Chapter 4 Quadratic Equations

Solution (i)

To Find: Mathematical representation of the situation.

Solution:

Let the number of marbles John had initially be $x$.

Since John and Jivanti together had 45 marbles, the number of marbles Jivanti had initially = $45 - x$.

After losing 5 marbles each:

Number of marbles John has = $x - 5$.

Number of marbles Jivanti has = $(45 - x) - 5 = 40 - x$.

According to the problem, the product of the number of marbles they now have is 124.

Therefore, $(x - 5)(40 - x) = 124$.

Expanding the expression:

$x(40 - x) - 5(40 - x) = 124$

$40x - x^2 - 200 + 5x = 124$

Combining like terms:

$-x^2 + (40x + 5x) - 200 = 124$

$-x^2 + 45x - 200 = 124$

Bringing all terms to one side:

$-x^2 + 45x - 200 - 124 = 0$

$-x^2 + 45x - 324 = 0$

Multiplying the entire equation by -1 to make the coefficient of $x^2$ positive:

$x^2 - 45x + 324 = 0$.

This is the required quadratic equation representing the situation mathematically.

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Solution (ii)

To Find: Mathematical representation of the situation.

Solution:

Let the number of toys produced on that day be $x$.

The cost of production of each toy on that day is given as 55 minus the number of toys produced.

So, the cost of production of each toy = $\textsf{₹} (55 - x)$.

The total cost of production is the product of the number of toys produced and the cost of production of each toy.

Total cost of production = (Number of toys produced) $\times$ (Cost of production of each toy)

Total Cost = $x \times (55 - x)$.

It is given that the total cost of production on that particular day was $\textsf{₹} 750$.

Therefore, $x(55 - x) = 750$.

Expanding the expression:

$55x - x^2 = 750$.

Bringing all terms to one side:

$-x^2 + 55x - 750 = 0$.

Multiplying the entire equation by -1 to make the coefficient of $x^2$ positive:

$x^2 - 55x + 750 = 0$.

This is the required quadratic equation representing the situation mathematically.

General Concept:

A polynomial equation of degree 2 is called a quadratic equation. The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

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

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Solution (i)

Given equation: $(x – 2)^2 + 1 = 2x – 3$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have the equation $(x – 2)^2 + 1 = 2x – 3$.

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

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

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

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

Now, move all terms to the left-hand side (LHS):

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

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 $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $(x – 2)^2 + 1 = 2x – 3$ is a quadratic equation.

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Solution (ii)

Given equation: $x(x + 1) + 8 = (x + 2)(x – 2)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have the equation $x(x + 1) + 8 = (x + 2)(x – 2)$.

Expand the LHS: $x^2 + x + 8$.

Expand the RHS using the identity $(a+b)(a-b) = a^2 - b^2$.

RHS = $x^2 - 2^2 = x^2 - 4$.

So the equation becomes: $x^2 + x + 8 = x^2 - 4$.

Now, move all terms to the LHS:

$x^2 + x + 8 - x^2 + 4 = 0$

Combine like terms:

$(x^2 - x^2) + x + (8 + 4) = 0$

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

$x + 12 = 0$

This equation is in the form $ax + b = 0$, where $a = 1$ and $b = 12$. The degree of the polynomial is 1.

It cannot be written in the form $ax^2 + bx + c = 0$ with $a \neq 0$ (here $a=0$).

Conclusion: Therefore, the given equation $x(x + 1) + 8 = (x + 2)(x – 2)$ is not a quadratic equation. It is a linear equation.

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Solution (iii)

Given equation: $x(2x + 3) = x^2 + 1$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have the equation $x(2x + 3) = x^2 + 1$.

Expand the LHS:

$2x^2 + 3x = x^2 + 1$.

Now, move all terms to the LHS:

$2x^2 + 3x - x^2 - 1 = 0$

Combine like terms:

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

$x^2 + 3x - 1 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 3$, and $c = -1$.

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

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

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Solution (iv)

Given equation: $(x + 2)^3 = x^3 – 4$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have the equation $(x + 2)^3 = x^3 – 4$.

Expand the term $(x + 2)^3$ using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

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

$x^3 + 6x^2 + 3x(4) + 8 = x^3 - 4$

$x^3 + 6x^2 + 12x + 8 = x^3 - 4$

Now, move all terms to the LHS:

$x^3 + 6x^2 + 12x + 8 - x^3 + 4 = 0$

Combine like terms:

$(x^3 - x^3) + 6x^2 + 12x + (8 + 4) = 0$

$0x^3 + 6x^2 + 12x + 12 = 0$

$6x^2 + 12x + 12 = 0$

We can divide the entire equation by 6:

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

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

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $(x + 2)^3 = x^3 – 4$ is a quadratic equation.

General Concept:

A polynomial equation of degree 2 is called a quadratic equation. The standard form of a quadratic equation is $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

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

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Solution (i)

Given equation: $(x + 1)^2 = 2(x – 3)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $(x + 1)^2 = 2(x – 3)$.

Expand the LHS using $(a+b)^2 = a^2 + 2ab + b^2$:

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

$x^2 + 2x + 1 = 2x - 6$

Move all terms to the LHS:

$x^2 + 2x + 1 - 2x + 6 = 0$

Combine like terms:

$x^2 + (2x - 2x) + (1 + 6) = 0$

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

$x^2 + 7 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = 0$, and $c = 7$.

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $(x + 1)^2 = 2(x – 3)$ is a quadratic equation.

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Solution (ii)

Given equation: $x^2 – 2x = (–2) (3 – x)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $x^2 – 2x = (–2) (3 – x)$.

Expand the RHS:

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

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

Move all terms to the LHS:

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

Combine like terms:

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

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

This equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -4$, and $c = 6$.

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $x^2 – 2x = (–2) (3 – x)$ is a quadratic equation.

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Solution (iii)

Given equation: $(x – 2)(x + 1) = (x – 1)(x + 3)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $(x – 2)(x + 1) = (x – 1)(x + 3)$.

Expand both sides:

LHS: $x(x+1) - 2(x+1) = x^2 + x - 2x - 2 = x^2 - x - 2$

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

So the equation is: $x^2 - x - 2 = x^2 + 2x - 3$.

Move all terms to the LHS:

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

Combine like terms:

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

$0x^2 - 3x + 1 = 0$

$-3x + 1 = 0$

This equation is a linear equation ($ax + b = 0$ with $a=-3, b=1$). The highest power of x is 1.

It cannot be written in the form $ax^2 + bx + c = 0$ with $a \neq 0$ (here $a=0$).

Conclusion: Therefore, the given equation $(x – 2)(x + 1) = (x – 1)(x + 3)$ is not a quadratic equation.

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Solution (iv)

Given equation: $(x – 3)(2x +1) = x(x + 5)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $(x – 3)(2x +1) = x(x + 5)$.

Expand both sides:

LHS: $x(2x+1) - 3(2x+1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$

RHS: $x(x+5) = x^2 + 5x$

So the equation is: $2x^2 - 5x - 3 = x^2 + 5x$.

Move all terms to the LHS:

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

Combine like terms:

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

$x^2 - 10x - 3 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -10$, and $c = -3$.

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $(x – 3)(2x +1) = x(x + 5)$ is a quadratic equation.

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Solution (v)

Given equation: $(2x – 1)(x – 3) = (x + 5)(x – 1)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $(2x – 1)(x – 3) = (x + 5)(x – 1)$.

Expand both sides:

LHS: $2x(x-3) - 1(x-3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$

RHS: $x(x-1) + 5(x-1) = x^2 - x + 5x - 5 = x^2 + 4x - 5$

So the equation is: $2x^2 - 7x + 3 = x^2 + 4x - 5$.

Move all terms to the LHS:

$2x^2 - 7x + 3 - x^2 - 4x + 5 = 0$

Combine like terms:

$(2x^2 - x^2) + (-7x - 4x) + (3 + 5) = 0$

$x^2 - 11x + 8 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -11$, and $c = 8$.

Since $a = 1 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $(2x – 1)(x – 3) = (x + 5)(x – 1)$ is a quadratic equation.

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Solution (vi)

Given equation: $x^2 + 3x + 1 = (x – 2)^2$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $x^2 + 3x + 1 = (x – 2)^2$.

Expand the RHS using $(a-b)^2 = a^2 - 2ab + b^2$:

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

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

Move all terms to the LHS:

$x^2 + 3x + 1 - x^2 + 4x - 4 = 0$

Combine like terms:

$(x^2 - x^2) + (3x + 4x) + (1 - 4) = 0$

$0x^2 + 7x - 3 = 0$

$7x - 3 = 0$

This equation is a linear equation ($ax + b = 0$ with $a=7, b=-3$). The highest power of x is 1.

It cannot be written in the form $ax^2 + bx + c = 0$ with $a \neq 0$ (here $a=0$).

Conclusion: Therefore, the given equation $x^2 + 3x + 1 = (x – 2)^2$ is not a quadratic equation.

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Solution (vii)

Given equation: $(x + 2)^3 = 2x (x^2 – 1)$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $(x + 2)^3 = 2x (x^2 – 1)$.

Expand the LHS using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

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

$x^3 + 6x^2 + 3x(4) + 8 = 2x^3 - 2x$

$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$

Move all terms to the LHS:

$x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0$

Combine like terms:

$(x^3 - 2x^3) + 6x^2 + (12x + 2x) + 8 = 0$

$-x^3 + 6x^2 + 14x + 8 = 0$

Multiply by -1 (optional, for standard form):

$x^3 - 6x^2 - 14x - 8 = 0$

The highest power of x in this equation is 3. Therefore, it is a cubic equation.

Conclusion: Therefore, the given equation $(x + 2)^3 = 2x (x^2 – 1)$ is not a quadratic equation. It is a cubic equation.

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Solution (viii)

Given equation: $x^3 – 4x^2 – x + 1 = (x – 2)^3$

To Check: Whether the given equation is a quadratic equation.

Solution:

We have $x^3 – 4x^2 – x + 1 = (x – 2)^3$.

Expand the RHS using $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:

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

$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 3x(4) - 8$

$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$

Move all terms from RHS to LHS (or notice the $x^3$ terms cancel):

$x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$

Combine like terms:

$(x^3 - x^3) + (-4x^2 + 6x^2) + (-x - 12x) + (1 + 8) = 0$

$0x^3 + 2x^2 - 13x + 9 = 0$

$2x^2 - 13x + 9 = 0$

This equation is in the form $ax^2 + bx + c = 0$, where $a = 2$, $b = -13$, and $c = 9$.

Since $a = 2 \neq 0$, the degree of the polynomial is 2.

Conclusion: Therefore, the given equation $x^3 – 4x^2 – x + 1 = (x – 2)^3$ is a quadratic equation.

Solution (i)

Given:

Area of the rectangular plot = 528 m².

The length of the plot (in metres) is one more than twice its breadth.

To Find:

Represent the situation in the form of a quadratic equation.

Solution:

Let the breadth of the rectangular plot be $x$ metres.

According to the condition, the length of the plot is one more than twice its breadth.

Length = $2 \times (\text{breadth}) + 1 = (2x + 1)$ metres.

We know that the area of a rectangle = Length $\times$ Breadth.

Given Area = 528 m².

Therefore, $(2x + 1) \times x = 528$.

$2x^2 + x = 528$.

Subtracting 528 from both sides to get the standard quadratic form:

$2x^2 + x - 528 = 0$.

This is the required quadratic equation representing the situation. Here, $x$ represents the breadth of the plot in metres. Since breadth must be positive, $x > 0$.

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Solution (ii)

Given:

The product of two consecutive positive integers is 306.

To Find:

Represent the situation in the form of a quadratic equation.

Solution:

Let the first positive integer be $x$.

Since the integers are consecutive, the next positive integer will be $x + 1$.

According to the condition, the product of these two integers is 306.

Therefore, $x \times (x + 1) = 306$.

Expanding the equation:

$x^2 + x = 306$.

Subtracting 306 from both sides to get the standard quadratic form:

$x^2 + x - 306 = 0$.

This is the required quadratic equation representing the situation. Here, $x$ represents the smaller of the two consecutive positive integers, so $x$ must be a positive integer.

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Solution (iii)

Given:

Rohan’s mother is 26 years older than him.

The product of their ages (in years) 3 years from now will be 360.

To Find:

Represent the situation in the form of a quadratic equation to find Rohan's present age.

Solution:

Let Rohan's present age be $x$ years.

Since his mother is 26 years older, her present age is $(x + 26)$ years.

Ages 3 years from now:

Rohan's age = $(x + 3)$ years.

Mother's age = $(x + 26) + 3 = (x + 29)$ years.

According to the condition, the product of their ages 3 years from now will be 360.

Therefore, $(x + 3)(x + 29) = 360$.

Expanding the equation:

$x(x + 29) + 3(x + 29) = 360$

$x^2 + 29x + 3x + 87 = 360$

$x^2 + 32x + 87 = 360$

Subtracting 360 from both sides to get the standard quadratic form:

$x^2 + 32x + 87 - 360 = 0$

$x^2 + 32x - 273 = 0$.

This is the required quadratic equation representing the situation. Here, $x$ represents Rohan's present age in years, and $x > 0$.

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Solution (iv)

Given:

Distance travelled by the train = 480 km.

The train travels at a uniform speed.

If the speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance.

To Find:

Represent the situation in the form of a quadratic equation to find the speed of the train.

Solution:

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

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Original time taken to cover 480 km ($T_1$) = $\frac{480}{x}$ hours.

If the speed had been 8 km/h less, the new speed would be $(x - 8)$ km/h.

Time taken to cover 480 km with the new speed ($T_2$) = $\frac{480}{x - 8}$ hours.

According to the condition, the new time ($T_2$) is 3 hours more than the original time ($T_1$).

$T_2 = T_1 + 3$

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

Rearranging the terms:

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

Taking 480 common from the LHS:

$480 \left( \frac{1}{x - 8} - \frac{1}{x} \right) = 3$

Finding a common denominator inside the parenthesis:

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

$480 \left( \frac{x - x + 8}{x^2 - 8x} \right) = 3$

$480 \left( \frac{8}{x^2 - 8x} \right) = 3$

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

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

Cross-multiplying (or multiplying both sides by $x^2 - 8x$):

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

Divide both sides by 3:

$1280 = x^2 - 8x$

Rearranging to get the standard quadratic form:

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

This is the required quadratic equation representing the situation. Here, $x$ represents the uniform speed of the train in km/h. Since speed must be positive and the reduced speed $(x-8)$ must also be positive, we have the condition $x > 8$.

