Quadratic Equations — Chapter Test

Solve the following equations by factorisation:

x² + 6x - 16 = 0

Answer

Given,

$\Rightarrow x^2 + 6x - 16 = 0 \\[1em] \Rightarrow x^2 + 8x - 2x - 16 = 0 \\[1em] \Rightarrow x(x + 8) - 2(x + 8) = 0 \\[1em] \Rightarrow (x - 2)(x + 8) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x + 8 = 0 \\[1em] \Rightarrow x = 2 \text{ or } x = -8$

Hence, roots of given equation are 2, -8.

Solve the following equations by factorisation:

3x² + 11x + 10 = 0

Answer

Given,

$\Rightarrow 3x^2 + 11x + 10 = 0 \\[1em] \Rightarrow 3x^2 + 6x + 5x + 10 = 0 \\[1em] \Rightarrow 3x(x + 2) + 5(x + 2) = 0 \\[1em] \Rightarrow (x + 2)(3x + 5) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } 3x + 5 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = -\dfrac{5}{3}$

Hence, roots of given equation are -2, -$\dfrac{5}{3}$.

Solve the following equations by factorisation:

2x² + ax - a² = 0

Answer

Given,

$\Rightarrow 2x^2 + ax - a^2 = 0 \\[1em] \Rightarrow 2x^2 + 2ax - ax - a^2 = 0 \\[1em] \Rightarrow 2x(x + a) - a(x + a) = 0 \\[1em] \Rightarrow (x + a)(2x - a) = 0 \\[1em] \Rightarrow x + a = 0 \text{ or } 2x - a = 0 \\[1em] \Rightarrow x = -a \text{ or } x = \dfrac{a}{2}$

Hence, roots of given equation are $-a, \dfrac{a}{2}$.

Solve the following equations by factorisation:

$\sqrt{3}x^2 + 10x + 7\sqrt{3}$ = 0

Answer

Given,

$\Rightarrow \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \\[1em] \Rightarrow \sqrt{3}x^2 + 7x + 3x + 7\sqrt{3} = 0 \\[1em] \Rightarrow x(\sqrt{3}x + 7) + \sqrt{3}(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow (x + \sqrt{3})(\sqrt{3}x + 7) = 0 \\[1em] \Rightarrow x + \sqrt{3} = 0 \text{ or } \sqrt{3}x + 7 = 0 \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\dfrac{7}{\sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = -\dfrac{7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow x = -\sqrt{3} \text{ or } x = - \dfrac{7\sqrt{3}}{3}$

Hence, roots of given equation are $-\sqrt{3}, -\dfrac{7\sqrt{3}}{3}$.

Solve the following equations by factorisation:

x(x + 1) +(x + 2)(x + 3) = 42

Answer

Given,

$\Rightarrow x(x + 1) + (x + 2)(x + 3) = 42 \\[1em] \Rightarrow x^2 + x + x^2 + 3x + 2x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 = 42 \\[1em] \Rightarrow 2x^2 + 6x + 6 - 42 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 36 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 18) = 0 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] x = 3 \text{ or } x = -6$

Hence, roots of given equation are 3 , -6.

Solve the following equations by factorisation:

$\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2}$

Answer

Given,

$\dfrac{6}{x} - \dfrac{2}{x - 1} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6(x - 1) - 2x}{x(x - 1)} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{6x - 6 - 2x}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow \dfrac{4x - 6}{x^2 - x} = \dfrac{1}{x - 2} \\[1em] \Rightarrow (4x - 6)(x - 2) = x^2 - x \\[1em] \Rightarrow (4x^2 - 8x - 6x + 12) = x^2 - x \\[1em] \Rightarrow 4x^2 - x^2 - 14x + x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 3 = 0 \\[1em] x = \dfrac{4}{3} \text{ or } x = 3.$

Hence, roots of given equation are $3 ,\dfrac{4}{3}$

Solve the following equations by factorisation:

