = \frac{b^2}{(2a)^2} = \frac{b^2}{4a^2}$.

Add $\frac{b^2}{4a^2}$ to both sides of the equation:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$

Step 4: Factor the perfect square trinomial on the left side.

The left side is now in the form $p^2 + 2pq + q^2 = (p+q)^2$, where $p=x$ and $q=\frac{b}{2a}$.

So, the left side factors as $\left(x + \frac{b}{2a}\right)^2$.

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}$

Step 5: Simplify the right side.

Combine the terms on the right side by finding a common denominator, which is $4a^2$.

$\frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2}{4a^2} - \frac{c \times 4a}{a \times 4a} = \frac{b^2 - 4ac}{4a^2}$

So, the equation becomes:

$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$

Step 6: Take the square root of both sides.

Taking the square root of both sides introduces the $\pm$ sign on the right side:

$x + \frac{b}{2a} = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}}$

$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}}$

Since $\sqrt{4a^2} = \sqrt{4}\sqrt{a^2} = 2|a|$, we have:

$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2|a|}$

However, since the $\pm$ sign is already present, dividing by $2a$ covers both positive and negative values of $a$ in the denominator symmetrically. Thus, we can write:

$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

Step 7: Solve for $x$.

Subtract $\frac{b}{2a}$ from both sides to isolate $x$:

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

Since the terms on the right side have a common denominator ($2a$), we can combine them:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This is the quadratic formula, which gives the roots of the equation $ax^2 + bx + c = 0$. The term $b^2 - 4ac$ is the discriminant, which determines the nature of the roots.

The quadratic formula is:

$\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

Given: The quadratic equation $5x^2 - 8x - 4 = 0$.

To Solve: Find the roots of the equation using the quadratic formula.

Solution:

The given quadratic equation is $5x^2 - 8x - 4 = 0$.

This is in the standard form $ax^2 + bx + c = 0$.

Comparing the coefficients, we have:

$a = 5$

$b = -8$

$c = -4$

The quadratic formula for finding the roots of $ax^2 + bx + c = 0$ is:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(-4)}}{2(5)}$

First, calculate the discriminant, $\Delta = b^2 - 4ac$:

$\Delta = (-8)^2 - 4(5)(-4)$

$\Delta = 64 - (20)(-4)$

$\Delta = 64 - (-80)$

$\Delta = 64 + 80$

$\Delta = 144$

Now, substitute the value of the discriminant back into the quadratic formula:

$x = \frac{8 \pm \sqrt{144}}{10}$

Evaluate the square root of 144:

$\sqrt{144} = 12$

Substitute this value back into the formula:

$x = \frac{8 \pm 12}{10}$

This gives two possible values for $x$, corresponding to the '+' and '-' signs:

Case 1: Using the '+' sign

$x_1 = \frac{8 + 12}{10} = \frac{20}{10} = 2$

Case 2: Using the '-' sign

$x_2 = \frac{8 - 12}{10} = \frac{-4}{10} = -\frac{\cancel{4}^2}{\cancel{10}_5} = -\frac{2}{5}$

The roots of the quadratic equation $5x^2 - 8x - 4 = 0$ are $x = 2$ and $x = -\frac{2}{5}$.

The roots are 2 and $-\frac{2}{5}$.

For a standard quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$), the roots are given by the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The term inside the square root, $b^2 - 4ac$, is called the discriminant, denoted by $D$ or $\Delta$.

$D = b^2 - 4ac$

The value of the discriminant determines the nature of the roots of the quadratic equation. There are three cases:

Case 1: The discriminant is greater than zero ($D > 0$).

Condition: $b^2 - 4ac > 0$

In this case, $\sqrt{D}$ is a real and non-zero number. The quadratic formula gives two distinct values for $x$ because of the $\pm$ sign.

The roots are given by $x_1 = \frac{-b + \sqrt{D}}{2a}$ and $x_2 = \frac{-b - \sqrt{D}}{2a}$.

