Chapter 4 Quadratic Equations

Mathematical Representation for (i):

Given:

John and Jivanti together have 45 marbles.

Both lost 5 marbles each.

The product of the number of marbles they now have is 124.

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To Represent:

Represent the situation using a single mathematical equation.

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Solution:

Let the number of marbles John had to start with be $x$.

Since the total number of marbles they had together was 45, the number of marbles Jivanti had to start with must be $45 - x$.

After both lost 5 marbles:

The number of marbles John has now is $x - 5$.

The number of marbles Jivanti has now is $(45 - x) - 5$, which simplifies to $40 - x$.

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

So, we can write the equation:

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

This is the required mathematical representation using a single variable.

(Expanding this equation gives $40x - x^2 - 200 + 5x = 124$, which simplifies to $-x^2 + 45x - 200 = 124$. Rearranging to the standard quadratic form gives $x^2 - 45x + 324 = 0$).

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Mathematical Representation for (ii):

Given:

Cost of production of each toy = $55 -$ (number of toys produced in a day).

Total cost of production on a particular day = $\textsf{₹} 750$.

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To Represent:

Represent the situation using a single mathematical equation.

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Solution:

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

According to the problem, the cost of production of each toy (in $\textsf{₹}$) is 55 minus the number of toys produced in a day.

Cost of production of each toy $= 55 - t$ rupees.

The total cost of production on a particular day is given by 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)

We are given that the total cost of production was $\textsf{₹} 750$.

$750 = t \times (55 - t)$

This equation relates the number of toys produced ($t$) to the total cost.

Expand the right side of the equation:

$750 = 55t - t^2$

Rearrange the terms to form a standard quadratic equation $at^2 + bt + c = 0$:

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

This is the required mathematical representation using a single variable.

(i) $(x – 2)^2 + 1 = 2x – 3$

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

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

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

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

Move all terms to one side to get the standard form $ax^2 + bx + c = 0$:

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

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

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

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(ii) $x(x + 1) + 8 = (x + 2) (x – 2)$

Expand both sides:

Left side: $x(x + 1) + 8 = x^2 + x + 8$

Right side: $(x + 2)(x – 2) = x^2 - 2^2$ (using the identity $(a+b)(a-b) = a^2 - b^2$)

Right side: $x^2 - 4$

So the equation is:

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

Move all terms to one side:

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

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

$x + 12 = 0$

This equation is in the form $ax^2 + bx + c = 0$, but the coefficient of $x^2$ is $a=0$. Since $a=0$, this is not a quadratic equation; it is a linear equation.

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

Expand the left side:

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

Move all terms to one side:

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

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

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

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(iv) $(x + 2)^3 = x^3 – 4$

Expand the left side using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:

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

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

Move all terms to one side:

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

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

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

This equation is in the form $ax^2 + bx + c = 0$, where $a=6$, $b=12$, and $c=12$. Since $a \neq 0$, this is a quadratic equation.

To check if an equation is quadratic, we need to simplify it and write it in the standard form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$.

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(i) $(x + 1)^2 = 2(x – 3)$

Expand both sides:

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

Move all terms to one side:

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

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

$x^2 + 7 = 0$

This is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=0$, $c=7$. Since $a=1 \neq 0$, it is a quadratic equation.

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(ii) $x^2 – 2x = (–2) (3 – x)$

Expand the right side:

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

Move all terms to one side:

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

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

This is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=-4$, $c=6$. Since $a=1 \neq 0$, it is a quadratic equation.

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(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$

Expand both sides:

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

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

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

Move all terms to one side:

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

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

$-3x + 1 = 0$

This is not in the form $ax^2 + bx + c = 0$ with $a \neq 0$. The coefficient of $x^2$ is 0. It is a linear equation.

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(iv) $(x – 3)(2x +1) = x(x + 5)$

Expand both sides:

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

$2x^2 + x - 6x - 3 = x^2 + 5x$

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

Move all terms to one side:

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

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

This is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=-10$, $c=-3$. Since $a=1 \neq 0$, it is a quadratic equation.

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(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$

Expand both sides:

$2x(x-3) - 1(x-3) = x(x-1) + 5(x-1)$

$2x^2 - 6x - x + 3 = x^2 - x + 5x - 5$

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

Move all terms to one side:

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

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

This is in the form $ax^2 + bx + c = 0$ with $a=1$, $b=-11$, $c=8$. Since $a=1 \neq 0$, it is a quadratic equation.

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(vi) $x^2 + 3x + 1 = (x – 2)^2$

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

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

Move all terms to one side:

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

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

$7x - 3 = 0$

This is not in the form $ax^2 + bx + c = 0$ with $a \neq 0$. The coefficient of $x^2$ is 0. It is a linear equation.