$\sqrt{x + 15} = x + 3$

Answer

Given,

$\Rightarrow \sqrt{x + 15} = x + 3 \\[1em] \Rightarrow x + 15 = (x + 3)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow x + 15 = x^2 + 9 + 6x \\[1em] \Rightarrow x^2 + 6x - x + 9 - 15 = 0 \\[1em] \Rightarrow x^2 + 5x - 6 = 0 \\[1em] \Rightarrow x^2 + 6x - x - 6 = 0 \\[1em] \Rightarrow x(x + 6) - 1(x + 6) = 0 \\[1em] \Rightarrow (x - 1)(x + 6) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 1 \text{ or } x = -6$

Since we squared the equation, so roots need to be checked

Putting x = -6 in equation

$\Rightarrow \sqrt{-6 + 15} = -6 + 3 \\[1em] \Rightarrow \sqrt{ 9 } = -3 \\[1em]$

L.H.S. = 3 and R.H.S. = -3

Since, L.H.S. ≠ R.H.S. hence, x = -6 is not root of the equation

Putting x = 1 in equation

$\Rightarrow \sqrt{1 + 15} = 1 + 3 \\[1em] \Rightarrow \sqrt{ 16 } = 4 \\[1em]$

L.H.S. = R.H.S. = 4

Hence, root of the equation is 1.

Solve the following equations by factorisation:

$\sqrt{3x^2 - 2x - 1} = 2x - 2$

Answer

Given,

$\Rightarrow \sqrt{3x^2 - 2x - 1} = 2x - 2 \\[1em] \Rightarrow 3x^2 - 2x - 1 = (2x - 2)^2 \text{ (On squaring both sides) } \\[1em] \Rightarrow 3x^2 - 2x - 1 = 4x^2 + 4 - 8x \\[1em] \Rightarrow 3x^2 - 4x^2 - 2x + 8x - 1 - 4 = 0 \\[1em] \Rightarrow -x^2 + 6x - 5 = 0 \\[1em] \Rightarrow x^2 - 6x + 5 = 0 \text{ (On multiplying equation by -1) }\\[1em] \Rightarrow x^2 - 5x - x + 5 = 0 \\[1em] \Rightarrow x(x - 5) - 1(x - 5) = 0 \\[1em] \Rightarrow (x - 1)(x - 5) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 5 = 0 \\[1em] x = 1 \text{ or } x = 5$

Since we squared the equation, so roots need to be checked

Putting x = 1 in equation

$\Rightarrow \sqrt{3(1)^2 - 2(1) - 1} = 2(1) - 2 \\[1em] \Rightarrow \sqrt{ 0 } = 0 \\[1em]$

L.H.S. = R.H.S. = 0

Putting x = 5 in equation

$\Rightarrow \sqrt{3(5)^2 - 2(5) - 1} = 2(5) - 2 \\[1em] \Rightarrow \sqrt{ 64 } = 8 \\[1em]$

L.H.S. = R.H.S. = 8

Hence, roots of the equation are 1, 5.

Solve the following equations by using formula:

2x² - 3x - 1 = 0

Answer

The given equation is 2x² - 3x - 1 = 0

Comparing it with ax² + bx + c = 0
a = 2, b = -3, c = -1

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-3) ± \sqrt{(-3)^2 - 4 \times 2 \times -1 }}{2 \times 2} \\[1em] \Rightarrow \dfrac{3 ± \sqrt{9 + 8}}{4} \\[1em] \Rightarrow \dfrac{3 + \sqrt{17}}{4} \text{ or } \dfrac{3 - \sqrt{17}}{4} \\[1em]$

Hence, roots of the equation are $\dfrac{3 + \sqrt{17}}{4} , \dfrac{3 - \sqrt{17}}{4}$.

Solve the following equations by using formula:

$x \Big(3x + \dfrac{1}{2} \Big)$ = 6

Answer

The given equation is $x \Big(3x + \dfrac{1}{2} \Big)$ = 6

$\Rightarrow 3x^2 + \dfrac{1}{2}x = 6 \\[1em] \Rightarrow 6x^2 + x = 12 \\[1em] \Rightarrow 6x^2 + x - 12 = 0$

Comparing it with ax² + bx + c = 0
a= 6, b = 1, c = -12

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 6 \times -12 }}{2 \times 6} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 288}}{12} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{289}}{12} \text{ or } \dfrac{-1 - \sqrt{289}}{12} \\[1em] \Rightarrow \dfrac{-1 + 17}{12} \text{ or } \dfrac{-1 - 17}{12} \\[1em] \Rightarrow \dfrac{16}{12} \text{ or } -\dfrac{18}{12} \\[1em] \Rightarrow \dfrac{4}{3} \text{ or } -\dfrac{3}{2}$

Hence, roots of the equation are $\dfrac{4}{3} , -\dfrac{3}{2}$.