Nature of Roots: The equation has two distinct real roots.

Case 2: The discriminant is equal to zero ($D = 0$).

Condition: $b^2 - 4ac = 0$

In this case, $\sqrt{D} = \sqrt{0} = 0$. The quadratic formula simplifies to $x = \frac{-b \pm 0}{2a} = \frac{-b}{2a}$.

Both values of $x$ from the $\pm$ sign are the same.

Nature of Roots: The equation has two equal real roots (sometimes referred to as one real root with multiplicity 2).

Case 3: The discriminant is less than zero ($D < 0$).

Condition: $b^2 - 4ac < 0$

In this case, $\sqrt{D}$ involves taking the square root of a negative number, which is not a real number. The roots cannot be expressed as real numbers.

Nature of Roots: The equation has no real roots. The roots are complex (conjugate pair).

In summary:

• $D > 0$: Two distinct real roots
• $D = 0$: Two equal real roots
• $D < 0$: No real roots

Given: The quadratic equation $(k + 1)x^2 - 2(k - 1)x + 1 = 0$ has real and equal roots.

To Find: The value(s) of $k$ satisfying this condition.

Solution:

The given equation is $(k + 1)x^2 - 2(k - 1)x + 1 = 0$.

This equation is in the standard quadratic form $Ax^2 + Bx + C = 0$, where:

$A = k + 1$

$B = -2(k - 1) = -2k + 2$

$C = 1$

For the equation to be a quadratic equation, the coefficient of $x^2$ must be non-zero. Thus, $A \neq 0$, which means:

$k + 1 \neq 0$

$k \neq -1$

A quadratic equation has real and equal roots if and only if its discriminant ($\Delta$) is equal to zero.

The discriminant is given by the formula $\Delta = B^2 - 4AC$.

Set the discriminant equal to zero:

$\Delta = 0$

$B^2 - 4AC = 0$

Substitute the identified coefficients $A = k + 1$, $B = -2(k - 1)$, and $C = 1$ into the discriminant equation:

$(-2(k - 1))^2 - 4(k + 1)(1) = 0$

Simplify the equation:

$(-2)^2 (k - 1)^2 - 4(k + 1) = 0$

$4(k^2 - 2k + 1) - 4k - 4 = 0$

$4k^2 - 8k + 4 - 4k - 4 = 0$

Combine the like terms:

$4k^2 + (-8k - 4k) + (4 - 4) = 0$

$4k^2 - 12k + 0 = 0$

$4k^2 - 12k = 0$

Factor out the common term $4k$ from the left side:

$4k(k - 3) = 0$

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1:

$4k = 0$

$k = \frac{0}{4}$

$k = 0$

Case 2:

$k - 3 = 0$

$k = 3$

We obtained two possible values for $k$: $0$ and $3$. We must check if these values satisfy the condition for the equation to be quadratic, which is $k \neq -1$.

Both $k=0$ and $k=3$ are not equal to $-1$. Therefore, for both these values, the given equation is a quadratic equation and has real and equal roots.

The values of $k$ for which the quadratic equation has real and equal roots are 0 and 3.

Given: The sum of the squares of two consecutive odd positive integers is 290.

To Find: The two consecutive odd positive integers.

Solution:

Let the first consecutive odd positive integer be $x$.

Since the integers are consecutive and odd, the next consecutive odd integer will be $x + 2$.

(For example, if the first odd integer is 3, the next is $3+2=5$. If the first is 11, the next is $11+2=13$).

The problem states that both integers are positive, so $x > 0$. Since $x$ is an odd integer, $x \ge 1$.

According to the problem, the sum of the squares of these two integers is 290.