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(vii) $(x + 2)^3 = 2x (x^2 – 1)$

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

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

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

Move all terms to one side:

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

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

The highest power of $x$ is 3. This is a cubic equation, not a quadratic equation.

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(viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$

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

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

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

Move all terms to one side:

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

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

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

This is in the form $ax^2 + bx + c = 0$ with $a=2$, $b=-13$, $c=9$. Since $a=2 \neq 0$, it is a quadratic equation.

(i) Rectangular Plot Area

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

The length is one more than twice the breadth, so the length is $(2x + 1)$ metres.

The area of the rectangle is given by Length $\times$ Breadth.

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

Given area = 528 m$^2$.

So, $(2x + 1)(x) = 528$

Expanding the equation:

$2x^2 + x = 528$

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

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

This is the required quadratic equation. Here, $x$ represents the breadth of the plot.

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(ii) Product of Consecutive Integers

Let the first positive integer be $x$.

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

The product of the two consecutive positive integers is 306.

$x(x + 1) = 306$

Expanding the equation:

$x^2 + x = 306$

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

$x^2 + x - 306 = 0$

This is the required quadratic equation. Here, $x$ represents the first of the two consecutive positive integers.

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(iii) Ages Problem

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

Rohan's mother's present age is $(x + 26)$ years.

Ages 3 years from now:

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

Rohan's mother's age will be $(x + 26 + 3) = (x + 29)$ years.

The product of their ages 3 years from now is 360.

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

Expanding the left side:

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

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

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

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

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

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

This is the required quadratic equation. Here, $x$ represents Rohan's present age.

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(iv) Train Speed Problem

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

The distance traveled is 480 km.

Time taken = $\frac{\text{Distance}}{\text{Speed}} = \frac{480}{v}$ hours.

If the speed had been 8 km/h less, the new speed is $(v - 8)$ km/h.

The time taken at the reduced speed would be $\frac{480}{v-8}$ hours.

According to the problem, the time taken at the reduced speed is 3 hours more than the original time.

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

To eliminate the denominators, multiply both sides by $v(v-8)$ (assuming $v \neq 0$ and $v \neq 8$):

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

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

$480v = (v-8)(480) + 3v(v-8)$

$480v = 480v - 3840 + 3v^2 - 24v$

Move all terms to one side:

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

Divide the entire equation by 3 to simplify:

$\frac{0}{3} = \frac{3v^2}{3} - \frac{24v}{3} - \frac{3840}{3}$

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

Rearranging into the standard quadratic form $ax^2 + bx + c = 0$ (using $v$ as the variable):

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

This is the required quadratic equation. Here, $v$ represents the uniform speed of the train.

Solution:

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

We will solve this by the factorisation method, which involves splitting the middle term.

We need to find two numbers whose sum is the coefficient of $x$ (-5) and whose product is the product of the coefficient of $x^2$ and the constant term ($2 \times 3 = 6$).

The two numbers are -2 and -3, because $(-2) + (-3) = -5$ and $(-2) \times (-3) = 6$.

Split the middle term $-5x$ as $-2x - 3x$:

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

Group the terms and factor by grouping:

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

Factor out the common factors from each group:

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

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

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

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

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 $x = 1$ and $x = \frac{3}{2}$.

Solution:

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

We will find the roots by the factorisation method (splitting the middle term).

We need to find two numbers whose sum is the coefficient of $x$ (-1) and whose product is the product of the coefficient of $x^2$ and the constant term ($6 \times -2 = -12$).

The numbers are 3 and -4, because $3 + (-4) = -1$ and $3 \times (-4) = -12$.

Split the middle term $-x$ as $3x - 4x$:

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

Group the terms and factor by grouping:

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

Factor out the common factors from each group:

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

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

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

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

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 quadratic equation $6x^2 - x - 2 = 0$ are $x = -\frac{1}{2}$ and $x = \frac{2}{3}$.

Solution:

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

We will find the roots by the factorisation method (splitting the middle term).

We need to find two numbers whose sum is the coefficient of $x$, which is $-2\sqrt{6}$, and whose product is the product of the coefficient of $x^2$ and the constant term, which is $3 \times 2 = 6$.

The two numbers are $-\sqrt{6}$ and $-\sqrt{6}$, because $(-\sqrt{6}) + (-\sqrt{6}) = -2\sqrt{6}$ and $(-\sqrt{6}) \times (-\sqrt{6}) = 6$.

Split the middle term $-2\sqrt{6}x$ as $-\sqrt{6}x - \sqrt{6}x$:

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

Group the terms and factor by grouping. Note that $3 = \sqrt{3} \times \sqrt{3}$ and $2 = \sqrt{2} \times \sqrt{2}$ and $\sqrt{6} = \sqrt{3} \times \sqrt{2}$.