Given:

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

To Find:

The roots of the given equation by the factorisation method.

Solution:

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

To factorise the quadratic expression $ax^2 + bx + c$, we need to find two numbers whose product is $a \times c$ and whose sum is $b$.

In this equation, $a = 2$, $b = -5$, and $c = 3$.

We need two numbers whose product is $a \times c = 2 \times 3 = 6$ and whose sum is $b = -5$.

Let's find pairs of factors of 6:

• 1 and 6 (Sum = 1 + 6 = 7)
• -1 and -6 (Sum = -1 + (-6) = -7)
• 2 and 3 (Sum = 2 + 3 = 5)
• -2 and -3 (Sum = -2 + (-3) = -5)

The pair of numbers that satisfies both conditions (product = 6, sum = -5) is -2 and -3.

Now, we split the middle term ($-5x$) using these two numbers:

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

Next, we factor by grouping the terms:

Group the first two terms and the last two terms:

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

Factor out the greatest common factor (GCF) from each group:

$2x(x - 1) - 3(x - 1) = 0$

Now, factor out the common binomial factor $(x - 1)$:

$(x - 1)(2x - 3) = 0$

According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero.

So, either $x - 1 = 0$ or $2x - 3 = 0$.

Case 1: $x - 1 = 0$

$x = 1$

Case 2: $2x - 3 = 0$

$2x = 3$

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

Thus, the roots of the equation $2x^2 - 5x + 3 = 0$ are $1$ and $\frac{3}{2}$.

Final Answer:

The roots of the equation $2x^2 - 5x + 3 = 0$ are $\mathbf{1}$ and $\mathbf{\frac{3}{2}}$.

Given:

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

To Find:

The roots of the given equation.

Solution:

We will find the roots using the factorisation method.

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

To factorise the quadratic expression $ax^2 + bx + c$, we need to find two numbers whose product is $a \times c$ and whose sum is $b$.

In this equation, $a = 6$, $b = -1$, and $c = -2$.

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

Let's find pairs of factors of -12 and their sums:

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

The pair of numbers that satisfies both conditions (product = -12, sum = -1) is 3 and -4.

Now, we split the middle term ($-x$) using these two numbers:

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

Next, we factor by grouping the terms:

Group the first two terms and the last two terms:

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

Factor out the greatest common factor (GCF) from each group:

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

Now, factor out the common binomial factor $(2x + 1)$:

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

According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero.

So, either $2x + 1 = 0$ or $3x - 2 = 0$.

Case 1: $2x + 1 = 0$

$2x = -1$

$x = -\frac{1}{2}$

Case 2: $3x - 2 = 0$

$3x = 2$

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

Thus, the roots of the equation $6x^2 - x - 2 = 0$ are $-\frac{1}{2}$ and $\frac{2}{3}$.

Final Answer:

The roots of the equation $6x^2 - x - 2 = 0$ are $\mathbf{-\frac{1}{2}}$ and $\mathbf{\frac{2}{3}}$.

Given:

The quadratic equation is $3x^2 - 2\sqrt{6}x + 2 = 0$.

To Find:

The roots of the given equation.

Solution:

We will find the roots using the factorisation method.

The given quadratic equation is $3x^2 - 2\sqrt{6}x + 2 = 0$.

To factorise the quadratic expression $ax^2 + bx + c$, we need to find two numbers whose product is $a \times c$ and whose sum is $b$.

In this equation, $a = 3$, $b = -2\sqrt{6}$, and $c = 2$.

We need two numbers whose product is $a \times c = 3 \times 2 = 6$ and whose sum is $b = -2\sqrt{6}$.

Let's consider splitting the middle term coefficient, $-2\sqrt{6}$. We are looking for two numbers $p$ and $q$ such that $p+q = -2\sqrt{6}$ and $p \times q = 6$.

The numbers are $-\sqrt{6}$ and $-\sqrt{6}$.

Check:

Sum: $(-\sqrt{6}) + (-\sqrt{6}) = -2\sqrt{6}$ (Matches $b$)

Product: $(-\sqrt{6}) \times (-\sqrt{6}) = 6$ (Matches $ac$)

Now, we split the middle term ($-2\sqrt{6}x$) using these two numbers:

$3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0$

Next, we factor by grouping the terms. We can rewrite the coefficients to make factoring easier:

Note that $3 = (\sqrt{3})^2$, $\sqrt{6} = \sqrt{3} \times \sqrt{2}$, and $2 = (\sqrt{2})^2$.

So the equation becomes: $(\sqrt{3})^2 x^2 - (\sqrt{3}\sqrt{2})x - (\sqrt{3}\sqrt{2})x + (\sqrt{2})^2 = 0$

Group the first two terms and the last two terms:

$[(\sqrt{3})^2 x^2 - (\sqrt{3}\sqrt{2})x] + [-(\sqrt{3}\sqrt{2})x + (\sqrt{2})^2] = 0$

Factor out the greatest common factor (GCF) from each group:

From the first group, the GCF is $\sqrt{3}x$:

$\sqrt{3}x (\sqrt{3}x - \sqrt{2})$

From the second group, the GCF is $-\sqrt{2}$ (factoring out the negative sign):

$-\sqrt{2} (\sqrt{3}x - \sqrt{2})$

Substitute back into the equation:

$\sqrt{3}x (\sqrt{3}x - \sqrt{2}) - \sqrt{2} (\sqrt{3}x - \sqrt{2}) = 0$

Now, factor out the common binomial factor $(\sqrt{3}x - \sqrt{2})$:

$(\sqrt{3}x - \sqrt{2})(\sqrt{3}x - \sqrt{2}) = 0$

This can be written as $(\sqrt{3}x - \sqrt{2})^2 = 0$.

According to the Zero Product Property, we set the factor equal to zero:

$\sqrt{3}x - \sqrt{2} = 0$

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

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

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$:

$x = \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{\sqrt{6}}{3}$

Since the factor $(\sqrt{3}x - \sqrt{2})$ is repeated, the quadratic equation has two equal real roots.

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

Final Answer:

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

Given:

A rectangular prayer hall.

Carpet Area = 300 m².

Length = 1 metre more than twice its breadth.

To Find:

The length and breadth of the prayer hall.

Solution:

Let the breadth of the prayer hall be $x$ metres.

According to the given condition, the length is one metre more than twice its breadth.

Length = $2 \times (\text{breadth}) + 1 = (2x + 1)$ metres.

The area of a rectangle is given by the formula: Area = Length $\times$ Breadth.

We are given that the area is 300 m².

Therefore, $(2x + 1) \times x = 300$.

Expanding the equation:

$2x^2 + x = 300$.

Rearranging the terms to form a standard quadratic equation:

$2x^2 + x - 300 = 0$.

Now, we need to solve this quadratic equation for $x$. We can use the factorisation method.

We need to find two numbers whose product is $a \times c = 2 \times (-300) = -600$ and whose sum is $b = 1$.

Let's find pairs of factors of -600 whose sum is 1.

Consider the factors 25 and -24:

Product = $25 \times (-24) = -600$.

Sum = $25 + (-24) = 1$.

These numbers satisfy the conditions.

Now, split the middle term ($x$) using these numbers:

$2x^2 + 25x - 24x - 300 = 0$.

Factor by grouping:

$(2x^2 + 25x) + (-24x - 300) = 0$.

Factor out the GCF from each group:

$x(2x + 25) - 12(2x + 25) = 0$.

Factor out the common binomial factor $(2x + 25)$:

$(2x + 25)(x - 12) = 0$.

Using the Zero Product Property, set each factor to zero:

Either $2x + 25 = 0$ or $x - 12 = 0$.

Case 1: $2x + 25 = 0$

$2x = -25$

$x = -\frac{25}{2} = -12.5$.

Case 2: $x - 12 = 0$

$x = 12$.

Since $x$ represents the breadth of the hall, it must be a positive value. Therefore, $x = -12.5$ is not a valid solution.

The breadth of the hall is $x = 12$ metres.

Now, calculate the length using the expression Length = $2x + 1$:

Length = $2(12) + 1 = 24 + 1 = 25$ metres.

Final Answer:

The dimensions of the prayer hall are:

Breadth = 12 metres

Length = 25 metres

(Check: Area = $12 \times 25 = 300$ m². Length $25 = 2 \times 12 + 1$. Conditions are satisfied.)

Solution (i)

Given equation: $x^2 – 3x – 10 = 0$

To Find: Roots by factorisation.

Solution:

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

The pairs of factors of -10 are (1, -10), (-1, 10), (2, -5), (-2, 5).

The pair whose sum is -3 is 2 and -5.

Splitting the middle term:

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

Grouping the terms:

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

Factoring out the GCF from each group:

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

Factoring out the common binomial factor $(x + 2)$:

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

By the Zero Product Property:

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

Or $x - 5 = 0 \implies x = 5$

Conclusion: The roots of the equation $x^2 – 3x – 10 = 0$ are $\mathbf{-2}$ and $\mathbf{5}$.

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Solution (ii)

Given equation: $2x^2 + x – 6 = 0$

To Find: Roots by factorisation.

Solution:

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

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

The pair whose sum is 1 is -3 and 4.

Splitting the middle term:

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

Grouping the terms:

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

Factoring out the GCF from each group:

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

Factoring out the common binomial factor $(2x - 3)$:

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

By the Zero Product Property:

Either $2x - 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$

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

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

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Solution (iii)

Given equation: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

To Find: Roots by factorisation.

Solution:

We need to find two numbers whose product is $\sqrt{2} \times 5\sqrt{2} = 5 \times (\sqrt{2})^2 = 5 \times 2 = 10$ and whose sum is $7$.

The pairs of factors of 10 are (1, 10), (-1, -10), (2, 5), (-2, -5).

The pair whose sum is 7 is 2 and 5.

Splitting the middle term:

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

Grouping the terms. Note that $2 = \sqrt{2} \times \sqrt{2}$.

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

Factoring out the GCF from each group:

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

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

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

By the Zero Product Property:

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

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

Rationalizing the denominator for the second root: $x = -\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$

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

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Solution (iv)

Given equation: $2x^2 - x + \frac{1}{8} = 0$

To Find: Roots by factorisation.

Solution:

First, multiply the entire equation by 8 to eliminate the fraction:

$8 \times (2x^2 - x + \frac{1}{8}) = 8 \times 0$

$16x^2 - 8x + 1 = 0$

Now, factorise $16x^2 - 8x + 1 = 0$. We need two numbers whose product is $16 \times 1 = 16$ and whose sum is $-8$.

The pair of numbers is -4 and -4.

Splitting the middle term:

$16x^2 - 4x - 4x + 1 = 0$

Grouping the terms:

$(16x^2 - 4x) + (-4x + 1) = 0$

Factoring out the GCF from each group:

$4x(4x - 1) - 1(4x - 1) = 0$

Factoring out the common binomial factor $(4x - 1)$:

$(4x - 1)(4x - 1) = 0$

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

By the Zero Product Property:

$4x - 1 = 0 \implies 4x = 1 \implies x = \frac{1}{4}$

Since the factor is repeated, we have equal roots.

Conclusion: The roots of the equation $2x^2 - x + \frac{1}{8} = 0$ are $\mathbf{\frac{1}{4}}$ and $\mathbf{\frac{1}{4}}$.

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Solution (v)

Given equation: $100x^2 - 20x + 1 = 0$

To Find: Roots by factorisation.

Solution:

We need to find two numbers whose product is $100 \times 1 = 100$ and whose sum is $-20$.

The pair of numbers is -10 and -10.

Splitting the middle term:

$100x^2 - 10x - 10x + 1 = 0$

Grouping the terms:

$(100x^2 - 10x) + (-10x + 1) = 0$

Factoring out the GCF from each group:

$10x(10x - 1) - 1(10x - 1) = 0$

Factoring out the common binomial factor $(10x - 1)$:

$(10x - 1)(10x - 1) = 0$

$(10x - 1)^2 = 0$

By the Zero Product Property:

$10x - 1 = 0 \implies 10x = 1 \implies x = \frac{1}{10}$

Since the factor is repeated, we have equal roots.

Conclusion: The roots of the equation $100x^2 - 20x + 1 = 0$ are $\mathbf{\frac{1}{10}}$ and $\mathbf{\frac{1}{10}}$.

Solution (i)

From Example 1(i), the mathematical representation of the situation is the quadratic equation:

$x^2 - 45x + 324 = 0$

where $x$ is the number of marbles John had initially.

To Find: The initial number of marbles John and Jivanti had.

Solution:

We solve the quadratic equation $x^2 - 45x + 324 = 0$ by factorisation.

We need two numbers whose product is $1 \times 324 = 324$ and whose sum is $-45$.

Since the product is positive and the sum is negative, both numbers must be negative.

Let's find factors of 324. We find that $36 \times 9 = 324$ and $36 + 9 = 45$.

So, the two numbers are $-36$ and $-9$.

Splitting the middle term:

$x^2 - 9x - 36x + 324 = 0$

Grouping the terms:

$(x^2 - 9x) + (-36x + 324) = 0$

Factoring out the GCF from each group:

$x(x - 9) - 36(x - 9) = 0$

Factoring out the common binomial factor $(x - 9)$:

$(x - 9)(x - 36) = 0$

By the Zero Product Property:

Either $x - 9 = 0 \implies x = 9$

Or $x - 36 = 0 \implies x = 36$

So, the possible number of marbles John had initially is 9 or 36.

Case 1: If John had $x=9$ marbles, then Jivanti had $45 - x = 45 - 9 = 36$ marbles.

Case 2: If John had $x=36$ marbles, then Jivanti had $45 - x = 45 - 36 = 9$ marbles.

In both cases, one person had 9 marbles and the other had 36 marbles initially.

Check: If they had 9 and 36 marbles. After losing 5 each, they have $9-5=4$ and $36-5=31$ marbles. Product = $4 \times 31 = 124$. This matches the given condition.

Conclusion: The number of marbles they had to start with were $\mathbf{9}$ and $\mathbf{36}$.

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Solution (ii)

From Example 1(ii), the mathematical representation of the situation is the quadratic equation:

$x^2 - 55x + 750 = 0$

where $x$ is the number of toys produced on that day.