So, we can write the equation:

$(x)^2 + (x + 2)^2 = 290$

Now, we frame the quadratic equation by expanding and simplifying the equation:

$x^2 + (x^2 + 2(x)(2) + 2^2) = 290 \quad$ (Using $(a+b)^2 = a^2 + 2ab + b^2$)

$x^2 + x^2 + 4x + 4 = 290$

$2x^2 + 4x + 4 = 290$

Move all terms to one side to get the standard form $ax^2 + bx + c = 0$:

$2x^2 + 4x + 4 - 290 = 0$

$2x^2 + 4x - 286 = 0$

Divide the entire equation by 2 to simplify (as all coefficients are divisible by 2):

$\frac{2x^2}{2} + \frac{4x}{2} - \frac{286}{2} = \frac{0}{2}$

$x^2 + 2x - 143 = 0$

The quadratic equation is $x^2 + 2x - 143 = 0$.

Now, we solve this quadratic equation to find the value of $x$. We will use the factorisation method.

We need to find two numbers whose product is $1 \times (-143) = -143$ and whose sum is the coefficient of $x$, which is 2.

Let's find the factors of 143. We know that $143 = 11 \times 13$.

To get a product of -143 and a sum of 2, the two numbers must be 13 and -11.

$13 \times (-11) = -143$

$13 + (-11) = 13 - 11 = 2$

Split the middle term $2x$ using 13 and -11:

$x^2 + 13x - 11x - 143 = 0$

Group the terms and factor out common factors:

$(x^2 + 13x) + (-11x - 143) = 0$

$x(x + 13) - 11(x + 13) = 0$

Factor out the common binomial term $(x + 13)$:

$(x + 13)(x - 11) = 0$

Set each factor equal to zero to find the possible values of $x$:

$x + 13 = 0 \quad$ or $\quad x - 11 = 0$

$x = -13 \quad$ or $\quad x = 11$

We defined $x$ as a positive odd integer. Therefore, we discard the solution $x = -13$ as it is not positive.

The valid value for the first odd positive integer is $x = 11$.

The first integer is $x = 11$.

The second consecutive odd integer is $x + 2 = 11 + 2 = 13$.

Check the answer: The integers are 11 and 13. Both are positive and odd. The sum of their squares is $11^2 + 13^2 = 121 + 169 = 290$, which matches the given information.

The two consecutive odd positive integers are 11 and 13.

Given:

Distance covered by the train = 360 km.

If speed increases by 5 km/hr, time taken decreases by 1 hour.

To Find: The uniform speed of the train.

Solution:

Let the uniform speed of the train be $v$ km/hr.

Using the formula Time = Distance / Speed:

Original time taken for the journey = $\frac{360}{v}$ hours.

If the speed had been 5 km/hr more, the new speed would be $(v + 5)$ km/hr.

The time taken with the increased speed would be $\frac{360}{v + 5}$ hours.

According to the problem, the new time is 1 hour less than the original time.

So, we can write the equation:

$\frac{360}{v + 5} = \frac{360}{v} - 1$

Now, we frame the quadratic equation by rearranging this equation into the standard form $ax^2 + bx + c = 0$.

Add 1 to both sides:

$\frac{360}{v + 5} + 1 = \frac{360}{v}$

Combine the terms on the left side by finding a common denominator $(v+5)$:

$\frac{360}{v + 5} + \frac{v + 5}{v + 5} = \frac{360}{v}$

$\frac{360 + v + 5}{v + 5} = \frac{360}{v}$

$\frac{v + 365}{v + 5} = \frac{360}{v}$

Cross-multiply:

$v(v + 365) = 360(v + 5)$

$v^2 + 365v = 360v + 1800$

Move all terms to the left side to set the equation to zero:

$v^2 + 365v - 360v - 1800 = 0$

$v^2 + 5v - 1800 = 0$

The quadratic equation is $v^2 + 5v - 1800 = 0$.

Now, we solve this quadratic equation for $v$. We can use the factorisation method.

We need to find two numbers whose product is $1 \times (-1800) = -1800$ and whose sum is the coefficient of $v$, which is 5.