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

Factor out the common factors from each group:

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

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:

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

For the square of an expression to be zero, the expression itself must be 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 $\sqrt{3}$:

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

Since the quadratic equation is a perfect square, it has two equal roots.

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

Solution:

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

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

Length = $2x + 1$ metres.

The carpet area of the hall is given as 300 square metres.

Area of a rectangle = Length $\times$ Breadth

So, $(2x + 1)(x) = 300$

Expand the equation:

$2x^2 + x = 300$

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

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

We need to find the roots of this quadratic equation by factorisation.

We look for two numbers whose sum is the coefficient of $x$ (which is 1) and whose product is the product of the coefficient of $x^2$ and the constant term ($2 \times -300 = -600$).

We need two numbers $p$ and $q$ such that $p+q = 1$ and $pq = -600$.

Let's find the factors of 600 and look for pairs that have a difference of 1:

Factors of 600: (1, 600), (2, 300), (3, 200), (4, 150), (5, 120), (6, 100), (8, 75), (10, 60), (12, 50), (15, 40), (20, 30), (24, 25).

The pair (24, 25) has a difference of 1. To get a sum of +1 and a product of -600, the numbers should be +25 and -24.

Split the middle term $x$ as $25x - 24x$:

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

Group the terms and factor by grouping:

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

Factor out common factors from each group:

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

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

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

Set each factor to zero to find the possible values for $x$:

Case 1: $2x + 25 = 0$

$2x = -25$

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

Case 2: $x - 12 = 0$

$x = 12$

Since $x$ represents the breadth of the hall, it must be a positive value. Therefore, $x = -\frac{25}{2}$ is not a valid solution in this context.

The valid breadth is $x = 12$ metres.

Now, find the length using the expression for length:

Length $= 2x + 1$

Length $= 2(12) + 1$

Length $= 24 + 1$

Length $= 25$ metres.

Check the area: Length $\times$ Breadth = $25 \times 12 = 300$ m$^2$, which matches the given area.

Thus, the breadth of the prayer hall is 12 metres and the length is 25 metres.

(i) $x^2 – 3x – 10 = 0$

We need to split the middle term $-3x$ into two terms such that their sum is $-3$ and their product is $1 \times (-10) = -10$.

The numbers are $-5$ and $2$ (since $(-5) + 2 = -3$ and $(-5) \times 2 = -10$).

Rewrite the equation:

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

Group the terms and factor by grouping:

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

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

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

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

Set each factor equal to zero to find the roots:

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

$x = 5$ or $x = -2$

The roots are $x = 5$ and $x = -2$.

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(ii) $2x^2 + x – 6 = 0$

We need to split the middle term $x$ into two terms such that their sum is $1$ and their product is $2 \times (-6) = -12$.

The numbers are $4$ and $-3$ (since $4 + (-3) = 1$ and $4 \times (-3) = -12$).

Rewrite the equation:

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

Group the terms and factor by grouping:

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

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

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

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

Set each factor equal to zero to find the roots:

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

$x = -2$ or $2x = 3$

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

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

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(iii) $\sqrt{2} x^2 + 7x + 5\sqrt{2} = 0$

We need to split the middle term $7x$ into two terms such that their sum is $7$ and their product is $\sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10$.

The numbers are $5$ and $2$ (since $5 + 2 = 7$ and $5 \times 2 = 10$).

Rewrite the equation:

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

Group the terms and factor by grouping. Note that $2 = \sqrt{2} \times \sqrt{2}$.

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

Factor out common factors from each group:

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

Factor out the common binomial factor $(\sqrt{2}x + 5)$:

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

Set each factor equal to zero to find the roots:

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

$\sqrt{2}x = -5$ or $x = -\sqrt{2}$

$x = -\frac{5}{\sqrt{2}}$ or $x = -\sqrt{2}$

We can rationalize the first root:

$x = -\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$

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

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

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

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

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

We need to split the middle term $-8x$ into two terms such that their sum is $-8$ and their product is $16 \times 1 = 16$.

The numbers are $-4$ and $-4$ (since $(-4) + (-4) = -8$ and $(-4) \times (-4) = 16$).

Rewrite the equation:

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

Group the terms and factor by grouping:

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

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

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

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

This is a perfect square:

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

Set the factor equal to zero to find the root:

$4x - 1 = 0$

$4x = 1$

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

This quadratic equation has repeated roots.

The roots are $x = \frac{1}{4}$ and $x = \frac{1}{4}$.