Solve the following equations by using formula:

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

Answer

Given,

$\dfrac{2x + 5}{3x + 4} = \dfrac{x + 1}{x + 3} \\[1em] \Rightarrow (2x + 5)(x + 3) = (x + 1)(3x + 4) \\[1em] \Rightarrow (2x^2 + 6x + 5x + 15) = (3x^2 + 4x + 3x + 4) \\[1em] \Rightarrow 2x^2 - 3x^2 + 11x - 7x + 15 - 4 = 0 \\[1em] \Rightarrow -x^2 + 4x + 11 = 0 \\[1em] \Rightarrow x^2 - 4x - 11 = 0 \text{ ( On multiplying equation by -1) }$

Comparing it with ax² + bx + c = 0
a= 1, b = -4, c = -11

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-4) ± \sqrt{(-4)^2 - 4 \times 1 \times -11 }}{2 \times 1} \\[1em] \Rightarrow \dfrac{4 ± \sqrt{16 + 44}}{2} \\[1em] \Rightarrow \dfrac{4 + \sqrt{60}}{2} \text{ or } \dfrac{4 - \sqrt{60}}{2} \\[1em] \Rightarrow \dfrac{4 + 2\sqrt{15}}{2} \text{ or } \dfrac{4 - 2\sqrt{15}}{2} \\[1em] \Rightarrow 2 + \sqrt{15} \text{ or } 2 - \sqrt{15}$

Hence, roots of the equation are $2 + \sqrt{15} , 2 - \sqrt{15}$.

Solve the following equations by using formula:

$\dfrac{2}{x + 2} - \dfrac{1}{x + 1} = \dfrac{4}{x + 4} - \dfrac{3}{x + 3}$

Answer

The given equation is $\dfrac{2}{x + 2} - \dfrac{1}{x + 1} = \dfrac{4}{x + 4} - \dfrac{3}{x + 3}$

$\Rightarrow \dfrac{2(x + 1) - (x + 2)}{(x + 2)(x + 1)} = \dfrac{4(x + 3) - 3(x + 4)}{(x + 4)(x + 3)} \\[1em] \Rightarrow \dfrac{2x + 2 - x - 2}{x^2 + x + 2x + 2} = \dfrac{4x + 12 - 3x - 12}{x^2 + 3x + 4x + 12} \\[1em] \Rightarrow \dfrac{x}{x^2 + 3x + 2} = \dfrac{x}{x^2 + 7x + 12} \\[1em] \Rightarrow x(x^2 + 7x + 12) = x(x^2 + 3x + 2) \\[1em] \Rightarrow (x^3 + 7x^2 + 12x) = (x^3 + 3x^2 + 2x) \\[1em] \Rightarrow x^3 + 7x^2 + 12x - x^3 - 3x^2 - 2x = 0 \\[1em] \Rightarrow 4x^2 + 10x = 0 \\[1em] \Rightarrow 2x(2x + 5) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } 2x + 5 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -\dfrac{5}{2}$

Hence, roots of the equation are $0 ,-\dfrac{5}{2}$.