Consider factors of 1800 that are close to each other. Let's test pairs:

• $40 \times 45 = 1800$

To get a product of -1800 and a sum of +5, the numbers must be 45 and -40.

$45 \times (-40) = -1800$

$45 + (-40) = 5$

Split the middle term $5v$ using 45 and -40:

$v^2 + 45v - 40v - 1800 = 0$

Group the terms and factor out common factors:

$(v^2 + 45v) + (-40v - 1800) = 0$

$v(v + 45) - 40(v + 45) = 0$

Factor out the common binomial term $(v + 45)$:

$(v + 45)(v - 40) = 0$

Set each factor equal to zero to find the possible values of $v$:

$v + 45 = 0 \quad$ or $\quad v - 40 = 0$

$v = -45 \quad$ or $\quad v = 40$

Since speed must be a positive value, we discard the solution $v = -45$.

The valid speed of the train is $v = 40$ km/hr.

The speed of the train is 40 km/hr.

Given:

Total amount spent by the merchant = $\textsf{₹}1200$.

Condition: If the cost per article were $\textsf{₹}3$ less, he could buy 20 more articles for the same amount.

To Find: The number of articles the merchant bought.

Solution:

Let the number of articles the merchant bought be $n$. Since the number of articles must be a positive integer, $n > 0$.

The original cost per article is given by the total amount divided by the number of articles:

Original cost per article $= \frac{\text{Total Amount}}{\text{Number of articles}} = \frac{1200}{n} \textsf{₹}$.

According to the problem, if the cost per article were $\textsf{₹}3$ less, the new cost per article would be $\left(\frac{1200}{n} - 3\right) \textsf{₹}$.

With this reduced cost, the merchant could buy 20 more articles. So, the new number of articles would be $(n + 20)$.

The total cost for the new scenario (new number of articles $\times$ new cost per article) is still $\textsf{₹}1200$.

So, we can write the equation:

$(n + 20) \times \left(\frac{1200}{n} - 3\right) = 1200$

Frame the quadratic equation:

Expand and simplify the equation to frame the quadratic equation in standard form $an^2 + bn + c = 0$.

$(n + 20)\left(\frac{1200 - 3n}{n}\right) = 1200$

Multiply both sides by $n$ (assuming $n \neq 0$, which is true as $n$ is the number of articles):

$(n + 20)(1200 - 3n) = 1200n$

Expand the left side using the distributive property:

$n(1200) + n(-3n) + 20(1200) + 20(-3n) = 1200n$

$1200n - 3n^2 + 24000 - 60n = 1200n$

Combine like terms on the left side:

$-3n^2 + (1200n - 60n) + 24000 = 1200n$

$-3n^2 + 1140n + 24000 = 1200n$

Move all terms to one side to set the equation to zero. Subtract $1200n$ from both sides:

$-3n^2 + 1140n - 1200n + 24000 = 0$

$-3n^2 - 60n + 24000 = 0$

To make the leading coefficient positive and simplify the equation, multiply or divide the entire equation by -1:

$3n^2 + 60n - 24000 = 0$

Divide the entire equation by the common factor 3:

$\frac{3n^2}{3} + \frac{60n}{3} - \frac{24000}{3} = \frac{0}{3}$

$n^2 + 20n - 8000 = 0$

The quadratic equation is $n^2 + 20n - 8000 = 0$.

Solve the equation:

We solve the quadratic equation $n^2 + 20n - 8000 = 0$ for $n$. We can use the factorisation method by splitting the middle term.

We need to find two numbers whose product is $1 \times (-8000) = -8000$ and whose sum is the coefficient of $n$, which is 20.

Consider pairs of factors of 8000. We look for a pair that has a difference of 20. The pair (100, 80) has a product of 8000 and a difference of 20.

To get a product of -8000 and a sum of +20, the numbers must be 100 and -80.