To Find: The number of toys produced on that day.

Solution:

We solve the quadratic equation $x^2 - 55x + 750 = 0$ by factorisation.

We need two numbers whose product is $1 \times 750 = 750$ and whose sum is $-55$.

Since the product is positive and the sum is negative, both numbers must be negative.

Let's find factors of 750. We find that $25 \times 30 = 750$ and $25 + 30 = 55$.

So, the two numbers are $-25$ and $-30$.

Splitting the middle term:

$x^2 - 25x - 30x + 750 = 0$

Grouping the terms:

$(x^2 - 25x) + (-30x + 750) = 0$

Factoring out the GCF from each group:

$x(x - 25) - 30(x - 25) = 0$

Factoring out the common binomial factor $(x - 25)$:

$(x - 25)(x - 30) = 0$

By the Zero Product Property:

Either $x - 25 = 0 \implies x = 25$

Or $x - 30 = 0 \implies x = 30$

Both values are possible numbers for the toys produced.

Case 1: If $x=25$ toys were produced, the cost of each toy is $55 - 25 = \textsf{₹} 30$. Total cost = $25 \times 30 = \textsf{₹} 750$.

Case 2: If $x=30$ toys were produced, the cost of each toy is $55 - 30 = \textsf{₹} 25$. Total cost = $30 \times 25 = \textsf{₹} 750$.

Both scenarios fit the problem description.

Conclusion: The number of toys produced on that day was either $\mathbf{25}$ or $\mathbf{30}$.

Given:

The sum of two numbers is 27.

The product of the two numbers is 182.

To Find:

The two numbers.

Solution:

Let the two numbers be $x$ and $y$.

According to the problem statement, we have the following equations:

From equation (i), we can express $y$ in terms of $x$:

$y = 27 - x$

Now substitute this expression for $y$ into equation (ii):

$x(27 - x) = 182$

Expand the equation:

$27x - x^2 = 182$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$-x^2 + 27x - 182 = 0$

Multiply the entire equation by -1 to make the leading coefficient positive:

$x^2 - 27x + 182 = 0$

Now, we solve this quadratic equation by factorisation.

We need to find two numbers whose product is $1 \times 182 = 182$ and whose sum is $-27$.

Since the product is positive and the sum is negative, both numbers must be negative.

Let's find the factors of 182:

$182 = 2 \times 91 = 2 \times 7 \times 13$.

The pairs of factors are (1, 182), (2, 91), (7, 26), (13, 14).

The sum of 13 and 14 is $13 + 14 = 27$. Since we need a sum of -27, the required numbers are -13 and -14.

Split the middle term ($-27x$) using -13 and -14:

$x^2 - 13x - 14x + 182 = 0$

Factor by grouping:

$(x^2 - 13x) + (-14x + 182) = 0$

$x(x - 13) - 14(x - 13) = 0$

Factor out the common binomial factor $(x - 13)$:

$(x - 13)(x - 14) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x - 13 = 0 \implies x = 13$

Or $x - 14 = 0 \implies x = 14$

Now, find the corresponding value of $y$ using $y = 27 - x$.

If $x = 13$, then $y = 27 - 13 = 14$.

If $x = 14$, then $y = 27 - 14 = 13$.

In both cases, the two numbers are 13 and 14.

Check:

Sum: $13 + 14 = 27$. (Correct)

Product: $13 \times 14 = 182$. (Correct)

Final Answer:

The two numbers are $\mathbf{13}$ and $\mathbf{14}$.

Given:

Two consecutive positive integers.

The sum of the squares of these integers is 365.

To Find:

The two consecutive positive integers.

Solution:

Let the first positive integer be $x$.

Since the integers are consecutive, the next positive integer is $x + 1$.

According to the problem statement, the sum of the squares of these two integers is 365.

Square of the first integer = $x^2$.

Square of the second integer = $(x + 1)^2$.

The equation representing the condition is:

$x^2 + (x + 1)^2 = 365$

Expand the term $(x + 1)^2$ using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

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

Substitute this back into the equation:

$x^2 + (x^2 + 2x + 1) = 365$

Combine like terms:

$2x^2 + 2x + 1 = 365$

Subtract 365 from both sides to set the equation to zero:

$2x^2 + 2x + 1 - 365 = 0$

$2x^2 + 2x - 364 = 0$

Divide the entire equation by 2 to simplify:

$\frac{2x^2}{2} + \frac{2x}{2} - \frac{364}{2} = \frac{0}{2}$

$x^2 + x - 182 = 0$

Now, we solve this quadratic equation by factorisation.

We need to find two numbers whose product is $1 \times (-182) = -182$ and whose sum is $1$.

Let's find the factors of 182: $182 = 2 \times 91 = 2 \times 7 \times 13$.

We look for two factors whose difference is 1. These are 13 and 14.

Since the sum is positive (+1), the larger factor (14) must be positive, and the smaller factor (13) must be negative. The two numbers are 14 and -13.

Check: Product = $14 \times (-13) = -182$. Sum = $14 + (-13) = 1$.

Split the middle term ($x$) using 14 and -13:

$x^2 + 14x - 13x - 182 = 0$

Factor by grouping:

$(x^2 + 14x) + (-13x - 182) = 0$

$x(x + 14) - 13(x + 14) = 0$

Factor out the common binomial factor $(x + 14)$:

$(x + 14)(x - 13) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x + 14 = 0 \implies x = -14$

Or $x - 13 = 0 \implies x = 13$

The problem asks for positive integers. Therefore, $x = -14$ is not a valid solution.

The first positive integer is $x = 13$.

The second consecutive positive integer is $x + 1 = 13 + 1 = 14$.

Check:

The integers are 13 and 14. They are consecutive and positive.

Sum of their squares = $13^2 + 14^2 = 169 + 196 = 365$. (Correct)

Final Answer:

The two consecutive positive integers are $\mathbf{13}$ and $\mathbf{14}$.

Given:

A right-angled triangle.

The altitude is 7 cm less than its base.

The hypotenuse is 13 cm.

To Find:

The lengths of the other two sides (base and altitude).

Solution:

Let the base of the right triangle be $x$ cm.

According to the problem statement, the altitude is 7 cm less than the base.

So, Altitude = $(x - 7)$ cm.

The hypotenuse is given as 13 cm.

By the Pythagorean theorem, for a right-angled triangle:

$(\text{Base})^2 + (\text{Altitude})^2 = (\text{Hypotenuse})^2$

Substituting the values:

$x^2 + (x - 7)^2 = 13^2$

Expand $(x - 7)^2$ using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$(x - 7)^2 = x^2 - 2(x)(7) + 7^2 = x^2 - 14x + 49$

Substitute this back into the Pythagorean equation:

$x^2 + (x^2 - 14x + 49) = 169$

Combine like terms:

$2x^2 - 14x + 49 = 169$

Move all terms to one side to form a quadratic equation:

$2x^2 - 14x + 49 - 169 = 0$

$2x^2 - 14x - 120 = 0$

Divide the entire equation by 2 to simplify:

$\frac{2x^2}{2} - \frac{14x}{2} - \frac{120}{2} = \frac{0}{2}$

$x^2 - 7x - 60 = 0$

Now, solve this quadratic equation by factorisation.

We need to find two numbers whose product is $1 \times (-60) = -60$ and whose sum is $-7$.

The factors of -60 whose sum is -7 are 5 and -12. ($5 \times -12 = -60$ and $5 + (-12) = -7$)

Split the middle term ($-7x$):

$x^2 + 5x - 12x - 60 = 0$

Factor by grouping:

$(x^2 + 5x) + (-12x - 60) = 0$

$x(x + 5) - 12(x + 5) = 0$

Factor out the common binomial factor $(x + 5)$:

$(x + 5)(x - 12) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x + 5 = 0 \implies x = -5$

Or $x - 12 = 0 \implies x = 12$

Since $x$ represents the length of the base of the triangle, it cannot be negative. Therefore, $x = -5$ is rejected.

The base of the triangle is $x = 12$ cm.

The altitude of the triangle is $x - 7 = 12 - 7 = 5$ cm.

Check:

Base = 12 cm, Altitude = 5 cm, Hypotenuse = 13 cm.

Altitude (5 cm) is 7 cm less than Base (12 cm). (Correct)

Check Pythagorean theorem: $12^2 + 5^2 = 144 + 25 = 169$. And $13^2 = 169$. The theorem holds true.

Final Answer:

The other two sides of the right triangle are its base = 12 cm and its altitude = 5 cm.

Given:

A cottage industry produces pottery articles.

Cost of production of each article = 3 more than twice the number of articles produced that day.

Total cost of production on that day = $\textsf{₹} 90$.

To Find:

The number of articles produced on that day.

The cost of each article.

Solution:

Let the number of pottery articles produced on that day be $x$.

According to the problem statement, the cost of production of each article (in rupees) was 3 more than twice the number of articles produced.

Cost of each article = $2 \times (\text{Number of articles}) + 3 = (2x + 3)$ rupees.

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)

Given that the total cost of production was $\textsf{₹} 90$.

Therefore, $x \times (2x + 3) = 90$.

Expand the equation:

$2x^2 + 3x = 90$.

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$2x^2 + 3x - 90 = 0$.

Now, we solve this quadratic equation by factorisation.

We need to find two numbers whose product is $a \times c = 2 \times (-90) = -180$ and whose sum is $b = 3$.

Since the product is negative, the two numbers must have opposite signs. Since the sum is positive, the larger number (in magnitude) must be positive.

Factors of 180 are: (1, 180), (2, 90), (3, 60), (4, 45), (5, 36), (6, 30), (9, 20), (10, 18), (12, 15).

We need a pair whose difference is 3. The pair is 12 and 15.

To get a sum of +3, the numbers must be 15 and -12.

Check: Product = $15 \times (-12) = -180$. Sum = $15 + (-12) = 3$.

Split the middle term ($3x$) using 15 and -12:

$2x^2 + 15x - 12x - 90 = 0$.

Factor by grouping:

$(2x^2 + 15x) + (-12x - 90) = 0$.

$x(2x + 15) - 6(2x + 15) = 0$.

Factor out the common binomial factor $(2x + 15)$:

$(2x + 15)(x - 6) = 0$.

Using the Zero Product Property, set each factor equal to zero:

Either $2x + 15 = 0 \implies 2x = -15 \implies x = -\frac{15}{2}$.

Or $x - 6 = 0 \implies x = 6$.

Since $x$ represents the number of articles produced, it must be a positive integer. Therefore, $x = -\frac{15}{2}$ is not a valid solution.

The number of articles produced on that day is $x = 6$.

Now, find the cost of each article using the expression Cost = $2x + 3$:

Cost of each article = $2(6) + 3 = 12 + 3 = 15$ rupees.

Check:

Number of articles = 6.

Cost of each article = $\textsf{₹} 15$.

Total cost = $6 \times 15 = \textsf{₹} 90$. This matches the given information.

Also, the cost ($\textsf{₹} 15$) is 3 more than twice the number of articles ($2 \times 6 + 3 = 12 + 3 = 15$). This condition is also satisfied.

Final Answer:

The number of articles produced on that day is $\mathbf{6}$.

The cost of each article is $\mathbf{\textsf{₹} 15}$.

Given:

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

To Find:

The roots of the given equation by the method of completing the square.

Solution:

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

Step 1: Divide the entire equation by the coefficient of $x^2$, which is 2, to make the coefficient of $x^2$ equal to 1.

$\frac{2x^2}{2} - \frac{5x}{2} + \frac{3}{2} = \frac{0}{2}$

$x^2 - \frac{5}{2}x + \frac{3}{2} = 0$

Step 2: Move the constant term ($\frac{3}{2}$) to the right-hand side (RHS).

$x^2 - \frac{5}{2}x = -\frac{3}{2}$

Step 3: Take half of the coefficient of the $x$ term, square it, and add it to both sides.

The coefficient of $x$ is $-\frac{5}{2}$.

Half of the coefficient is $\frac{1}{2} \times (-\frac{5}{2}) = -\frac{5}{4}$.

Squaring this value: $(-\frac{5}{4})^2 = \frac{25}{16}$.

Add $\frac{25}{16}$ to both sides of the equation:

$x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16}$

Step 4: Factor the left-hand side (LHS) as a perfect square.

The LHS is now $(x - \frac{5}{4})^2$.

$(x - \frac{5}{4})^2 = -\frac{3}{2} + \frac{25}{16}$

Step 5: Simplify the RHS.

Find a common denominator for the RHS, which is 16.

$-\frac{3}{2} = -\frac{3 \times 8}{2 \times 8} = -\frac{24}{16}$

$(x - \frac{5}{4})^2 = -\frac{24}{16} + \frac{25}{16}$

$(x - \frac{5}{4})^2 = \frac{-24 + 25}{16}$

$(x - \frac{5}{4})^2 = \frac{1}{16}$

Step 6: Take the square root of both sides.

$x - \frac{5}{4} = \pm \sqrt{\frac{1}{16}}$

$x - \frac{5}{4} = \pm \frac{1}{4}$

Step 7: Solve for $x$.

$x = \frac{5}{4} \pm \frac{1}{4}$

We have two possible values for $x$:

Case 1: Using the '+' sign

$x = \frac{5}{4} + \frac{1}{4} = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2}$

Case 2: Using the '-' sign

$x = \frac{5}{4} - \frac{1}{4} = \frac{5-1}{4} = \frac{4}{4} = 1$

Final Answer:

The roots of the equation $2x^2 - 5x + 3 = 0$, found by completing the square, are $\mathbf{1}$ and $\mathbf{\frac{3}{2}}$.

Given:

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

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To Find:

The roots of the given equation by the method of completing the square.

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Solution:

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

Step 1: Divide the entire equation by the coefficient of $x^2$, which is 5, to make the coefficient of $x^2$ equal to 1.

$\frac{5x^2}{5} - \frac{6x}{5} - \frac{2}{5} = \frac{0}{5}$

$x^2 - \frac{6}{5}x - \frac{2}{5} = 0$

Step 2: Move the constant term ($-\frac{2}{5}$) to the right-hand side (RHS).

$x^2 - \frac{6}{5}x = \frac{2}{5}$

Step 3: Take half of the coefficient of the $x$ term, square it, and add it to both sides.