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(v) $100x^2 - 20x + 1 = 0$

We need to split the middle term $-20x$ into two terms such that their sum is $-20$ and their product is $100 \times 1 = 100$.

The numbers are $-10$ and $-10$ (since $(-10) + (-10) = -20$ and $(-10) \times (-10) = 100$).

Rewrite the equation:

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

Group the terms and factor by grouping:

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

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

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

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

This is a perfect square:

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

Set the factor equal to zero to find the root:

$10x - 1 = 0$

$10x = 1$

$x = \frac{1}{10}$

This quadratic equation has repeated roots.

The roots are $x = \frac{1}{10}$ and $x = \frac{1}{10}$.

We need to solve the quadratic equations formulated in Example 1 using the factorisation method.

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(i) John and Jivanti's Marbles:

Let the number of marbles John had be $x$. Since John and Jivanti together have 45 marbles, the number of marbles Jivanti had is $45-x$.

After losing 5 marbles each, John has $x-5$ marbles and Jivanti has $(45-x)-5 = 40-x$ marbles.

The product of the number of marbles they now have is 124.

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

Expanding the left side:

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

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

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

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

Multiplying by -1:

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

We need to find two numbers whose sum is -45 and whose product is $1 \times 324 = 324$. The numbers are -9 and -36.

Split the middle term:

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

Factor by grouping:

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

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

Setting each factor to zero:

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

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

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

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

In either case, the numbers of marbles they had to start with are 9 and 36.

Thus, one person had 9 marbles and the other had 36 marbles.

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(ii) Cottage Industry Toy Production:

Let the number of toys produced in a day be $t$.

The cost of production of each toy is $55 - t$ (in $\textsf{₹}$).

The total cost of production on a particular day was $\textsf{₹}750$.

Total cost = (Number of toys) $\times$ (Cost per toy)

$750 = t(55 - t)$

Expanding and rearranging into the standard quadratic form:

$750 = 55t - t^2$

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

We need to find two numbers whose sum is -55 and whose product is $1 \times 750 = 750$. The numbers are -25 and -30.

Split the middle term:

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

Factor by grouping:

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

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

Setting each factor to zero:

$t - 25 = 0 \implies t = 25$

$t - 30 = 0 \implies t = 30$

Both 25 and 30 are positive integers, which is a valid number of toys produced.

If 25 toys are produced, cost per toy = $55-25 = 30$. Total cost = $25 \times 30 = 750$.

If 30 toys are produced, cost per toy = $55-30 = 25$. Total cost = $30 \times 25 = 750$.

Thus, the number of toys produced on that day could be 25 or 30.

Solution:

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

According to the first condition, their sum is 27.

According to the second condition, their product is 182.

From equation (1), we can express $y$ in terms of $x$:

$y = 27 - x$

Substitute the value of $y$ from equation (3) into equation (2):

$x(27 - x) = 182$

$27x - x^2 = 182$

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

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

We need to find the roots of this quadratic equation by factorisation (splitting the middle term).

We look for two numbers whose sum is the coefficient of $x$ (-27) and whose product is the product of the coefficient of $x^2$ and the constant term ($1 \times 182 = 182$).

We list factors of 182 to find a pair that sums to 27:

$182 = 1 \times 182 \implies 1+182 = 183$

$182 = 2 \times 91 \implies 2+91 = 93$

$182 = 7 \times 26 \implies 7+26 = 33$

$182 = 13 \times 14 \implies 13+14 = 27$

The numbers are 13 and 14. Since the sum required is -27, the numbers must be -13 and -14 (as $(-13) + (-14) = -27$ and $(-13) \times (-14) = 182$).

Split the middle term $-27x$ as $-13x - 14x$:

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

Group the terms and factor by grouping:

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

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

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

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

Set each factor equal to zero to find the possible values for $x$:

Case 1: $x - 13 = 0 \implies x = 13$

Case 2: $x - 14 = 0 \implies x = 14$

If $x = 13$, substitute this value into equation (3) to find $y$:

$y = 27 - x = 27 - 13 = 14$

The numbers are 13 and 14.

If $x = 14$, substitute this value into equation (3) to find $y$:

$y = 27 - x = 27 - 14 = 13$

The numbers are 14 and 13.

In both cases, the pair of numbers is {13, 14}.

Thus, the two numbers are 13 and 14.

Solution:

Let the first positive integer be $x$.

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

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

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

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

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

Combine like terms:

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

Move all terms to one side to get the standard quadratic form:

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

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

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 (splitting the middle term).

We need to find two numbers whose sum is the coefficient of $x$ (which is 1) and whose product is the product of the coefficient of $x^2$ and the constant term ($1 \times -182 = -182$).