$100 \times (-80) = -8000$

$100 + (-80) = 20$

Split the middle term $20n$ as $+100n - 80n$:

$n^2 + 100n - 80n - 8000 = 0$

Group the terms and factor out common factors:

$(n^2 + 100n) + (-80n - 8000) = 0$

$n(n + 100) - 80(n + 100) = 0$

Factor out the common binomial term $(n + 100)$:

$(n + 100)(n - 80) = 0$

Set each factor equal to zero to find the possible values of $n$:

$n + 100 = 0 \quad$ or $\quad n - 80 = 0$

$n = -100 \quad$ or $\quad n = 80$

Since the number of articles must be a positive integer, the solution $n = -100$ is not valid in this context.

The valid value for the number of articles bought is $n = 80$.

Check the answer: If the merchant bought 80 articles for $\textsf{₹}1200$, the original cost per article was $\frac{1200}{80} = \textsf{₹}15$. If the cost were $\textsf{₹}3$ less, the new cost would be $15 - 3 = \textsf{₹}12$. For $\textsf{₹}1200$ at $\textsf{₹}12$ per article, he could buy $\frac{1200}{12} = 100$ articles. This is $100 - 80 = 20$ more articles than originally, which matches the problem statement.

The number of articles the merchant bought is 80.

Given:

The sum of the areas of two squares is 468 m$^2$.

The difference of their perimeters is 24 m.

To Find: The sides of the two squares.

Solution:

Let the side length of the larger square be $x$ meters.

Let the side length of the smaller square be $y$ meters.

The area of the larger square is $x^2$ m$^2$.

The area of the smaller square is $y^2$ m$^2$.

The sum of their areas is given as 468 m$^2$.

So, we have the equation:

$x^2 + y^2 = 468$

The perimeter of the larger square is $4x$ meters.

The perimeter of the smaller square is $4y$ meters.

The difference of their perimeters is given as 24 m. Assuming the larger square has the larger perimeter:

$4x - 4y = 24$

We can simplify the equation for the difference in perimeters by dividing by 4:

$\frac{4x}{4} - \frac{4y}{4} = \frac{24}{4}$

$x - y = 6$

From this linear equation, we can express one variable in terms of the other. Let's express $x$ in terms of $y$:

$x = y + 6$

Now, we use this relationship to form a quadratic equation in one variable by substituting the expression for $x$ into the equation for the sum of areas:

$(y + 6)^2 + y^2 = 468$

Expand the term $(y + 6)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$y^2 + 2(y)(6) + 6^2 + y^2 = 468$

$y^2 + 12y + 36 + y^2 = 468$

Combine like terms:

$(y^2 + y^2) + 12y + 36 = 468$

$2y^2 + 12y + 36 = 468$

Move the constant term from the right side to the left side to get the standard quadratic form:

$2y^2 + 12y + 36 - 468 = 0$

$2y^2 + 12y - 432 = 0$

Divide the entire equation by the common factor 2 to simplify:

$\frac{2y^2}{2} + \frac{12y}{2} - \frac{432}{2} = \frac{0}{2}$

$y^2 + 6y - 216 = 0$

The quadratic equation to find the side $y$ is $y^2 + 6y - 216 = 0$.

Now, we solve this quadratic equation for $y$. We can use the factorisation method or the quadratic formula. Let's use factorisation.

We need to find two numbers whose product is $1 \times (-216) = -216$ and whose sum is the coefficient of $y$, which is 6.

Consider pairs of factors of 216. We look for a pair that has a difference of 6. Factors of 216 include $(1, 216), (2, 108), (3, 72), (4, 54), (6, 36), (8, 27), (9, 24), (12, 18)$.

The pair (12, 18) has a product of 216 and a difference of 6.

To get a product of -216 and a sum of +6, the two numbers must be 18 and -12.