The coefficient of $x$ is $-\frac{6}{5}$.

Half of the coefficient is $\frac{1}{2} \times (-\frac{6}{5}) = -\frac{3}{5}$.

Squaring this value: $(-\frac{3}{5})^2 = \frac{9}{25}$.

Add $\frac{9}{25}$ to both sides of the equation:

$x^2 - \frac{6}{5}x + \frac{9}{25} = \frac{2}{5} + \frac{9}{25}$

Step 4: Factor the left-hand side (LHS) as a perfect square.

The LHS is now $(x - \frac{3}{5})^2$.

$(x - \frac{3}{5})^2 = \frac{2}{5} + \frac{9}{25}$

Step 5: Simplify the RHS.

Find a common denominator for the RHS, which is 25.

$\frac{2}{5} = \frac{2 \times 5}{5 \times 5} = \frac{10}{25}$

$(x - \frac{3}{5})^2 = \frac{10}{25} + \frac{9}{25}$

$(x - \frac{3}{5})^2 = \frac{10 + 9}{25}$

$(x - \frac{3}{5})^2 = \frac{19}{25}$

Step 6: Take the square root of both sides.

$x - \frac{3}{5} = \pm \sqrt{\frac{19}{25}}$

$x - \frac{3}{5} = \pm \frac{\sqrt{19}}{\sqrt{25}}$

$x - \frac{3}{5} = \pm \frac{\sqrt{19}}{5}$

Step 7: Solve for $x$.

$x = \frac{3}{5} \pm \frac{\sqrt{19}}{5}$

$x = \frac{3 \pm \sqrt{19}}{5}$

The two roots are $x = \frac{3 + \sqrt{19}}{5}$ and $x = \frac{3 - \sqrt{19}}{5}$.

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Final Answer:

The roots of the equation $5x^2 - 6x - 2 = 0$, found by completing the square, are $\mathbf{\frac{3 + \sqrt{19}}{5}}$ and $\mathbf{\frac{3 - \sqrt{19}}{5}}$.

Given:

The quadratic equation is $4x^2 + 3x + 5 = 0$.

---

To Find:

The roots of the given equation by the method of completing the square.

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Solution:

The given equation is $4x^2 + 3x + 5 = 0$.

Step 1: Divide the entire equation by the coefficient of $x^2$, which is 4, to make the coefficient of $x^2$ equal to 1.

$\frac{4x^2}{4} + \frac{3x}{4} + \frac{5}{4} = \frac{0}{4}$

$x^2 + \frac{3}{4}x + \frac{5}{4} = 0$

Step 2: Move the constant term ($\frac{5}{4}$) to the right-hand side (RHS).

$x^2 + \frac{3}{4}x = -\frac{5}{4}$

Step 3: Take half of the coefficient of the $x$ term, square it, and add it to both sides.

The coefficient of $x$ is $\frac{3}{4}$.

Half of the coefficient is $\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$.

Squaring this value: $(\frac{3}{8})^2 = \frac{9}{64}$.

Add $\frac{9}{64}$ to both sides of the equation:

$x^2 + \frac{3}{4}x + \frac{9}{64} = -\frac{5}{4} + \frac{9}{64}$

Step 4: Factor the left-hand side (LHS) as a perfect square.

The LHS is now $(x + \frac{3}{8})^2$.

$(x + \frac{3}{8})^2 = -\frac{5}{4} + \frac{9}{64}$

Step 5: Simplify the RHS.

Find a common denominator for the RHS, which is 64.

$-\frac{5}{4} = -\frac{5 \times 16}{4 \times 16} = -\frac{80}{64}$

$(x + \frac{3}{8})^2 = -\frac{80}{64} + \frac{9}{64}$

$(x + \frac{3}{8})^2 = \frac{-80 + 9}{64}$

$(x + \frac{3}{8})^2 = -\frac{71}{64}$

Step 6: Analyze the result.

The square of any real number cannot be negative. However, the expression $(x + \frac{3}{8})^2$ is equal to a negative number ($-\frac{71}{64}$).

This means that there is no real value of $x$ that satisfies the equation $(x + \frac{3}{8})^2 = -\frac{71}{64}$.

Therefore, the original quadratic equation $4x^2 + 3x + 5 = 0$ has no real roots.

---

Final Answer:

The equation $4x^2 + 3x + 5 = 0$ has no real roots because the square of a real number cannot be negative.

Given:

A rectangular plot with Area = 528 m².

Length of the plot = $1 + 2 \times (\text{Breadth})$.

---

To Find:

The length and breadth of the plot using the quadratic formula.

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Solution:

First, let's set up the quadratic equation representing the situation, as derived in Exercise 4.1, Q.2(i).

Let the breadth of the plot be $x$ metres.

Then, the length of the plot is $(2x + 1)$ metres.

Area = Length $\times$ Breadth

$528 = (2x + 1)x$

$528 = 2x^2 + x$

Rearranging into the standard quadratic form $ax^2 + bx + c = 0$:

$2x^2 + x - 528 = 0$

Now, we will solve this equation using the quadratic formula.

Compare the equation $2x^2 + x - 528 = 0$ with the standard form $ax^2 + bx + c = 0$.

We have: $a = 2$, $b = 1$, $c = -528$.

The quadratic formula is given by:

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

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

$D = (1)^2 - 4(2)(-528)$

$D = 1 - 8(-528)$

$D = 1 + 4224$

$D = 4225$

Since $D > 0$, the equation has two distinct real roots.

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

$x = \frac{-(1) \pm \sqrt{4225}}{2(2)}$

We need to find the square root of 4225. We know $60^2 = 3600$ and $70^2 = 4900$. Since it ends in 5, the square root must end in 5. Let's check $65^2 = 4225$.

So, $\sqrt{4225} = 65$.

$x = \frac{-1 \pm 65}{4}$

We have two possible values for $x$:

Case 1: Using the '+' sign

$x = \frac{-1 + 65}{4} = \frac{64}{4} = 16$

Case 2: Using the '-' sign

$x = \frac{-1 - 65}{4} = \frac{-66}{4} = -\frac{33}{2} = -16.5$

Since $x$ represents the breadth of the plot, it must be a positive quantity. Therefore, we reject $x = -16.5$.

The breadth of the plot is $x = 16$ metres.

Now, calculate the length:

Length = $2x + 1 = 2(16) + 1 = 32 + 1 = 33$ metres.

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Final Answer:

The dimensions of the rectangular plot are:

Breadth = 16 metres

Length = 33 metres

Given:

Two consecutive odd positive integers.

The sum of the squares of these integers is 290.

---

To Find:

The two consecutive odd positive integers.

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Solution:

Let the first odd positive integer be $x$.

Since the integers are consecutive odd integers, the difference between them is 2. Therefore, the next consecutive odd positive integer is $x + 2$.

According to the problem statement, the sum of the squares of these two integers is 290.

Square of the first integer = $x^2$.

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

The equation representing the condition is:

$x^2 + (x + 2)^2 = 290$

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 equation:

$x^2 + (x^2 + 4x + 4) = 290$

Combine like terms:

$2x^2 + 4x + 4 = 290$

Subtract 290 from both sides to set the equation to zero:

$2x^2 + 4x + 4 - 290 = 0$

$2x^2 + 4x - 286 = 0$

Divide the entire equation by 2 to simplify:

$\frac{2x^2}{2} + \frac{4x}{2} - \frac{286}{2} = \frac{0}{2}$

$x^2 + 2x - 143 = 0$

Now, we solve this quadratic equation. We can use either factorisation or the quadratic formula.

Method 1: Factorisation

We need to find two numbers whose product is $1 \times (-143) = -143$ and whose sum is $2$.

Let's find factors of 143. $143 = 11 \times 13$.

The pair of factors whose difference is 2 is 11 and 13.

Since the sum is positive (+2), the numbers must be 13 and -11.

Split the middle term ($2x$):

$x^2 + 13x - 11x - 143 = 0$

Factor by grouping:

$(x^2 + 13x) + (-11x - 143) = 0$

$x(x + 13) - 11(x + 13) = 0$

Factor out the common binomial factor $(x + 13)$:

$(x + 13)(x - 11) = 0$

Using the Zero Product Property:

Either $x + 13 = 0 \implies x = -13$

Or $x - 11 = 0 \implies x = 11$

Method 2: Quadratic Formula

For the equation $x^2 + 2x - 143 = 0$, we have $a=1, b=2, c=-143$.

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (2)^2 - 4(1)(-143) = 4 + 572 = 576$.

$x = \frac{-2 \pm \sqrt{576}}{2(1)}$

$x = \frac{-2 \pm 24}{2}$

Case 1: $x = \frac{-2 + 24}{2} = \frac{22}{2} = 11$

Case 2: $x = \frac{-2 - 24}{2} = \frac{-26}{2} = -13$

Interpreting the result:

The problem asks for positive odd integers. Therefore, $x = -13$ is rejected.

The first positive odd integer is $x = 11$.

The second consecutive odd positive integer is $x + 2 = 11 + 2 = 13$.

Check:

The integers are 11 and 13. They are consecutive, odd, and positive.

Sum of their squares = $11^2 + 13^2 = 121 + 169 = 290$. (Correct)

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Final Answer:

The two consecutive odd positive integers are $\mathbf{11}$ and $\mathbf{13}$.

Given:

A rectangular park.

Breadth of the rectangular park ($b$) = Length ($l$) - 3 m.

An isosceles triangular park.

Base of the triangular park = Breadth ($b$) of the rectangular park.

Altitude of the triangular park = 12 m.

Area of the rectangular park ($A_r$) = Area of the triangular park ($A_t$) + 4 m².

---

To Find:

The length ($l$) and breadth ($b$) of the rectangular park.

---

Solution:

Let the length of the rectangular park be $l$ metres.

Given that the breadth is 3 m less than its length, the breadth $b = (l - 3)$ metres.

Area of the rectangular park, $A_r = \text{Length} \times \text{Breadth} = l \times (l - 3)$.

$A_r = l^2 - 3l$.

Now, consider the isosceles triangular park.

Base of the triangle = Breadth of the rectangle = $b = (l - 3)$ metres.

Altitude of the triangle = 12 m.

Area of the triangular park, $A_t = \frac{1}{2} \times \text{Base} \times \text{Altitude}$.

$A_t = \frac{1}{2} \times (l - 3) \times 12$

$A_t = 6(l - 3)$

$A_t = 6l - 18$.

According to the problem, the area of the rectangular park is 4 square metres more than the area of the triangular park:

$A_r = A_t + 4$

Substitute the expressions for $A_r$ and $A_t$:

$l^2 - 3l = (6l - 18) + 4$

$l^2 - 3l = 6l - 14$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$l^2 - 3l - 6l + 14 = 0$

$l^2 - 9l + 14 = 0$

Now, solve this quadratic equation for $l$. We can use the factorisation method.

We need two numbers whose product is $1 \times 14 = 14$ and whose sum is $-9$.

The pairs of factors of 14 are (1, 14), (2, 7), (-1, -14), (-2, -7).

The pair whose sum is -9 is -2 and -7.

Split the middle term ($-9l$):

$l^2 - 2l - 7l + 14 = 0$

Factor by grouping:

$(l^2 - 2l) + (-7l + 14) = 0$

$l(l - 2) - 7(l - 2) = 0$

Factor out the common binomial factor $(l - 2)$:

$(l - 2)(l - 7) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $l - 2 = 0 \implies l = 2$

Or $l - 7 = 0 \implies l = 7$

We need to check if both values of $l$ are valid by calculating the breadth $b = l - 3$.

Case 1: If $l = 2$ m

Breadth $b = l - 3 = 2 - 3 = -1$ m. Since the breadth cannot be negative, $l=2$ is not a valid solution.

Case 2: If $l = 7$ m

Breadth $b = l - 3 = 7 - 3 = 4$ m. This is a valid positive dimension.

So, the length of the rectangular park is 7 m and the breadth is 4 m.

Check:

Length = 7 m, Breadth = 4 m. (Breadth = 7 - 3 = 4 m).

Area of rectangle $A_r = 7 \times 4 = 28$ m².

Base of triangle = 4 m, Altitude = 12 m.

Area of triangle $A_t = \frac{1}{2} \times 4 \times 12 = 24$ m².

Is $A_r = A_t + 4$? Yes, $28 = 24 + 4$. The conditions are satisfied.

---

Final Answer:

The dimensions of the rectangular park are:

Length = 7 metres

Breadth = 4 metres

General Concept: Quadratic Formula

For a quadratic equation in the standard form $ax^2 + bx + c = 0$ (where $a \neq 0$), the roots are given by the quadratic formula:

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

The expression $D = b^2 - 4ac$ is called the discriminant.

• If $D > 0$, the equation has two distinct real roots.
• If $D = 0$, the equation has two equal real roots (or one real root).
• If $D < 0$, the equation has no real roots.

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Solution (i): $3x^2 - 5x + 2 = 0$

Given equation: $3x^2 - 5x + 2 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 3$, $b = -5$, $c = 2$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-5)^2 - 4(3)(2)$

$D = 25 - 24$

$D = 1$

Since $D = 1 > 0$, the equation has two distinct real roots.

Now, apply the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{1}}{2(3)}$

$x = \frac{5 \pm 1}{6}$

The two roots are:

$x_1 = \frac{5 + 1}{6} = \frac{6}{6} = 1$

$x_2 = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3}$

Conclusion: The roots of the equation $3x^2 - 5x + 2 = 0$ are $\mathbf{1}$ and $\mathbf{\frac{2}{3}}$.

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Solution (ii): $x^2 + 4x + 5 = 0$

Given equation: $x^2 + 4x + 5 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 1$, $b = 4$, $c = 5$.

Calculate the discriminant $D = b^2 - 4ac$:

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

$D = 16 - 20$

$D = -4$

Since $D = -4 < 0$, the discriminant is negative.

Conclusion: The equation $x^2 + 4x + 5 = 0$ has no real roots.

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Solution (iii): $2x^2 - 2\sqrt{2}x + 1 = 0$

Given equation: $2x^2 - 2\sqrt{2}x + 1 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = -2\sqrt{2}$, $c = 1$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-2\sqrt{2})^2 - 4(2)(1)$

$D = (4 \times 2) - 8$

$D = 8 - 8$

$D = 0$

Since $D = 0$, the equation has two equal real roots (or one real root).