We look for factors of 182 that differ by 1. The factors 13 and 14 satisfy this condition ($14 - 13 = 1$). To get a sum of +1 and a product of -182, the numbers are +14 and -13.

Split the middle term $x$ as $14x - 13x$:

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

Group the terms and factor by grouping:

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

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

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

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

Set each factor equal to zero to find the possible values for $x$:

Case 1: $x + 14 = 0 \implies x = -14$

Case 2: $x - 13 = 0 \implies x = 13$

The problem asks for positive integers. The value $x = -14$ is not a positive integer, so we discard it.

The valid value is $x = 13$.

The first positive integer is $x = 13$.

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

Check the sum of squares: $13^2 + 14^2 = 169 + 196 = 365$. This matches the given condition.

Thus, the two consecutive positive integers are 13 and 14.

Solution:

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

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

Altitude $= (x - 7)$ cm.

The hypotenuse is given as 13 cm.

For a right triangle, according to the Pythagorean theorem:

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

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

Expand the equation. Recall $(a-b)^2 = a^2 - 2ab + b^2$:

$x^2 + (x^2 - 2(x)(7) + 7^2) = 169$

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

Combine like terms:

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

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

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

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

Divide the entire equation by 2 to simplify:

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

Now, we solve this quadratic equation by factorisation (splitting the middle term).

We need to find two numbers whose sum is the coefficient of $x$ (-7) and whose product is the product of the coefficient of $x^2$ and the constant term ($1 \times -60 = -60$).

We look for factors of 60 that have a difference of 7. The numbers 12 and 5 fit this condition ($12 - 5 = 7$). To get a sum of -7 and a product of -60, the numbers are -12 and +5.

Split the middle term $-7x$ as $-12x + 5x$:

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

Group the terms and factor by grouping:

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

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

Factor out the common binomial factor $(x - 12)$:

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

Set each factor equal to zero to find the possible values for $x$:

Case 1: $x - 12 = 0 \implies x = 12$

Case 2: $x + 5 = 0 \implies x = -5$

Since $x$ represents the base length of a triangle, it must be a positive value. Therefore, $x = -5$ is not a valid solution.

The valid base is $x = 12$ cm.

Now, find the altitude using the expression for altitude:

Altitude $= x - 7$

Altitude $= 12 - 7$

Altitude $= 5$ cm.

Check: Base = 12 cm, Altitude = 5 cm, Hypotenuse = 13 cm. $12^2 + 5^2 = 144 + 25 = 169 = 13^2$. The dimensions satisfy the Pythagorean theorem.

Thus, the other two sides of the right triangle are 12 cm (base) and 5 cm (altitude).

Solution:

Let the number of pottery articles produced in a day be $x$.

According to the problem, the cost of production of each article is 3 more than twice the number of articles produced.

Cost of each article $= (2x + 3)$ rupees.

The total cost of production on that day was $\textsf{₹}90$.

Total Cost = (Number of articles produced) $\times$ (Cost of each article)

$90 = x(2x + 3)$

Expand the equation:

$90 = 2x^2 + 3x$

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

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

We need to find the roots of this quadratic equation by factorisation (splitting the middle term).

We need to find two numbers whose sum is the coefficient of $x$ (which is 3) and whose product is the product of the coefficient of $x^2$ and the constant term ($2 \times -90 = -180$).

We look for factors of 180 that have a difference of 3. The factors 15 and 12 fit this condition ($15 - 12 = 3$). To get a sum of +3 and a product of -180, the numbers are +15 and -12.

Split the middle term $3x$ as $15x - 12x$:

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

Group the terms and factor by grouping:

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

Factor out common factors from each group:

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

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

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

Set each factor equal to zero to find the possible values for $x$:

Case 1: $2x + 15 = 0 \implies 2x = -15 \implies x = -\frac{15}{2}$

Case 2: $x - 6 = 0 \implies x = 6$

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

The valid number of articles produced is $x = 6$.

Now, find the cost of each article using the expression for cost:

Cost of each article $= 2x + 3$

Cost of each article $= 2(6) + 3$

Cost of each article $= 12 + 3$

Cost of each article $= 15$

Check: Number of articles = 6, Cost per article = $\textsf{₹}15$. Total cost = $6 \times 15 = 90$. This matches the given total cost.

Thus, the number of articles produced on that day is 6 and the cost of each article is $\textsf{₹}15$.

Solution:

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

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

$a = 2$

$b = -4$

$c = 3$

The discriminant of a quadratic equation is given by the formula $D = b^2 - 4ac$.

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

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

$D = 16 - 4(6)$

$D = 16 - 24$

$D = -8$

Since the discriminant $D = -8$, which is less than zero ($D < 0$), the quadratic equation has no real roots.