$18 \times (-12) = -216$

$18 + (-12) = 6$

Split the middle term $6y$ as $+18y - 12y$:

$y^2 + 18y - 12y - 216 = 0$

Group the terms and factor out common factors:

$(y^2 + 18y) + (-12y - 216) = 0$

$y(y + 18) - 12(y + 18) = 0$

Factor out the common binomial term $(y + 18)$:

$(y + 18)(y - 12) = 0$

Set each factor equal to zero to find the possible values of $y$:

$y + 18 = 0 \quad$ or $\quad y - 12 = 0$

$y = -18 \quad$ or $\quad y = 12$

Since $y$ represents the side length of a square, it must be a positive value. Therefore, we discard the solution $y = -18$.

The valid value for the side of the smaller square is $y = 12$ meters.

Now find the side of the larger square $x$ using the relationship $x = y + 6$:

$x = 12 + 6$

$x = 18$ meters.

The sides of the two squares are 12 meters and 18 meters.

The sides of the two squares are 12 m and 18 m.

Given: $\alpha$ and $\beta$ are the roots of the quadratic equation $2x^2 - 6x + 3 = 0$.

The given quadratic equation is $2x^2 - 6x + 3 = 0$.

This is in the standard form $ax^2 + bx + c = 0$.

Comparing the coefficients, we have:

$a = 2$

$b = -6$

$c = 3$

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, the sum and product of the roots are given by:

Sum of roots: $\alpha + \beta = -\frac{b}{a}$

Product of roots: $\alpha \beta = \frac{c}{a}$

(a) Find the value of $\alpha + \beta$

Using the formula for the sum of roots:

$\alpha + \beta = -\frac{b}{a}$

Substitute the values $b = -6$ and $a = 2$:

$\alpha + \beta = -\frac{-6}{2}$

$\alpha + \beta = \frac{6}{2}$

$\alpha + \beta = 3$

The value of $\alpha + \beta$ is 3.

(b) Find the value of $\alpha \beta$

Using the formula for the product of roots:

$\alpha \beta = \frac{c}{a}$

Substitute the values $c = 3$ and $a = 2$:

$\alpha \beta = \frac{3}{2}$

The value of $\alpha \beta$ is $\frac{3}{2}$.

(c) Find the value of $\alpha^2 + \beta^2$

We know the algebraic identity: $(\alpha + \beta)^2 = \alpha^2 + 2\alpha \beta + \beta^2$.

From this, we can express $\alpha^2 + \beta^2$ as: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.

Substitute the values of $(\alpha + \beta)$ and $(\alpha \beta)$ calculated in parts (a) and (b):

$\alpha^2 + \beta^2 = (3)^2 - 2\left(\frac{3}{2}\right)$

$\alpha^2 + \beta^2 = 9 - \cancel{2} \times \frac{3}{\cancel{2}}$

$\alpha^2 + \beta^2 = 9 - 3$

$\alpha^2 + \beta^2 = 6$

The value of $\alpha^2 + \beta^2$ is 6.

(d) Find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$

First, combine the fractions on the left side by finding a common denominator, which is $\alpha \beta$:

$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta}{\alpha \beta} + \frac{\alpha}{\alpha \beta} = \frac{\alpha + \beta}{\alpha \beta}$

Substitute the values of $(\alpha + \beta)$ and $(\alpha \beta)$ calculated in parts (a) and (b):

$\frac{\alpha + \beta}{\alpha \beta} = \frac{3}{\frac{3}{2}}$

To divide by a fraction, multiply by its reciprocal:

$\frac{3}{\frac{3}{2}} = 3 \times \frac{2}{3}$

$\frac{1}{\alpha} + \frac{1}{\beta} = \cancel{3} \times \frac{2}{\cancel{3}} = 2$

The value of $\frac{1}{\alpha} + \frac{1}{\beta}$ is 2.

Given:

Distance between Mysore and Bangalore = 132 km.

Time taken by the express train is 1 hour less than the passenger train.

Average speed of the express train is 11 km/hr more than that of the passenger train.

To Find: The average speeds of the two trains.