Now, apply the quadratic formula:

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

$x = \frac{2\sqrt{2} \pm 0}{4}$

$x = \frac{2\sqrt{2}}{4}$

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

The roots are equal.

Conclusion: The roots of the equation $2x^2 - 2\sqrt{2}x + 1 = 0$ are $\mathbf{\frac{\sqrt{2}}{2}}$ and $\mathbf{\frac{\sqrt{2}}{2}}$ (or simply $\frac{\sqrt{2}}{2}$ as a repeated root).

Solution (i): $x + \frac{1}{x} = 3$, $x \neq 0$

Given equation: $x + \frac{1}{x} = 3$. The condition $x \neq 0$ is given because division by zero is undefined.

To Find: Roots of the equation.

Solution:

Multiply the entire equation by $x$ to eliminate the fraction:

$x \times (x + \frac{1}{x}) = x \times 3$

$x^2 + x \times \frac{1}{x} = 3x$

$x^2 + 1 = 3x$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$x^2 - 3x + 1 = 0$

Now, we use the quadratic formula to find the roots.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 1$, $b = -3$, $c = 1$.

Calculate the discriminant $D = b^2 - 4ac$:

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

$D = 9 - 4$

$D = 5$

Since $D = 5 > 0$, the equation has two distinct real roots.

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-3) \pm \sqrt{5}}{2(1)}$

$x = \frac{3 \pm \sqrt{5}}{2}$

The two roots are $x_1 = \frac{3 + \sqrt{5}}{2}$ and $x_2 = \frac{3 - \sqrt{5}}{2}$.

Neither of these roots is 0, so they are valid solutions.

Conclusion: The roots of the equation $x + \frac{1}{x} = 3$ are $\mathbf{\frac{3 + \sqrt{5}}{2}}$ and $\mathbf{\frac{3 - \sqrt{5}}{2}}$.

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Solution (ii): $\frac{1}{x} - \frac{1}{x - 2} = 3$, $x \neq 0, 2$

Given equation: $\frac{1}{x} - \frac{1}{x - 2} = 3$. The conditions $x \neq 0$ and $x \neq 2$ are given because they would lead to division by zero.

To Find: Roots of the equation.

Solution:

Find a common denominator for the left-hand side (LHS), which is $x(x - 2)$:

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

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

Simplify the numerator:

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

Expand the denominator:

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

Multiply both sides by $(x^2 - 2x)$ to eliminate the fraction (assuming $x \neq 0, 2$):

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

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

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

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

Now, we use the quadratic formula to find the roots.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 3$, $b = -6$, $c = 2$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-6)^2 - 4(3)(2)$

$D = 36 - 24$

$D = 12$

Since $D = 12 > 0$, the equation has two distinct real roots.

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-6) \pm \sqrt{12}}{2(3)}$

$x = \frac{6 \pm \sqrt{4 \times 3}}{6}$

$x = \frac{6 \pm 2\sqrt{3}}{6}$

Factor out 2 from the numerator:

$x = \frac{2(3 \pm \sqrt{3})}{6}$

Simplify by cancelling the common factor 2:

$x = \frac{3 \pm \sqrt{3}}{3}$

The two roots are $x_1 = \frac{3 + \sqrt{3}}{3}$ and $x_2 = \frac{3 - \sqrt{3}}{3}$.

Neither of these roots is 0 or 2, so they are valid solutions.

Conclusion: The roots of the equation $\frac{1}{x} - \frac{1}{x - 2} = 3$ are $\mathbf{\frac{3 + \sqrt{3}}{3}}$ and $\mathbf{\frac{3 - \sqrt{3}}{3}}$.

Given:

• Speed of the motor boat in still water = 18 km/h.
• Distance to travel upstream and downstream = 24 km.
• Time taken for the upstream journey is 1 hour more than the time taken for the downstream journey.

---

To Find:

The speed of the stream.

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Solution:

Let the speed of the stream be $x$ km/h.

The speed of the boat in still water is 18 km/h.

Speed Downstream: When the boat travels downstream (in the direction of the stream), the speed of the stream adds to the speed of the boat.

Speed downstream = (Speed of boat in still water) + (Speed of stream) = $(18 + x)$ km/h.

Speed Upstream: When the boat travels upstream (against the direction of the stream), the speed of the stream subtracts from the speed of the boat.

Speed upstream = (Speed of boat in still water) - (Speed of stream) = $(18 - x)$ km/h.

We must have $18 - x > 0$ for the boat to travel upstream, so $x < 18$. Also, speed $x$ must be positive.

The distance for both journeys is 24 km.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Time taken for the downstream journey ($T_{down}$) = $\frac{24}{18 + x}$ hours.

Time taken for the upstream journey ($T_{up}$) = $\frac{24}{18 - x}$ hours.

According to the problem statement, the time taken for the upstream journey is 1 hour more than the time taken for the downstream journey:

$T_{up} = T_{down} + 1$

$\frac{24}{18 - x} = \frac{24}{18 + x} + 1$

Rearrange the equation to solve for $x$:

$\frac{24}{18 - x} - \frac{24}{18 + x} = 1$

Factor out 24 from the left-hand side:

$24 \left( \frac{1}{18 - x} - \frac{1}{18 + x} \right) = 1$

Find a common denominator inside the parentheses, which is $(18 - x)(18 + x) = 18^2 - x^2 = 324 - x^2$.

$24 \left( \frac{(18 + x) - (18 - x)}{(18 - x)(18 + x)} \right) = 1$

$24 \left( \frac{18 + x - 18 + x}{324 - x^2} \right) = 1$

$24 \left( \frac{2x}{324 - x^2} \right) = 1$

$\frac{48x}{324 - x^2} = 1$

Multiply both sides by $(324 - x^2)$:

$48x = 324 - x^2$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$x^2 + 48x - 324 = 0$

Now, solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 1$, $b = 48$, $c = -324$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (48)^2 - 4(1)(-324)$

$D = 2304 + 1296$

$D = 3600$

Since $D > 0$, there are two distinct real roots.

$x = \frac{-48 \pm \sqrt{3600}}{2(1)}$

$x = \frac{-48 \pm 60}{2}$

The two possible values for $x$ are:

$x_1 = \frac{-48 + 60}{2} = \frac{12}{2} = 6$

$x_2 = \frac{-48 - 60}{2} = \frac{-108}{2} = -54$

Since the speed of the stream ($x$) cannot be negative, we discard the solution $x = -54$.

Therefore, the speed of the stream is $x = 6$ km/h.

This value satisfies the condition $0 < x < 18$.

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Final Answer:

The speed of the stream is $\mathbf{6}$ km/h.

Solution (i): $2x^2 - 7x + 3 = 0$

Given equation: $2x^2 - 7x + 3 = 0$.

To Find: Roots by completing the square.

Solution:

First, check if real roots exist using the discriminant $D = b^2 - 4ac$.

Here $a=2$, $b=-7$, $c=3$.

$D = (-7)^2 - 4(2)(3) = 49 - 24 = 25$.

Since $D = 25 > 0$, real and distinct roots exist.

Step 1: Divide by the coefficient of $x^2$ (which is 2).

$x^2 - \frac{7}{2}x + \frac{3}{2} = 0$

Step 2: Move the constant term to the RHS.

$x^2 - \frac{7}{2}x = -\frac{3}{2}$

Step 3: Take half of the coefficient of $x$, square it, and add to both sides.

Coefficient of $x$ is $-\frac{7}{2}$. Half is $\frac{1}{2} \times (-\frac{7}{2}) = -\frac{7}{4}$.

Square of half the coefficient is $(-\frac{7}{4})^2 = \frac{49}{16}$.

Add $\frac{49}{16}$ to both sides:

$x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}$

Step 4: Factor the LHS as a perfect square.

$(x - \frac{7}{4})^2 = -\frac{3 \times 8}{2 \times 8} + \frac{49}{16}$

$(x - \frac{7}{4})^2 = -\frac{24}{16} + \frac{49}{16}$

$(x - \frac{7}{4})^2 = \frac{-24 + 49}{16}$

$(x - \frac{7}{4})^2 = \frac{25}{16}$

Step 5: Take the square root of both sides.

$x - \frac{7}{4} = \pm \sqrt{\frac{25}{16}}$

$x - \frac{7}{4} = \pm \frac{5}{4}$

Step 6: Solve for $x$.

$x = \frac{7}{4} \pm \frac{5}{4}$

Two possible roots:

$x_1 = \frac{7}{4} + \frac{5}{4} = \frac{7+5}{4} = \frac{12}{4} = 3$

$x_2 = \frac{7}{4} - \frac{5}{4} = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$

Conclusion: The roots of the equation $2x^2 - 7x + 3 = 0$ are $\mathbf{3}$ and $\mathbf{\frac{1}{2}}$.

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Solution (ii): $2x^2 + x - 4 = 0$

Given equation: $2x^2 + x - 4 = 0$.

To Find: Roots by completing the square.

Solution:

First, check if real roots exist using the discriminant $D = b^2 - 4ac$.

Here $a=2$, $b=1$, $c=-4$.

$D = (1)^2 - 4(2)(-4) = 1 - (-32) = 1 + 32 = 33$.

Since $D = 33 > 0$, real and distinct roots exist.

Step 1: Divide by the coefficient of $x^2$ (which is 2).

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

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

Step 2: Move the constant term to the RHS.

$x^2 + \frac{1}{2}x = 2$

Step 3: Take half of the coefficient of $x$, square it, and add to both sides.

Coefficient of $x$ is $\frac{1}{2}$. Half is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Square of half the coefficient is $(\frac{1}{4})^2 = \frac{1}{16}$.

Add $\frac{1}{16}$ to both sides:

$x^2 + \frac{1}{2}x + \frac{1}{16} = 2 + \frac{1}{16}$

Step 4: Factor the LHS as a perfect square.

$(x + \frac{1}{4})^2 = \frac{2 \times 16}{1 \times 16} + \frac{1}{16}$

$(x + \frac{1}{4})^2 = \frac{32}{16} + \frac{1}{16}$

$(x + \frac{1}{4})^2 = \frac{33}{16}$

Step 5: Take the square root of both sides.

$x + \frac{1}{4} = \pm \sqrt{\frac{33}{16}}$

$x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4}$

Step 6: Solve for $x$.

$x = -\frac{1}{4} \pm \frac{\sqrt{33}}{4}$

$x = \frac{-1 \pm \sqrt{33}}{4}$

Two possible roots:

$x_1 = \frac{-1 + \sqrt{33}}{4}$

$x_2 = \frac{-1 - \sqrt{33}}{4}$

Conclusion: The roots of the equation $2x^2 + x - 4 = 0$ are $\mathbf{\frac{-1 + \sqrt{33}}{4}}$ and $\mathbf{\frac{-1 - \sqrt{33}}{4}}$.

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Solution (iii): $4x^2 + 4\sqrt{3}x + 3 = 0$

Given equation: $4x^2 + 4\sqrt{3}x + 3 = 0$.

To Find: Roots by completing the square.

Solution:

First, check if real roots exist using the discriminant $D = b^2 - 4ac$.

Here $a=4$, $b=4\sqrt{3}$, $c=3$.

$D = (4\sqrt{3})^2 - 4(4)(3) = (16 \times 3) - 48 = 48 - 48 = 0$.

Since $D = 0$, real and equal roots exist.

Step 1: Divide by the coefficient of $x^2$ (which is 4).

$x^2 + \frac{4\sqrt{3}}{4}x + \frac{3}{4} = 0$

$x^2 + \sqrt{3}x + \frac{3}{4} = 0$

Step 2: Move the constant term to the RHS.

$x^2 + \sqrt{3}x = -\frac{3}{4}$

Step 3: Take half of the coefficient of $x$, square it, and add to both sides.

Coefficient of $x$ is $\sqrt{3}$. Half is $\frac{\sqrt{3}}{2}$.

Square of half the coefficient is $(\frac{\sqrt{3}}{2})^2 = \frac{3}{4}$.

Add $\frac{3}{4}$ to both sides:

$x^2 + \sqrt{3}x + \frac{3}{4} = -\frac{3}{4} + \frac{3}{4}$

Step 4: Factor the LHS as a perfect square.

$(x + \frac{\sqrt{3}}{2})^2 = 0$

Step 5: Take the square root of both sides.

$x + \frac{\sqrt{3}}{2} = \pm \sqrt{0}$

$x + \frac{\sqrt{3}}{2} = 0$

Step 6: Solve for $x$.

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

Since the discriminant was 0, we get two equal roots.

Conclusion: The roots of the equation $4x^2 + 4\sqrt{3}x + 3 = 0$ are $\mathbf{-\frac{\sqrt{3}}{2}}$ and $\mathbf{-\frac{\sqrt{3}}{2}}$.

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Solution (iv): $2x^2 + x + 4 = 0$

Given equation: $2x^2 + x + 4 = 0$.

To Find: Roots by completing the square.

Solution:

First, check if real roots exist using the discriminant $D = b^2 - 4ac$.

Here $a=2$, $b=1$, $c=4$.

$D = (1)^2 - 4(2)(4) = 1 - 32 = -31$.

Since $D = -31 < 0$, the discriminant is negative.

Therefore, the equation has no real roots.

Attempting to complete the square (for illustration):

Divide by 2: $x^2 + \frac{1}{2}x + 2 = 0$

Move constant term: $x^2 + \frac{1}{2}x = -2$

Half of coefficient of $x$ is $\frac{1}{4}$. Square is $\frac{1}{16}$. Add to both sides:

$x^2 + \frac{1}{2}x + \frac{1}{16} = -2 + \frac{1}{16}$

Factor LHS: $(x + \frac{1}{4})^2 = -\frac{32}{16} + \frac{1}{16}$

$(x + \frac{1}{4})^2 = -\frac{31}{16}$

As shown here, the square of a real expression $(x + \frac{1}{4})^2$ is equal to a negative number $(-\frac{31}{16})$. This is impossible for real numbers.

Conclusion: The equation $2x^2 + x + 4 = 0$ has no real roots.

General Concept: Quadratic Formula

For a quadratic equation in the standard form $ax^2 + bx + c = 0$ (where $a \neq 0$), the roots are given by the quadratic formula:

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

The expression $D = b^2 - 4ac$ is the discriminant, which determines the nature of the roots:

• If $D > 0$, there are two distinct real roots.
• If $D = 0$, there are two equal real roots (one real root).
• If $D < 0$, there are no real roots.