---

Nature of Roots:

Since the discriminant $D < 0$, the roots of the equation $2x^2 - 4x + 3 = 0$ are non-real (complex) and distinct.

Thus, the equation has no real roots.

Given:

A circular park with diameter AB = 13 metres. A and B are two diametrically opposite gates on the boundary.

A pole is to be erected at a point P on the boundary such that the difference of its distances from gates A and B is 7 metres.

---

To Check / To Find:

1. Is it possible to erect the pole under the given conditions?

2. If yes, at what distances from the two gates should the pole be erected?

---

Solution:

Let P be the point on the boundary of the circular park where the pole is to be erected.

The gates A and B are diametrically opposite, so the line segment AB is the diameter of the circular park, with length 13 metres.

Since P is on the boundary of the circle and AB is the diameter, the angle subtended by the diameter at point P on the circumference is $90^\circ$. This means $\angle APB = 90^\circ$.

Therefore, the triangle APB is a right-angled triangle with AB as the hypotenuse.

Let the distance of the pole P from gate A be $AP = x$ metres.

Let the distance of the pole P from gate B be $BP = y$ metres.

According to the Pythagorean theorem in the right-angled triangle APB:

Substituting the distances:

So, $x^2 + y^2 = 169$.

According to the second condition given in the problem, the difference of the distances from the two gates is 7 metres. This means the absolute difference between $x$ and $y$ is 7.

$|x - y| = 7$

This implies either $x - y = 7$ or $y - x = 7$. Since $x$ and $y$ represent distances, they must be non-negative. Without loss of generality, let's assume $x \ge y$. Then $x - y = 7$.

From $x - y = 7$, we can express $x$ in terms of $y$:

$x = y + 7$

Now, substitute this expression for $x$ into equation (1):

$(y + 7)^2 + y^2 = 169$

Expand the term $(y + 7)^2$:

$(y^2 + 2 \times y \times 7 + 7^2) + y^2 = 169$

$y^2 + 14y + 49 + y^2 = 169$}

Combine the like terms ($y^2$ terms) and move the constant term to the left side:

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

$2y^2 + 14y - 120 = 0$}

Now, divide the entire equation by 2 to simplify the coefficients:

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

This is a quadratic equation in the variable $y$. We can solve this equation by factorisation.

We need to find two numbers that multiply to the constant term (-60) and add up to the coefficient of $y$ (which is 7). These numbers are 12 and -5 ($12 \times -5 = -60$ and $12 + (-5) = 7$).

Split the middle term $7y$ into $12y - 5y$:

$y^2 + 12y - 5y - 60 = 0$}

Group the terms and factor by grouping:

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

$y(y + 12) - 5(y + 12) = 0$}

Factor out the common binomial factor $(y + 12)$:

$(y + 12)(y - 5) = 0$}

Set each factor equal to zero to find the possible values for $y$:

• Case 1: $y + 12 = 0 \implies y = -12$
• Case 2: $y - 5 = 0 \implies y = 5$

Since $y$ represents a distance, it must be a non-negative value. Therefore, $y = -12$ is not a valid solution.

The valid distance from gate B is $y = 5$ metres.

Now, substitute the value of $y = 5$ back into the equation $x = y + 7$ to find $x$:

$x = 5 + 7$

$x = 12$

The distance from gate A is $x = 12$ metres.

The distances are 12 metres and 5 metres. Their difference is $|12 - 5| = 7$ metres, which matches the given condition. Also, $12^2 + 5^2 = 144 + 25 = 169 = 13^2$, which satisfies the Pythagorean theorem and the diameter condition.

Since we found real and positive values for the distances, it is possible to erect the pole at such a point on the boundary.

Note: If we had started with the assumption $y - x = 7$, leading to $y = x + 7$, substituting into equation (1) would give $x^2 + (x+7)^2 = 169$, which also simplifies to the quadratic equation $x^2 + 7x - 60 = 0$. The positive solution for $x$ would be $x=5$, which would then give $y = 5+7 = 12$. The pair of distances remains {5, 12}.

---

Conclusion:

Yes, it is possible to erect the pole at a point on the boundary satisfying the given conditions.

The distances from the two gates should be 12 metres and 5 metres.

Solution:

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

To work with integer coefficients, we can multiply the entire equation by 3:

$3(3x^2 - 2x + \frac{1}{3}) = 3(0)$

$9x^2 - 6x + 1 = 0$

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

$a = 9$

$b = -6$

$c = 1$

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 discriminant formula:

$D = (-6)^2 - 4(9)(1)$

$D = 36 - 36$

$D = 0$

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Nature of Roots:

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

---

Finding the Roots:

Since the roots are real, we can find them using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$.