Solution:

Let the average speed of the passenger train be $v$ km/hr.

Since the speed must be positive, we have $v > 0$.

According to the problem, the average speed of the express train is 11 km/hr more than that of the passenger train.

So, the average speed of the express train is $(v + 11)$ km/hr.

For the speed to be positive, $v+11 > 0$, which implies $v > -11$. Since $v>0$, this condition is already satisfied.

Using the formula Time = Distance / Speed:

Time taken by the passenger train ($T_p$) $= \frac{\text{Distance}}{\text{Speed of passenger train}} = \frac{132}{v}$ hours.

Time taken by the express train ($T_e$) $= \frac{\text{Distance}}{\text{Speed of express train}} = \frac{132}{v + 11}$ hours.

The problem states that the express train takes 1 hour less than the passenger train for the same journey.

So, the relationship between their times is:

$T_e = T_p - 1$

Substitute the expressions for $T_e$ and $T_p$:

$\frac{132}{v + 11} = \frac{132}{v} - 1$

Frame the quadratic equation:

Rearrange the equation to bring all terms to one side and express it in the standard form $ax^2 + bx + c = 0$.

Add 1 to both sides:

$\frac{132}{v + 11} + 1 = \frac{132}{v}$

Combine the terms on the left side by finding a common denominator $(v+11)$:

$\frac{132}{v + 11} + \frac{v + 11}{v + 11} = \frac{132}{v}$

$\frac{132 + v + 11}{v + 11} = \frac{132}{v}$

$\frac{v + 143}{v + 11} = \frac{132}{v}$

Cross-multiply:

$v(v + 143) = 132(v + 11)$

$v^2 + 143v = 132v + 132 \times 11$

Calculate $132 \times 11$:

$132 \times 11 = 1452$

So, the equation is:

$v^2 + 143v = 132v + 1452$

Move all terms to the left side to set the equation to zero:

$v^2 + 143v - 132v - 1452 = 0$

$v^2 + (143 - 132)v - 1452 = 0$

$v^2 + 11v - 1452 = 0$

The quadratic equation is $v^2 + 11v - 1452 = 0$.

Solve the equation:

We solve the quadratic equation $v^2 + 11v - 1452 = 0$ for $v$ using the factorisation method by splitting the middle term.

We need to find two numbers whose product is $1 \times (-1452) = -1452$ and whose sum is the coefficient of $v$, which is 11.

We are looking for two factors of 1452 that have a difference of 11. Let's try factoring 1452:

The prime factors are $2, 2, 3, 11, 11$. We need to group these factors into two numbers that differ by 11.

Let's try combinations:

• $(2 \times 2 \times 3) = 12$, $(11 \times 11) = 121$. Difference is $121 - 12 = 109$ (not 11).
• $(2 \times 2 \times 11) = 44$, $(3 \times 11) = 33$. Difference is $44 - 33 = 11$.

The two numbers are 44 and 33.

To get a product of -1452 and a sum of +11, the numbers must be 44 and -33.

$44 \times (-33) = -1452$

$44 + (-33) = 11$

Split the middle term $11v$ as $+44v - 33v$:

$v^2 + 44v - 33v - 1452 = 0$

Group the terms and factor out common factors:

$(v^2 + 44v) + (-33v - 1452) = 0$

$v(v + 44) - 33(v + 44) = 0$

Factor out the common binomial term $(v + 44)$:

$(v + 44)(v - 33) = 0$

Set each factor equal to zero to find the possible values of $v$:

$v + 44 = 0 \quad$ or $\quad v - 33 = 0$

$v = -44 \quad$ or $\quad v = 33$

Since speed must be a positive value, we discard the solution $v = -44$.

The valid value for the speed of the passenger train is $v = 33$ km/hr.

Now, calculate the speed of the express train, which is $v + 11$:

Speed of express train $= 33 + 11 = 44$ km/hr.