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Solution (i): $2x^2 - 7x + 3 = 0$

Given equation: $2x^2 - 7x + 3 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = -7$, $c = 3$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-7)^2 - 4(2)(3)$

$D = 49 - 24$

$D = 25$

Since $D = 25 > 0$, the equation has two distinct real roots.

Apply the quadratic formula:

$x = \frac{-(-7) \pm \sqrt{25}}{2(2)}$

$x = \frac{7 \pm 5}{4}$

The two roots are:

$x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3$

$x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$

Conclusion: The roots are $\mathbf{3}$ and $\mathbf{\frac{1}{2}}$.

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Solution (ii): $2x^2 + x - 4 = 0$

Given equation: $2x^2 + x - 4 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = 1$, $c = -4$.

Calculate the discriminant $D = b^2 - 4ac$:

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

$D = 1 - (-32)$

$D = 1 + 32 = 33$

Since $D = 33 > 0$, the equation has two distinct real roots.

Apply the quadratic formula:

$x = \frac{-(1) \pm \sqrt{33}}{2(2)}$

$x = \frac{-1 \pm \sqrt{33}}{4}$

The two roots are:

$x_1 = \frac{-1 + \sqrt{33}}{4}$

$x_2 = \frac{-1 - \sqrt{33}}{4}$

Conclusion: The roots are $\mathbf{\frac{-1 + \sqrt{33}}{4}}$ and $\mathbf{\frac{-1 - \sqrt{33}}{4}}$.

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Solution (iii): $4x^2 + 4\sqrt{3}x + 3 = 0$

Given equation: $4x^2 + 4\sqrt{3}x + 3 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 4$, $b = 4\sqrt{3}$, $c = 3$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (4\sqrt{3})^2 - 4(4)(3)$

$D = (16 \times 3) - 48$

$D = 48 - 48 = 0$

Since $D = 0$, the equation has two equal real roots.

Apply the quadratic formula:

$x = \frac{-(4\sqrt{3}) \pm \sqrt{0}}{2(4)}$

$x = \frac{-4\sqrt{3} \pm 0}{8}$

$x = \frac{-4\sqrt{3}}{8}$

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

The two equal roots are $x_1 = x_2 = -\frac{\sqrt{3}}{2}$.

Conclusion: The roots are $\mathbf{-\frac{\sqrt{3}}{2}}$ and $\mathbf{-\frac{\sqrt{3}}{2}}$.

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Solution (iv): $2x^2 + x + 4 = 0$

Given equation: $2x^2 + x + 4 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = 1$, $c = 4$.

Calculate the discriminant $D = b^2 - 4ac$:

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

$D = 1 - 32$

$D = -31$

Since $D = -31 < 0$, the discriminant is negative.

Conclusion: The equation $2x^2 + x + 4 = 0$ has no real roots.

Solution (i): $x - \frac{1}{x} = 3$, $x \neq 0$

Given equation: $x - \frac{1}{x} = 3$. The condition $x \neq 0$ ensures the expression is defined.

To Find: Roots of the equation.

Solution:

Multiply the entire equation by $x$ to eliminate the fraction:

$x \times (x - \frac{1}{x}) = x \times 3$

$x^2 - x \times \frac{1}{x} = 3x$

$x^2 - 1 = 3x$

Rearrange the terms into the standard quadratic form $ax^2 + bx + c = 0$:

$x^2 - 3x - 1 = 0$

We will use the quadratic formula to find the roots. Comparing with $ax^2 + bx + c = 0$, we have:

$a = 1$, $b = -3$, $c = -1$.

Calculate the discriminant $D = b^2 - 4ac$:

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

$D = 9 - (-4)$

$D = 9 + 4 = 13$

Since $D = 13 > 0$, the equation has two distinct real roots.

Apply the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-3) \pm \sqrt{13}}{2(1)}$

$x = \frac{3 \pm \sqrt{13}}{2}$

The two roots are $x_1 = \frac{3 + \sqrt{13}}{2}$ and $x_2 = \frac{3 - \sqrt{13}}{2}$.

These roots are not equal to 0, so they are valid.

Conclusion: The roots of the equation $x - \frac{1}{x} = 3$ are $\mathbf{\frac{3 + \sqrt{13}}{2}}$ and $\mathbf{\frac{3 - \sqrt{13}}{2}}$.

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Solution (ii): $\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$, $x \neq -4, 7$

Given equation: $\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$. The conditions $x \neq -4$ and $x \neq 7$ ensure the denominators are non-zero.

To Find: Roots of the equation.

Solution:

Find a common denominator for the left-hand side (LHS), which is $(x + 4)(x - 7)$:

$\frac{1(x - 7)}{(x + 4)(x - 7)} - \frac{1(x + 4)}{(x + 4)(x - 7)} = \frac{11}{30}$

$\frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30}$

Simplify the numerator and expand the denominator:

$\frac{x - 7 - x - 4}{x(x - 7) + 4(x - 7)} = \frac{11}{30}$

$\frac{-11}{x^2 - 7x + 4x - 28} = \frac{11}{30}$

$\frac{-11}{x^2 - 3x - 28} = \frac{11}{30}$

Divide both sides by 11 (since $11 \neq 0$):

$\frac{-1}{x^2 - 3x - 28} = \frac{1}{30}$

Cross-multiply:

$-1 \times 30 = 1 \times (x^2 - 3x - 28)$

$-30 = x^2 - 3x - 28$

Rearrange the terms into the standard quadratic form $ax^2 + bx + c = 0$:

$x^2 - 3x - 28 + 30 = 0$

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

We can solve this quadratic equation by factorisation.

We need two numbers whose product is $1 \times 2 = 2$ and whose sum is $-3$. The numbers are -1 and -2.

Split the middle term:

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

Factor by grouping:

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

Factor out the common binomial factor $(x - 1)$:

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

Using the Zero Product Property, set each factor equal to zero:

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

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

The roots are 1 and 2. Both are valid as they are not equal to -4 or 7.

Conclusion: The roots of the equation $\frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}$ are $\mathbf{1}$ and $\mathbf{2}$.

Given:

The sum of the reciprocals of Rehman's age 3 years ago and his age 5 years from now is $\frac{1}{3}$.

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To Find:

Rehman's present age.

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Solution:

Let Rehman's present age be $x$ years.

Rehman's age 3 years ago = $(x - 3)$ years.

Rehman's age 5 years from now = $(x + 5)$ years.

The reciprocals of these ages are:

Reciprocal of age 3 years ago = $\frac{1}{x - 3}$. (We must have $x > 3$ for age to be positive).

Reciprocal of age 5 years from now = $\frac{1}{x + 5}$.

According to the problem statement, the sum of these reciprocals is $\frac{1}{3}$:

$\frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3}$

To solve this equation, find a common denominator for the left-hand side (LHS), which is $(x - 3)(x + 5)$:

$\frac{1(x + 5)}{(x - 3)(x + 5)} + \frac{1(x - 3)}{(x - 3)(x + 5)} = \frac{1}{3}$

$\frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3}$

Simplify the numerator and expand the denominator:

$\frac{2x + 2}{x^2 + 5x - 3x - 15} = \frac{1}{3}$

$\frac{2x + 2}{x^2 + 2x - 15} = \frac{1}{3}$

Cross-multiply:

$3(2x + 2) = 1(x^2 + 2x - 15)$

$6x + 6 = x^2 + 2x - 15$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$0 = x^2 + 2x - 15 - 6x - 6$

$x^2 - 4x - 21 = 0$

Now, solve this quadratic equation by factorisation.

We need two numbers whose product is $1 \times (-21) = -21$ and whose sum is $-4$. The numbers are $-7$ and $3$.

Split the middle term:

$x^2 - 7x + 3x - 21 = 0$

Factor by grouping:

$x(x - 7) + 3(x - 7) = 0$

Factor out the common binomial factor $(x - 7)$:

$(x - 7)(x + 3) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x - 7 = 0 \implies x = 7$

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

Since age ($x$) cannot be negative, we reject the solution $x = -3$.

Therefore, Rehman's present age is $x = 7$ years.

Check: If present age is 7, age 3 years ago was 4, age 5 years from now is 12. Sum of reciprocals = $\frac{1}{4} + \frac{1}{12} = \frac{3+1}{12} = \frac{4}{12} = \frac{1}{3}$. This matches the given condition.

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Final Answer:

Rehman's present age is $\mathbf{7}$ years.

Given:

• Sum of Shefali's marks in Mathematics and English = 30.
• If Mathematics marks were 2 more and English marks were 3 less, the product of these marks would be 210.

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To Find:

Shefali's marks in Mathematics and English.

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Solution:

Let Shefali's marks in Mathematics be $M$.

Let Shefali's marks in English be $E$.

From the first condition, the sum of marks is 30:

From this, we can express the marks in English in terms of marks in Mathematics:

$E = 30 - M$

Now consider the second condition:

If she got 2 marks more in Mathematics, her marks would be $M + 2$.

If she got 3 marks less in English, her marks would be $E - 3$.

Substituting $E = 30 - M$ into the modified English marks:

Modified English marks = $(30 - M) - 3 = 27 - M$.

The product of these modified marks is 210:

$(M + 2)(27 - M) = 210$

Expand the equation:

$M(27 - M) + 2(27 - M) = 210$

$27M - M^2 + 54 - 2M = 210$

Combine like terms:

$-M^2 + (27M - 2M) + 54 = 210$

$-M^2 + 25M + 54 = 210$

Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$-M^2 + 25M + 54 - 210 = 0$

$-M^2 + 25M - 156 = 0$

Multiply the entire equation by -1 to make the leading coefficient positive:

$M^2 - 25M + 156 = 0$

Now, solve this quadratic equation for $M$ by factorisation.

We need two numbers whose product is $1 \times 156 = 156$ and whose sum is $-25$. Since the product is positive and the sum is negative, both numbers must be negative.

Let's find factors of 156: $156 = 2 \times 78 = 2 \times 2 \times 39 = 2 \times 2 \times 3 \times 13$.

Possible pairs of factors: (1, 156), (2, 78), (3, 52), (4, 39), (6, 26), (12, 13).

The pair whose sum is 25 is 12 and 13. So, the required numbers are -12 and -13.

Split the middle term:

$M^2 - 12M - 13M + 156 = 0$

Factor by grouping:

$M(M - 12) - 13(M - 12) = 0$

Factor out the common binomial factor $(M - 12)$:

$(M - 12)(M - 13) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $M - 12 = 0 \implies M = 12$

Or $M - 13 = 0 \implies M = 13$

Now, find the corresponding marks in English ($E = 30 - M$) for each case:

Case 1: If Mathematics marks $M = 12$

English marks $E = 30 - 12 = 18$.

Case 2: If Mathematics marks $M = 13$

English marks $E = 30 - 13 = 17$.

Both sets of marks are possible solutions.

Check Case 1: Maths=12, English=18. Modified: Maths=14, English=15. Product=$14 \times 15 = 210$. (Correct)

Check Case 2: Maths=13, English=17. Modified: Maths=15, English=14. Product=$15 \times 14 = 210$. (Correct)

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Final Answer:

There are two possible sets of marks:

Either Shefali scored 12 marks in Mathematics and 18 marks in English.

Or Shefali scored 13 marks in Mathematics and 17 marks in English.

Given:

• A rectangular field.
• Diagonal = Shorter side + 60 metres.
• Longer side = Shorter side + 30 metres.

---

To Find:

The lengths of the sides of the rectangular field (shorter side and longer side).

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Solution:

Let the length of the shorter side of the rectangular field be $x$ metres.

According to the given conditions:

Longer side = $(x + 30)$ metres.

Diagonal = $(x + 60)$ metres.

In a rectangle, the two adjacent sides and the diagonal form a right-angled triangle, with the diagonal as the hypotenuse.

Applying the Pythagorean theorem:

$(\text{Shorter side})^2 + (\text{Longer side})^2 = (\text{Diagonal})^2$

Substitute the expressions in terms of $x$:

$x^2 + (x + 30)^2 = (x + 60)^2$

Expand the squared binomials using the identity $(a+b)^2 = a^2 + 2ab + b^2$:

$(x + 30)^2 = x^2 + 2(x)(30) + 30^2 = x^2 + 60x + 900$

$(x + 60)^2 = x^2 + 2(x)(60) + 60^2 = x^2 + 120x + 3600$

Substitute these expansions back into the Pythagorean equation:

$x^2 + (x^2 + 60x + 900) = x^2 + 120x + 3600$

Combine like terms on the left side:

$2x^2 + 60x + 900 = x^2 + 120x + 3600$

Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$(2x^2 - x^2) + (60x - 120x) + (900 - 3600) = 0$

$x^2 - 60x - 2700 = 0$

Now, solve this quadratic equation for $x$ by factorisation.

We need two numbers whose product is $-2700$ and whose sum is $-60$. The numbers are $-90$ and $30$.

Split the middle term:

$x^2 - 90x + 30x - 2700 = 0$

Factor by grouping:

$x(x - 90) + 30(x - 90) = 0$

Factor out the common binomial factor $(x - 90)$:

$(x - 90)(x + 30) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x - 90 = 0 \implies x = 90$

Or $x + 30 = 0 \implies x = -30$

Since $x$ represents the length of a side, it cannot be negative. Therefore, we reject $x = -30$.

The shorter side of the field is $x = 90$ metres.

Calculate the length of the longer side:

Longer side = $x + 30 = 90 + 30 = 120$ metres.

Check:

Shorter side = 90 m, Longer side = 120 m.

Diagonal = $x + 60 = 90 + 60 = 150$ m.

Check Pythagorean theorem: $90^2 + 120^2 = 8100 + 14400 = 22500$.

Diagonal$^2 = 150^2 = 22500$. The condition holds.

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Final Answer:

The sides of the rectangular field are $\mathbf{90}$ metres and $\mathbf{120}$ metres.

Given:

• Two numbers.
• The difference of their squares is 180.
• The square of the smaller number is 8 times the larger number.

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To Find:

The two numbers.

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Solution:

Let the larger number be $x$ and the smaller number be $y$.