As $D=0$, the formula simplifies to $x = \frac{-b}{2a}$.

$x = \frac{-(-6)}{2(9)}$

$x = \frac{6}{18}$

$x = \frac{\cancel{6}^1}{\cancel{18}_3}$

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

Both roots are the same.

Alternatively, since $D=0$, the quadratic equation is a perfect square. The equation $9x^2 - 6x + 1 = 0$ can be factored as $(3x - 1)^2 = 0$.

Setting the factor to zero:

$3x - 1 = 0$

$3x = 1$

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

The roots are $x = \frac{1}{3}$ and $x = \frac{1}{3}$.

Thus, the roots of the equation are $\frac{1}{3}$ and $\frac{1}{3}$.

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

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

If real roots exist ($D \ge 0$), the roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$. If $D=0$, the formula simplifies to $x = \frac{-b}{2a}$.

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(i) $2x^2 – 3x + 5 = 0$

Comparing this equation with $ax^2 + bx + c = 0$, we have $a = 2$, $b = -3$, and $c = 5$.

Calculate the discriminant $D$:

$D = b^2 - 4ac$

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

$D = 9 - 40$

$D = -31$

Since $D = -31 < 0$, the equation has no real roots.

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(ii) $3x^2 – 4\sqrt{3} x + 4 = 0$

Comparing this equation with $ax^2 + bx + c = 0$, we have $a = 3$, $b = -4\sqrt{3}$, and $c = 4$.

Calculate the discriminant $D$:

$D = b^2 - 4ac$

$D = (-4\sqrt{3})^2 - 4(3)(4)$

$D = (16 \times 3) - 48$

$D = 48 - 48$

$D = 0$

Since $D = 0$, the equation has two equal real roots.

Find the roots using the formula $x = \frac{-b}{2a}$:

$x = \frac{-(-4\sqrt{3})}{2(3)}$

$x = \frac{4\sqrt{3}}{6}$

$x = \frac{\cancel{4}^2\sqrt{3}}{\cancel{6}_3}$

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

The real roots are $x = \frac{2\sqrt{3}}{3}$ and $x = \frac{2\sqrt{3}}{3}$.

---

(iii) $2x^2 – 6x + 3 = 0$

Comparing this equation with $ax^2 + bx + c = 0$, we have $a = 2$, $b = -6$, and $c = 3$.

Calculate the discriminant $D$:

$D = b^2 - 4ac$

$D = (-6)^2 - 4(2)(3)$

$D = 36 - 24$

$D = 12$

Since $D = 12 > 0$, the equation has two distinct real roots.

Find the roots using 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:

$x = \frac{2(3 \pm \sqrt{3})}{4}$

Cancel the common factor 2:

$x = \frac{\cancel{2}^1(3 \pm \sqrt{3})}{\cancel{4}_2}$

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

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

A quadratic equation $ax^2 + bx + c = 0$ has two equal real roots if and only if its discriminant $D = b^2 - 4ac$ is equal to 0.

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(i) $2x^2 + kx + 3 = 0$

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

$a = 2$

$b = k$

$c = 3$

For equal roots, the discriminant must be zero ($D=0$).

$D = b^2 - 4ac = 0$

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

$k^2 - 24 = 0$

$k^2 = 24$

$k = \pm\sqrt{24}$

Simplify the square root:

$k = \pm\sqrt{4 \times 6}$

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

Thus, the values of $k$ for which the equation has two equal roots are $2\sqrt{6}$ and $-2\sqrt{6}$.

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(ii) $kx (x – 2) + 6 = 0$

First, rewrite the equation in the standard form $ax^2 + bx + c = 0$:

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

For this equation to be quadratic, the coefficient of $x^2$ cannot be zero, so $k \neq 0$.

Comparing the equation $kx^2 - 2kx + 6 = 0$ with $ax^2 + bx + c = 0$, we have:

$a = k$

$b = -2k$

$c = 6$

For equal roots, the discriminant must be zero ($D=0$).

$D = b^2 - 4ac = 0$

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

$4k^2 - 24k = 0$

Factor out the common term $4k$:

$4k(k - 6) = 0$

Setting each factor equal to zero:

$4k = 0$ or $k - 6 = 0$

$k = 0$ or $k = 6$

As noted earlier, for the equation to be quadratic, $k$ cannot be 0. Therefore, we discard $k=0$.

The only valid value of $k$ for which the quadratic equation has two equal roots is $k = 6$.

Solution:

Let the breadth of the rectangular mango grove be $x$ metres.

According to the problem, the length is twice its breadth.

Length = $2x$ metres.