Check the times: Passenger train time = $\frac{132}{33} = 4$ hours. Express train time = $\frac{132}{44} = 3$ hours. The express train takes $4 - 3 = 1$ hour less, which matches the problem statement.

The average speed of the passenger train is 33 km/hr and the average speed of the express train is 44 km/hr.

Given:

The cost of production of each article was $\textsf{₹}3$ more than twice the number of articles produced.

The total cost of production on that day was $\textsf{₹}90$.

To Find: The number of articles produced and the cost of each article.

Solution:

Let the number of pottery articles produced on that day be $n$.

Since the number of articles must be a non-negative integer, $n \ge 0$. However, to have a cost of production, $n$ must be positive, so $n > 0$ and $n$ is an integer.

According to the problem, the cost of production of each article is 3 more than twice the number of articles produced.

Cost of each article $= 2 \times (\text{Number of articles}) + 3$

Cost of each article $= (2n + 3)$ $\textsf{₹}$.

The total cost of production is the product of the number of articles and the cost of each article.

Total Cost = (Number of articles) $\times$ (Cost of each article)

$90 = n \times (2n + 3)$

Frame the quadratic equation:

Expand and simplify the equation to bring it into the standard quadratic form $an^2 + bn + c = 0$.

$90 = n(2n) + n(3)$

$90 = 2n^2 + 3n$

Move all terms to one side to set the equation to zero. Subtract 90 from both sides:

$0 = 2n^2 + 3n - 90$

Rearranging the terms:

$2n^2 + 3n - 90 = 0$

The quadratic equation is $2n^2 + 3n - 90 = 0$.

Solve the equation:

We solve the quadratic equation $2n^2 + 3n - 90 = 0$ for $n$ using the factorisation method by splitting the middle term.

We need to find two numbers whose product is $a \times c = 2 \times (-90) = -180$ and whose sum is the coefficient of $n$, which is $b = 3$.

We look for factors of 180 that have a difference of 3. Consider pairs of factors of 180: $(1, 180), (2, 90), (3, 60), (4, 45), (5, 36), (6, 30), (9, 20), (10, 18), (12, 15)$.

The pair $(12, 15)$ has a product of $12 \times 15 = 180$ and a difference of $15 - 12 = 3$.

To get a product of -180 and a sum of +3, the two numbers must be 15 and -12.

$15 \times (-12) = -180$

$15 + (-12) = 3$

Split the middle term $3n$ as $+15n - 12n$:

$2n^2 + 15n - 12n - 90 = 0$

Group the terms and factor out common factors:

$(2n^2 + 15n) + (-12n - 90) = 0$

Factor out $n$ from the first group and $-6$ from the second group:

$n(2n + 15) - 6(2n + 15) = 0$

Factor out the common binomial term $(2n + 15)$:

$(2n + 15)(n - 6) = 0$

Set each factor equal to zero to find the possible values of $n$:

$2n + 15 = 0 \quad$ or $\quad n - 6 = 0$

$2n = -15 \quad$ or $\quad n = 6$

$n = -\frac{15}{2} \quad$ or $\quad n = 6$

Since the number of articles produced must be a positive integer, the solution $n = -\frac{15}{2}$ is not valid in this context.

The valid value for the number of articles produced is $n = 6$.

Now, we find the cost of each article using the relationship Cost of each article $= (2n + 3)$ $\textsf{₹}$.

Substitute $n = 6$:

Cost of each article $= 2(6) + 3$

Cost of each article $= 12 + 3$

Cost of each article $= 15$ $\textsf{₹}$.

Check the answer: Number of articles = 6, Cost per article = $\textsf{₹}15$. Total cost = $6 \times 15 = \textsf{₹}90$. This matches the given total cost. The cost per article ($15$) is indeed 3 more than twice the number of articles ($2 \times 6 + 3 = 12 + 3 = 15$).

The number of articles produced is 6 and the cost of each article is $\textsf{₹}15$.