According to the first condition, the difference of their squares is 180:

According to the second condition, the square of the smaller number ($y^2$) is 8 times the larger number ($x$):

Now, substitute the expression for $y^2$ from equation (ii) into equation (i):

$x^2 - (8x) = 180$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$x^2 - 8x - 180 = 0$

Solve this quadratic equation for $x$. We can use the factorisation method.

We need two numbers whose product is $1 \times (-180) = -180$ and whose sum is $-8$.

Factors of 180 are: (1, 180), (2, 90), (3, 60), (4, 45), (5, 36), (6, 30), (9, 20), (10, 18), (12, 15).

We need a pair with a difference of 8, which is 10 and 18.

To get a sum of -8, the numbers must be $-18$ and $10$.

Split the middle term:

$x^2 - 18x + 10x - 180 = 0$

Factor by grouping:

$x(x - 18) + 10(x - 18) = 0$

Factor out the common binomial factor $(x - 18)$:

$(x - 18)(x + 10) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x - 18 = 0 \implies x = 18$

Or $x + 10 = 0 \implies x = -10$

Now, find the corresponding values of $y$ using equation (ii), $y^2 = 8x$.

Case 1: If the larger number $x = 18$.

$y^2 = 8(18) = 144$

$y = \pm \sqrt{144}$

$y = \pm 12$.

In this case, the smaller number $y$ can be 12 or -12. Since $18 > 12$ and $18 > -12$, $x=18$ is indeed the larger number. The possible pairs of numbers are (18, 12) and (18, -12).

Case 2: If the larger number $x = -10$.

$y^2 = 8(-10) = -80$.

The square of a real number cannot be negative. Therefore, there is no real value for $y$ when $x = -10$. This case yields no real solutions.

So, the only possible pairs of real numbers are (18, 12) and (18, -12).

Check:

For (18, 12): $18^2 - 12^2 = 324 - 144 = 180$. $12^2 = 144 = 8 \times 18$. Both conditions hold.

For (18, -12): $18^2 - (-12)^2 = 324 - 144 = 180$. $(-12)^2 = 144 = 8 \times 18$. Both conditions hold.

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Final Answer:

The two numbers are either $\mathbf{18}$ and $\mathbf{12}$, or $\mathbf{18}$ and $\mathbf{-12}$.

Given:

• Distance travelled by the train = 360 km.
• The train travels at a uniform speed.
• If the speed were 5 km/h more, the time taken would be 1 hour less.

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To Find:

The uniform speed of the train.

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Solution:

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

The distance to be covered is 360 km.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Time taken at the uniform speed ($T_1$) = $\frac{360}{x}$ hours.

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

Time taken at the new speed ($T_2$) = $\frac{360}{x + 5}$ hours.

According to the problem statement, the time taken at the new speed ($T_2$) is 1 hour less than the time taken at the original speed ($T_1$).

$T_2 = T_1 - 1$

Alternatively, $T_1 = T_2 + 1$.

Substituting the expressions for $T_1$ and $T_2$:

$\frac{360}{x} = \frac{360}{x + 5} + 1$

Rearrange the equation to solve for $x$:

$\frac{360}{x} - \frac{360}{x + 5} = 1$

Factor out 360 from the left-hand side:

$360 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 1$

Find a common denominator inside the parentheses, which is $x(x + 5)$.

$360 \left( \frac{(x + 5) - x}{x(x + 5)} \right) = 1$

$360 \left( \frac{x + 5 - x}{x^2 + 5x} \right) = 1$

$360 \left( \frac{5}{x^2 + 5x} \right) = 1$

$\frac{1800}{x^2 + 5x} = 1$

Multiply both sides by $(x^2 + 5x)$:

$1800 = x^2 + 5x$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$x^2 + 5x - 1800 = 0$

Now, solve this quadratic equation. We can use factorisation.

We need two numbers whose product is $-1800$ and whose sum is $5$. These numbers are $45$ and $-40$.

Split the middle term:

$x^2 + 45x - 40x - 1800 = 0$

Factor by grouping:

$x(x + 45) - 40(x + 45) = 0$

Factor out the common binomial factor $(x + 45)$:

$(x + 45)(x - 40) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $x + 45 = 0 \implies x = -45$

Or $x - 40 = 0 \implies x = 40$

Since the speed of the train ($x$) must be positive, we reject the solution $x = -45$.

Therefore, the uniform speed of the train is $x = 40$ km/h.

Check:

Original time = $360/40 = 9$ hours.

New speed = $40 + 5 = 45$ km/h.

New time = $360/45 = 8$ hours.

The new time is indeed 1 hour less than the original time ($8 = 9 - 1$).

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Final Answer:

The speed of the train is $\mathbf{40}$ km/h.

Given:

• Combined time taken by two water taps to fill the tank = $9\frac{3}{8}$ hours.
• The tap with the larger diameter takes 10 hours less than the smaller one to fill the tank alone.

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To Find:

The time taken by each tap to fill the tank separately.

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Solution:

Let the time taken by the smaller tap to fill the tank alone be $x$ hours.

Since the larger tap takes 10 hours less than the smaller one, the time taken by the larger tap to fill the tank alone is $(x - 10)$ hours.

For the solution to be physically possible, both times must be positive, so $x > 0$ and $x - 10 > 0$, which implies $x > 10$.

Now, consider the rate at which each tap fills the tank (the fraction of the tank filled per hour).

Rate of the smaller tap = $\frac{1}{x}$ tank per hour.

Rate of the larger tap = $\frac{1}{x - 10}$ tank per hour.

When both taps work together, their rates add up.

Combined rate = Rate of smaller tap + Rate of larger tap = $\frac{1}{x} + \frac{1}{x - 10}$ tank per hour.

The combined time taken is given as $9\frac{3}{8}$ hours. Let's convert this to an improper fraction:

$9\frac{3}{8} = \frac{9 \times 8 + 3}{8} = \frac{72 + 3}{8} = \frac{75}{8}$ hours.

The combined rate is the reciprocal of the combined time:

Combined rate = $\frac{1}{\text{Combined Time}} = \frac{1}{75/8} = \frac{8}{75}$ tank per hour.

Now, equate the sum of individual rates to the combined rate:

$\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$

Find a common denominator for the left-hand side:

$\frac{(x - 10) + x}{x(x - 10)} = \frac{8}{75}$

$\frac{2x - 10}{x^2 - 10x} = \frac{8}{75}$

Cross-multiply:

$75(2x - 10) = 8(x^2 - 10x)$

$150x - 750 = 8x^2 - 80x$

Rearrange the terms into the standard quadratic form $ax^2 + bx + c = 0$:

$0 = 8x^2 - 80x - 150x + 750$

$8x^2 - 230x + 750 = 0$

Divide the entire equation by 2 to simplify:

$4x^2 - 115x + 375 = 0$

Solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 4$, $b = -115$, $c = 375$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-115)^2 - 4(4)(375)$

$D = 13225 - 16(375)$

$D = 13225 - 6000$

$D = 7225$

Apply the quadratic formula:

$x = \frac{-(-115) \pm \sqrt{7225}}{2(4)}$

$x = \frac{115 \pm 85}{8}$ (Since $\sqrt{7225} = 85$)

The two possible values for $x$ are:

$x_1 = \frac{115 + 85}{8} = \frac{200}{8} = 25$

$x_2 = \frac{115 - 85}{8} = \frac{30}{8} = \frac{15}{4} = 3.75$

We established earlier that $x$ must be greater than 10 ($x > 10$).

Therefore, $x = 3.75$ is not a valid solution.

The only valid solution is $x = 25$.

Time taken by the smaller tap = $x = 25$ hours.

Time taken by the larger tap = $x - 10 = 25 - 10 = 15$ hours.

---

Final Answer:

The time taken by the smaller tap to fill the tank separately is $\mathbf{25}$ hours.

The time taken by the larger tap to fill the tank separately is $\mathbf{15}$ hours.

Given:

• Distance between Mysore and Bangalore = 132 km.
• The express train takes 1 hour less than the passenger train for the journey.
• Average speed of the express train = Average speed of the passenger train + 11 km/h.

---

To Find:

The average speed of the passenger train and the express train.

---

Solution:

Let the average speed of the passenger train be $x$ km/h.

Then, the average speed of the express train is $(x + 11)$ km/h.

The distance for the journey is 132 km.

We know that Time = $\frac{\text{Distance}}{\text{Speed}}$.

Time taken by the passenger train ($T_p$) = $\frac{132}{x}$ hours.

Time taken by the express train ($T_e$) = $\frac{132}{x + 11}$ hours.

According to the problem statement, the express train takes 1 hour less than the passenger train:

$T_e = T_p - 1$

Or, $T_p = T_e + 1$.

Substituting the time expressions:

$\frac{132}{x} = \frac{132}{x + 11} + 1$

Rearrange the equation to solve for $x$:

$\frac{132}{x} - \frac{132}{x + 11} = 1$

Factor out 132 from the left-hand side:

$132 \left( \frac{1}{x} - \frac{1}{x + 11} \right) = 1$

Find a common denominator inside the parentheses, which is $x(x + 11)$.

$132 \left( \frac{(x + 11) - x}{x(x + 11)} \right) = 1$

$132 \left( \frac{x + 11 - x}{x^2 + 11x} \right) = 1$

$132 \left( \frac{11}{x^2 + 11x} \right) = 1$

$\frac{132 \times 11}{x^2 + 11x} = 1$

$\frac{1452}{x^2 + 11x} = 1$

Multiply both sides by $(x^2 + 11x)$:

$1452 = x^2 + 11x$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$x^2 + 11x - 1452 = 0$

Now, solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Here, $a = 1$, $b = 11$, $c = -1452$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (11)^2 - 4(1)(-1452)$

$D = 121 + 5808$

$D = 5929$

Apply the quadratic formula:

$x = \frac{-11 \pm \sqrt{5929}}{2(1)}$

We find that $\sqrt{5929} = 77$.

$x = \frac{-11 \pm 77}{2}$

The two possible values for $x$ are:

$x_1 = \frac{-11 + 77}{2} = \frac{66}{2} = 33$

$x_2 = \frac{-11 - 77}{2} = \frac{-88}{2} = -44$

Since the speed ($x$) must be positive, we reject the solution $x = -44$.

Therefore, the average speed of the passenger train is $x = 33$ km/h.

The average speed of the express train is $x + 11 = 33 + 11 = 44$ km/h.

Check:

Time for passenger train = $132/33 = 4$ hours.

Time for express train = $132/44 = 3$ hours.

The express train takes $4 - 3 = 1$ hour less, which matches the given condition.

---

Final Answer:

The average speed of the passenger train is $\mathbf{33}$ km/h.

The average speed of the express train is $\mathbf{44}$ km/h.

Given:

• Two squares.
• Sum of their areas = 468 m².
• Difference of their perimeters = 24 m.

---

To Find:

The lengths of the sides of the two squares.

---

Solution:

Let the side length of the first square be $x$ metres and the side length of the second square be $y$ metres.

Area of the first square = $x^2$ m².

Area of the second square = $y^2$ m².

Perimeter of the first square = $4x$ metres.

Perimeter of the second square = $4y$ metres.

According to the first condition, the sum of their areas is 468 m²:

According to the second condition, the difference of their perimeters is 24 m. Let's assume $x$ is the side of the larger square (so $x > y$).

Difference of perimeters = Perimeter of first square - Perimeter of second square

$4x - 4y = 24$

Divide the equation by 4:

From equation (ii), we can express $x$ in terms of $y$:

$x = y + 6$

Now substitute this expression for $x$ into equation (i):

$(y + 6)^2 + y^2 = 468$

Expand $(y + 6)^2$ 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:

$2y^2 + 12y + 36 = 468$

Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$2y^2 + 12y + 36 - 468 = 0$

$2y^2 + 12y - 432 = 0$

Divide the entire equation by 2 to simplify:

$y^2 + 6y - 216 = 0$

Now, solve this quadratic equation for $y$ by factorisation.

We need two numbers whose product is $-216$ and whose sum is $6$. The numbers are $18$ and $-12$.

Split the middle term:

$y^2 + 18y - 12y - 216 = 0$

Factor by grouping:

$y(y + 18) - 12(y + 18) = 0$

Factor out the common binomial factor $(y + 18)$:

$(y + 18)(y - 12) = 0$

Using the Zero Product Property, set each factor equal to zero:

Either $y + 18 = 0 \implies y = -18$

Or $y - 12 = 0 \implies y = 12$

Since $y$ represents the side length of a square, it must be a positive value. Therefore, we reject $y = -18$.

The side length of the second (smaller) square is $y = 12$ metres.

Now find the side length of the first (larger) square using $x = y + 6$:

$x = 12 + 6 = 18$ metres.

Check:

Sides are 18 m and 12 m.

Sum of areas = $18^2 + 12^2 = 324 + 144 = 468$ m². (Correct)

Perimeters are $4 \times 18 = 72$ m and $4 \times 12 = 48$ m.

Difference of perimeters = $72 - 48 = 24$ m. (Correct)

---

Final Answer:

The sides of the two squares are $\mathbf{18}$ metres and $\mathbf{12}$ metres.

Given:

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

---

To Find:

1. The discriminant of the equation.

2. The nature of the roots of the equation.

---

Solution:

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

Compare this equation with the standard form of a quadratic equation, $ax^2 + bx + c = 0$.

We have:

$a = 2$

$b = -4$

$c = 3$

The discriminant ($D$) of a quadratic equation is given by the formula:

$D = b^2 - 4ac$

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

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

$D = 16 - 24$

$D = -8$

The discriminant of the equation is -8.

Now, we determine the nature of the roots based on the value of the discriminant:

• If $D > 0$, the equation has two distinct real roots.
• If $D = 0$, the equation has two equal real roots (one real root).
• If $D < 0$, the equation has no real roots.

In this case, $D = -8$, which is less than 0 ($D < 0$).

Therefore, the quadratic equation $2x^2 - 4x + 3 = 0$ has no real roots.

---

Final Answer:

The discriminant of the equation $2x^2 - 4x + 3 = 0$ is $\mathbf{-8}$.

Since the discriminant is negative ($D < 0$), the equation has no real roots.

Given:

• A circular park with diameter AB = 13 m.
• Gates A and B are diametrically opposite fixed points on the boundary.
• A pole is to be erected at a point P on the boundary.
• The difference of the distances from P to A and P to B is 7 m. That is, $|PA - PB| = 7$ m.