Solve the following equations by using formula:

$\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}, x$ ≠ $\dfrac{4}{3}$

Answer

The given equation is $\dfrac{3x - 4}{7} + \dfrac{7}{3x - 4} = \dfrac{5}{2}$

$\Rightarrow \dfrac{(3x - 4)^2 + 7^2}{7(3x - 4)} = \dfrac{5}{2} \text{ (On taking L.C.M. ) }\\[1em] \Rightarrow \dfrac{9x^2 + 16 - 24x + 49}{21x - 28} = \dfrac{5}{2} \\[1em] \Rightarrow 2(9x^2 - 24x + 65) = 5(21x - 28) \\[1em] \Rightarrow 18x^2 - 48x + 130 = 105x - 140 \\[1em] \Rightarrow 18x^2 - 48x - 105x + 130 + 140 = 0 \\[1em] \Rightarrow 18x^2 - 153x + 270 = 0 \\[1em] \Rightarrow 9(2x^2 - 17x + 30) = 0 \\[1em] \Rightarrow 2x^2 - 17x + 30 = 0$

Comparing it with ax² + bx + c = 0
a= 2, b = -17, c = 30

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-17) ± \sqrt{(-17)^2 - 4 \times 2 \times 30}}{2 \times 2} \\[1em] \Rightarrow \dfrac{17 ± \sqrt{289 - 240}}{4} \\[1em] \Rightarrow \dfrac{17 + \sqrt{49}}{4} \text{ or } \dfrac{17 - \sqrt{49}}{4} \\[1em] \Rightarrow \dfrac{17 + 7}{4} \text{ or } \dfrac{17 - 7}{4} \\[1em] \Rightarrow \dfrac{24}{4} \text{ or } \dfrac{10}{4} \\[1em] 6 \text{ or } \dfrac{5}{2}$

Hence, roots of the given equation are $6, \dfrac{5}{2}$.

Solve the following equations by using formula:

$\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3} , x$ ≠ $0, - \dfrac{3}{2}$.

Answer

The given equation is $\dfrac{4}{x} - 3 = \dfrac{5}{2x + 3}$

$\Rightarrow \dfrac{4 - 3x}{x} = \dfrac{5}{2x + 3} \text{ (On taking L.C.M.) } \\[1em] \Rightarrow (4 - 3x)(2x + 3) = 5x \text{ (On Cross multiplication) } \\[1em] \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \\[1em] \Rightarrow 6x^2 + 5x + 9x - 8x - 12 = 0 \\[1em] \Rightarrow 6x^2 + 6x - 12 = 0 \\[1em] \Rightarrow 6(x^2 + x - 2) = 0 \\[1em] x^2 + x - 2 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = 1, c = -2

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(1) ± \sqrt{(1)^2 - 4 \times 1 \times -2}}{2 \times 1} \\[1em] \Rightarrow \dfrac{-1 ± \sqrt{1 + 8}}{2} \\[1em] \Rightarrow \dfrac{-1 + \sqrt{9}}{2} \text{ or } \dfrac{-1 - \sqrt{9}}{2} \\[1em] \Rightarrow \dfrac{-1 + 3}{2} \text{ or } \dfrac{-1 - 3}{2} \\[1em] \Rightarrow \dfrac{2}{2} \text{ or } \dfrac{-4}{2} \\[1em] 1 \text{ or } -2$

Hence, roots of the given equation are 1, -2.

Solve the following equations by using formula:

x² + (4 - 3a)x - 12a = 0

Answer

The given equation is x² + (4 - 3a)x - 12a = 0

Comparing it with ax² + bx + c = 0
a= 1, b = (4 - 3a), c = -12a

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(4 - 3a) ± \sqrt{(4 - 3a)^2 - 4 \times 1 \times -12a}}{2 \times 1} \\[1em] \Rightarrow \dfrac{(3a - 4) ± \sqrt{16 + 9a^2 - 24a + 48a}}{2} \\[1em] \Rightarrow \dfrac{3a - 4 ± \sqrt{16 + 9a^2 + 24a}}{2} \\[1em] \Rightarrow \dfrac{3a - 4 ± \sqrt{(4 + 3a)^2}}{2}\\[1em] \Rightarrow \dfrac{3a - 4 + |4 + 3a|}{2} \text{ or } \dfrac{3a - 4 - |4 + 3a|}{2} \\[1em] \Rightarrow \dfrac{6a}{2} \text{ or } -\dfrac{8}{2} \\[1em] 3a \text{ or } -4$

Hence, roots of the given equation are 3a, -4.

Solve the following equations by using formula:

10ax² - 6x + 15ax - 9 = 0 , a ≠ 0.