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

Given that the area is 800 m$^2$, we can write the equation:

$(2x)(x) = 800$

$2x^2 = 800$

To determine if it is possible to design such a grove, we need to find if this equation has real roots for $x$. Rearrange the equation into the standard quadratic form $ax^2 + bx + c = 0$:

$2x^2 - 800 = 0$

This is a quadratic equation with $a=2$, $b=0$, and $c=-800$.

We can solve this equation for $x$:

$2x^2 = 800$

$x^2 = \frac{800}{2}$

$x^2 = 400$

Taking the square root of both sides:

$x = \pm\sqrt{400}$

$x = \pm 20$

Since $x$ represents the breadth of the rectangular grove, it must be a positive value. Therefore, we discard the negative solution $x = -20$.

The valid value for the breadth is $x = 20$ metres.

Since we obtained a real and positive value for the breadth, it is possible to design such a mango grove.

Now, we find the length using the relationship Length = $2x$:

Length $= 2 \times 20$

Length $= 40$ metres.

Check: Area = Length $\times$ Breadth = $40 \times 20 = 800$ m$^2$, which matches the given area.

Thus, it is possible to design the rectangular mango grove.

The length is 40 metres and the breadth is 20 metres.

Solution:

Let the present age of one friend be $x$ years.

Since the sum of the ages of two friends is 20 years, the present age of the other friend is $(20 - x)$ years.

Four years ago:

Age of the first friend = $(x - 4)$ years.

Age of the second friend = $(20 - x - 4) = (16 - x)$ years.

Note that for their ages to be meaningful in the past, we must have $x-4 \ge 0 \implies x \ge 4$ and $16-x \ge 0 \implies x \le 16$. Also, present ages must be non-negative, so $x \ge 0$ and $20-x \ge 0 \implies x \le 20$. Combining these, we must have $4 \le x \le 16$.

According to the problem, the product of their ages four years ago was 48.

$(x - 4)(16 - x) = 48$

Expand the left side:

$x(16 - x) - 4(16 - x) = 48$

$16x - x^2 - 64 + 4x = 48$

Combine like terms and rearrange into the standard quadratic form $ax^2 + bx + c = 0$:

$-x^2 + (16x + 4x) - 64 - 48 = 0$

$-x^2 + 20x - 112 = 0$

Multiply the 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. We use the discriminant $D = b^2 - 4ac$.

Comparing the equation $x^2 - 20x + 112 = 0$ with $ax^2 + bx + c = 0$, we have $a = 1$, $b = -20$, and $c = 112$.

Calculate the discriminant $D$:

$D = b^2 - 4ac$

$D = (-20)^2 - 4(1)(112)$

$D = 400 - 448$

$D = -48$

Since the discriminant $D = -48 < 0$, the quadratic equation has no real roots.

This means there are no real values of $x$ that satisfy the given conditions.

Thus, the situation described in the problem is not possible.

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 $2(l + b)$.

According to the problem, the perimeter is 80 m.

$2(l + b) = 80$

Divide by 2:

The area of a rectangle is given by $l \times b$.

According to the problem, the area is 400 m$^2$.

From equation (1), express $l$ in terms of $b$:

Substitute the value of $l$ from equation (3) into equation (2):

$(40 - b)b = 400$

$40b - b^2 = 400$

Rearrange into the standard quadratic form $ab^2 + bb + c = 0$:

$b^2 - 40b + 400 = 0$

This is a quadratic equation in the variable $b$. To determine if it is possible to design such a park, we need to check if this equation has real roots. We use the discriminant $D = B^2 - 4AC$, where $A=1$, $B=-40$, and $C=400$.

Calculate the discriminant $D$:

$D = (-40)^2 - 4(1)(400)$

$D = 1600 - 1600$

$D = 0$

Since the discriminant $D = 0$, the quadratic equation has two equal real roots. This means there is a real solution for the breadth $b$, so the situation is possible.

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Finding the length and breadth:

Since real roots exist ($D=0$), we can find the value of $b$ using the quadratic formula $b = \frac{-B \pm \sqrt{D}}{2A}$. As $D=0$, the formula simplifies to $b = \frac{-B}{2A}$.

$b = \frac{-(-40)}{2(1)}$

$b = \frac{40}{2}$

$b = 20$

The breadth of the park is 20 metres.

Now, substitute the value of $b = 20$ into equation (3) to find the length $l$:

$l = 40 - b$

$l = 40 - 20$

$l = 20$

The length of the park is 20 metres.

Check the dimensions: Perimeter = $2(20 + 20) = 2(40) = 80$ m. Area = $20 \times 20 = 400$ m$^2$. The values match the given conditions.

Thus, it is possible to design the rectangular park.

The length and breadth are both 20 metres.