---

To Find:

1. Whether it is possible to erect the pole under the given conditions.

2. If possible, the distances PA and PB.

---

Solution:

Let P be the point on the boundary of the circular park where the pole is erected.

Let A and B be the two diametrically opposite gates. Then AB is the diameter of the park.

Given, Diameter AB = 13 m.

Since P is a point on the circle and AB is the diameter, the angle subtended by the diameter at any point on the circumference is a right angle.

Therefore, $\angle APB = 90^\circ$.

Triangle APB is a right-angled triangle with hypotenuse AB.

By the Pythagorean theorem:

We are also given that the difference of the distances from P to A and P to B is 7 m.

$|PA - PB| = 7$

Let's assume PA > PB. Then:

From this, we can express PA in terms of PB:

Substitute equation (ii) into equation (i):

$(PB + 7)^2 + PB^2 = 169$

Expand $(PB + 7)^2$:

$(PB^2 + 14PB + 49) + PB^2 = 169$

Combine like terms:

$2PB^2 + 14PB + 49 = 169$

Rearrange into the standard quadratic form $ax^2 + bx + c = 0$ (letting $y = PB$ for simplicity):

$2y^2 + 14y + 49 - 169 = 0$

$2y^2 + 14y - 120 = 0$

Divide the entire equation by 2:

$y^2 + 7y - 60 = 0$

To check if it is possible to erect the pole, we need to check if this quadratic equation has real roots for $y$ (which represents the distance PB). We calculate the discriminant $D = b^2 - 4ac$.

Here, $a = 1$, $b = 7$, $c = -60$.

$D = (7)^2 - 4(1)(-60)$

$D = 49 + 240$

$D = 289$

Since the discriminant $D = 289 > 0$, the quadratic equation has real and distinct roots. Therefore, it is possible to erect the pole satisfying the given conditions.

Now, solve the quadratic equation $y^2 + 7y - 60 = 0$ using the quadratic formula or factorization.

Using factorization: We need two numbers whose product is -60 and sum is 7. The numbers are 12 and -5.

$y^2 + 12y - 5y - 60 = 0$

$y(y + 12) - 5(y + 12) = 0$

$(y + 12)(y - 5) = 0$

This gives two possible values for $y$ (which is PB):

$y = -12$ or $y = 5$.

Since $y$ represents the distance PB, it must be positive. Therefore, we reject $y = -12$.

So, the distance PB = 5 metres.

Now find the distance PA using equation (ii):

$PA = PB + 7 = 5 + 7 = 12$ metres.

The distances from the pole P to the gates A and B are 12 m and 5 m respectively.

---

Final Answer:

Yes, it is possible to erect the pole.

The distances from the two gates A and B to the pole P should be $\mathbf{12}$ metres and $\mathbf{5}$ metres.

Given:

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

---

To Find:

1. The discriminant of the equation.

2. The nature of its roots.

3. The roots, if they are real.

---

Solution:

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

Compare this equation with the standard form of a quadratic equation, $ax^2 + bx + c = 0$.

We have:

$a = 3$

$b = -2$

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

1. Finding the Discriminant:

The discriminant ($D$) is given by the formula $D = b^2 - 4ac$.

Substitute the values of $a$, $b$, and $c$:

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

$D = 4 - \frac{12}{3}$

$D = 4 - 4$

$D = 0$

The discriminant of the equation is 0.

2. Nature of Roots:

The nature of the roots is determined by the value of the discriminant ($D$):

• If $D > 0$, there are two distinct real roots.
• If $D = 0$, there are two equal real roots (or one repeated real root).
• If $D < 0$, there are no real roots.

Since $D = 0$, the equation has two equal real roots.

3. Finding the Roots:

Since the roots are real ($D=0$), we can find them using the quadratic formula:

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-2) \pm \sqrt{0}}{2(3)}$

$x = \frac{2 \pm 0}{6}$

$x = \frac{2}{6}$

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

The two equal real roots are $\frac{1}{3}$ and $\frac{1}{3}$.

---

Final Answer:

The discriminant of the equation $3x^2 - 2x + \frac{1}{3} = 0$ is $\mathbf{0}$.

The equation has two equal real roots.

The real roots are $\mathbf{\frac{1}{3}}$ and $\mathbf{\frac{1}{3}}$.

General Concept: Discriminant and Nature of Roots

For a quadratic equation $ax^2 + bx + c = 0$ ($a \neq 0$), the discriminant is given by $D = b^2 - 4ac$.

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

• If $D > 0$: Two distinct real roots exist.
• If $D = 0$: Two equal real roots exist (or one repeated real root).
• If $D < 0$: No real roots exist.

If real roots exist ($D \ge 0$), they can be found using the quadratic formula: $x = \frac{-b \pm \sqrt{D}}{2a}$.

---

Solution (i): $2x^2 - 3x + 5 = 0$

Given equation: $2x^2 - 3x + 5 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = -3$, $c = 5$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-3)^2 - 4(2)(5)$

$D = 9 - 40$

$D = -31$

Nature of Roots: Since $D = -31 < 0$, the discriminant is negative.

Conclusion: The equation $2x^2 - 3x + 5 = 0$ has no real roots.

---

Solution (ii): $3x^2 - 4\sqrt{3}x + 4 = 0$

Given equation: $3x^2 - 4\sqrt{3}x + 4 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 3$, $b = -4\sqrt{3}$, $c = 4$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-4\sqrt{3})^2 - 4(3)(4)$

$D = (16 \times 3) - 48$

$D = 48 - 48$

$D = 0$

Nature of Roots: Since $D = 0$, the equation has two equal real roots.

Finding the Roots: Since real roots exist, we use the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.

$x = \frac{-(-4\sqrt{3}) \pm \sqrt{0}}{2(3)}$

$x = \frac{4\sqrt{3} \pm 0}{6}$

$x = \frac{4\sqrt{3}}{6}$

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

Conclusion: The equation has two equal real roots, which are $\mathbf{\frac{2\sqrt{3}}{3}}$ and $\mathbf{\frac{2\sqrt{3}}{3}}$.

---

Solution (iii): $2x^2 - 6x + 3 = 0$

Given equation: $2x^2 - 6x + 3 = 0$.

Comparing with $ax^2 + bx + c = 0$, we have:

$a = 2$, $b = -6$, $c = 3$.

Calculate the discriminant $D = b^2 - 4ac$:

$D = (-6)^2 - 4(2)(3)$

$D = 36 - 24$

$D = 12$

Nature of Roots: Since $D = 12 > 0$, the equation has two distinct real roots.

Finding the Roots: Since real roots exist, we use the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.

$x = \frac{-(-6) \pm \sqrt{12}}{2(2)}$

$x = \frac{6 \pm \sqrt{4 \times 3}}{4}$

$x = \frac{6 \pm 2\sqrt{3}}{4}$

Factor out 2 from the numerator:

$x = \frac{2(3 \pm \sqrt{3})}{4}$

Simplify:

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

The two distinct real roots are $x_1 = \frac{3 + \sqrt{3}}{2}$ and $x_2 = \frac{3 - \sqrt{3}}{2}$.

Conclusion: The equation has two distinct real roots, which are $\mathbf{\frac{3 + \sqrt{3}}{2}}$ and $\mathbf{\frac{3 - \sqrt{3}}{2}}$.

General Concept: Condition for Equal Roots

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

The discriminant is calculated as $D = b^2 - 4ac$.

So, the condition for two equal roots is $b^2 - 4ac = 0$.

---

Solution (i): $2x^2 + kx + 3 = 0$

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

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

Solution:

Compare the given equation with the standard form $ax^2 + bx + c = 0$.

Here, $a = 2$, $b = k$, $c = 3$.

For the equation to have two equal roots, the discriminant $D$ must be zero.

$D = b^2 - 4ac = 0$

Substitute the values of $a$, $b$, and $c$:

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

$k^2 - 24 = 0$

$k^2 = 24$

Take the square root of both sides:

$k = \pm \sqrt{24}$

$k = \pm \sqrt{4 \times 6}$

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

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

---

Solution (ii): $kx(x - 2) + 6 = 0$

Given equation: $kx(x - 2) + 6 = 0$.

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

Solution:

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

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

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

Compare $kx^2 - 2kx + 6 = 0$ with the standard form $ax^2 + bx + c = 0$.

Here, $a = k$, $b = -2k$, $c = 6$.

For the equation to have two equal roots, the discriminant $D$ must be zero.

$D = b^2 - 4ac = 0$

Substitute the values of $a$, $b$, and $c$:

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

$4k^2 - 24k = 0$

Factor out $4k$:

$4k(k - 6) = 0$

This implies either $4k = 0$ or $k - 6 = 0$.

If $4k = 0$, then $k = 0$.

If $k - 6 = 0$, then $k = 6$.

However, we established that for the equation to be quadratic, $k \neq 0$. Therefore, we must reject the solution $k = 0$.

The only valid value for $k$ is $6$.

Conclusion: The value of $k$ for which the equation $kx(x - 2) + 6 = 0$ has two equal roots is $\mathbf{6}$.

Given:

• A rectangular mango grove.
• Length = 2 $\times$ Breadth.
• Area = 800 m².

---

To Find:

1. Whether it is possible to design such a mango grove.

2. If possible, find its length and breadth.

---

Solution:

Let the breadth of the rectangular mango grove be $b$ metres.

According to the given condition, the length ($l$) is twice its breadth:

$l = 2b$ metres.

The area of a rectangle is given by the formula: Area = Length $\times$ Breadth.

We are given that the Area = 800 m².

Substituting the expressions for length and breadth:

Area = $(2b) \times b = 800$

$2b^2 = 800$

Divide both sides by 2:

$b^2 = \frac{800}{2}$

$b^2 = 400$

To find the breadth $b$, take the square root of both sides:

$b = \pm \sqrt{400}$

$b = \pm 20$

Since the breadth ($b$) represents a physical dimension (length), it must be a positive value.

Therefore, $b = 20$ metres.

Since we found a positive real value for the breadth, it is possible to design such a rectangular mango grove.

Now, find the length using the relationship $l = 2b$:

$l = 2 \times 20$

$l = 40$ metres.

Check: Length (40 m) is twice the breadth (20 m). Area = $40 \times 20 = 800$ m². The conditions are satisfied.

---

Final Answer:

Yes, it is possible to design such a rectangular mango grove.

The length of the grove is $\mathbf{40}$ metres and the breadth is $\mathbf{20}$ metres.

Given:

• The sum of the present ages of two friends is 20 years.
• Four years ago, the product of their ages was 48.

---

To Find:

1. Whether the given situation is possible.

2. If possible, determine their present ages.

---

Solution:

Let the present age of the first friend be $x$ years.

Since the sum of their present ages is 20 years, the present age of the second friend is $(20 - x)$ years.

Ages four years ago:

First friend's age = $(x - 4)$ years.

Second friend's age = $(20 - x) - 4 = (16 - x)$ years.

According to the condition, the product of their ages four years ago was 48:

$(x - 4)(16 - x) = 48$

Expand the equation:

$x(16 - x) - 4(16 - x) = 48$

$16x - x^2 - 64 + 4x = 48$

Combine like terms:

$-x^2 + 20x - 64 = 48$

Move all terms to one side to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$-x^2 + 20x - 64 - 48 = 0$

$-x^2 + 20x - 112 = 0$

Multiply the entire equation by -1:

$x^2 - 20x + 112 = 0$

To determine if this situation is possible, we need to check if this quadratic equation has real roots for $x$. We examine the discriminant $D = b^2 - 4ac$.

Compare $x^2 - 20x + 112 = 0$ with $ax^2 + bx + c = 0$.

Here, $a = 1$, $b = -20$, $c = 112$.

Calculate the discriminant:

$D = (-20)^2 - 4(1)(112)$

$D = 400 - 448$

$D = -48$

Since the discriminant $D = -48$ is negative ($D < 0$), the quadratic equation $x^2 - 20x + 112 = 0$ has no real roots.

This means there is no real value of $x$ (the present age of the first friend) that satisfies the given conditions.

---

Final Answer:

No, the given situation is not possible because the corresponding quadratic equation has no real roots.

Given:

• A rectangular park.
• Perimeter = 80 m.
• Area = 400 m².

---

To Find:

1. Whether it is possible to design such a park.

2. If possible, find its length and breadth.

---

Solution:

Let the length of the rectangular park be $l$ metres and the breadth be $b$ metres.

The perimeter of a rectangle is given by $P = 2(l + b)$.

Given $P = 80$ m:

$2(l + b) = 80$

Divide by 2:

The area of a rectangle is given by $A = l \times b$.

Given $A = 400$ m²:

From equation (i), express length in terms of breadth (or vice versa):

$l = 40 - b$

Substitute this expression for $l$ into equation (ii):

$(40 - b) \times b = 400$

Expand the equation:

$40b - b^2 = 400$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$, using $b$ as the variable):

$0 = b^2 - 40b + 400$

$b^2 - 40b + 400 = 0$

To determine if this situation is possible, we check if the quadratic equation has real roots for $b$. We examine the discriminant $D = (\text{coefficient of } b)^2 - 4(\text{coefficient of } b^2)(\text{constant term})$.

Comparing with $ax^2 + bx + c = 0$, we have $a = 1$, $b = -40$, $c = 400$.

Calculate the discriminant:

$D = (-40)^2 - 4(1)(400)$

$D = 1600 - 1600$

$D = 0$

Since the discriminant $D = 0$, the quadratic equation has real and equal roots. This means it is possible to design such a rectangular park.

Now, find the roots using the quadratic formula $b = \frac{-(\text{coeff } b) \pm \sqrt{D}}{2(\text{coeff } b^2)}$ or by recognizing the perfect square.

The equation $b^2 - 40b + 400 = 0$ is equivalent to $(b - 20)^2 = 0$.

Taking the square root:

$b - 20 = 0$

$b = 20$

The breadth of the park is $b = 20$ metres.

Now find the length using $l = 40 - b$:

$l = 40 - 20 = 20$ metres.

Since the length and breadth are equal, the park is actually a square.

Check: Perimeter = $2(20 + 20) = 2(40) = 80$ m. Area = $20 \times 20 = 400$ m². Both conditions are satisfied.

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Final Answer:

Yes, it is possible to design such a rectangular park.

The length of the park is $\mathbf{20}$ metres and the breadth is $\mathbf{20}$ metres.