Answer

The given equation is 10ax² - 6x + 15ax - 9 = 0

Comparing it with ax² + bx + c = 0
a= 10a, b = (15a - 6), c = -9

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(15a - 6) ± \sqrt{(15a - 6)^2 - 4 \times 10a \times -9}}{2 \times 10a} \\[1em] \Rightarrow \dfrac{(6 - 15a) ± \sqrt{225a^2 + 36 - 180a + 360a}}{20a} \\[1em] \Rightarrow \dfrac{6 - 15a ± \sqrt{225a^2 + 36 + 180a}}{20a} \\[1em] \Rightarrow \dfrac{6 - 15a ± \sqrt{(15a + 6)^2}}{20a}\\[1em] \Rightarrow \dfrac{6 - 15a + |15a + 6|}{20a} \text{ or } \dfrac{6 - 15a - |15a + 6|}{20a} \\[1em] \Rightarrow \dfrac{12}{20a} \text{ or } -\dfrac{30a}{20a} \\[1em] \dfrac{3}{5a} \text{ or } -\dfrac{3}{2}$

Hence, roots of the given equation are $\dfrac{3}{5a}, -\dfrac{3}{2}$.

Solve for x using the quadratic formula. Write your answer correct to two significant figures : (x - 1)² - 3x + 4 = 0 .

Answer

Given,

$(x - 1)^2 - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 + 1 - 2x - 3x + 4 = 0 \\[0.5em] \Rightarrow x^2 - 5x + 5 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = -5, c = 5

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-5) ± \sqrt{(-5)^2 - 4 \times 1 \times 5}}{2 \times 1} \\[1em] \Rightarrow \dfrac{5 ± \sqrt{25 - 20}}{2} \\[1em] \Rightarrow \dfrac{5 + \sqrt{5}}{2} \text{ or } \dfrac{5 - \sqrt{5}}{2} \\[1em] \Rightarrow \dfrac{5 + 2.2}{2} \text{ or } \dfrac{5 - 2.2}{2} \\[1em] \Rightarrow \dfrac{7.2}{2} \text{ or } \dfrac{2.8}{2} \\[1em] 3.6 \text{ or } 1.4$

Hence, roots of the given equation are 3.6, 1.4.

Discuss the nature of the roots of the following equations :

(i) $3x^2 - 7x + 8 = 0$

(ii) $x^2 - \dfrac{1}{2}x - 4 = 0$

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

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

In case real roots exist , then find them.

Answer

(i) The given equation is 3x² - 7x + 8 = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -7, c = 8

$\therefore \text{Discriminant }= b^2 - 4ac \\[0.5em] = (-7)^2 - 4 \times 3 \times 8 \\[0.5em] = 49 - 96 \\[0.5em] = -47 \lt 0$

Since, Discriminant < 0 , hence equation has no real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x - 4 = 0$

Comparing it with ax² + bx + c = 0
a= 1, b = $-\dfrac{1}{2}$, c = -4

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-\dfrac{1}{2})^2 - 4 \times 1 \times -4 \\[1em] = \dfrac{1}{4} + 16 \\[1em] = \dfrac{65}{4} \gt 0$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-\dfrac{1}{2}) ± \sqrt{(-\dfrac{1}{2})^2 - 4 \times 1 \times -4}}{2 \times 1} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} ± \sqrt{\dfrac{1}{4} + 16}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \sqrt{\dfrac{65}{4}}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \sqrt{\dfrac{65}{4}}}{2} \\[1em] \Rightarrow \dfrac{\dfrac{1}{2} + \dfrac{\sqrt{65}}{2}}{2} \text{ or } \dfrac{\dfrac{1}{2} - \dfrac{\sqrt{65}}{2}}{2} \\[1em] \Rightarrow \dfrac{1 + \sqrt{65}}{4} \text{ or } \dfrac{1 - \sqrt{65}}{4}$

Hence, roots of the given equation are $\dfrac{1 + \sqrt{65}}{4} , \dfrac{1 - \sqrt{65}}{4}$.

(iii) $5x^2 - 6\sqrt{5}x + 9 = 0$

The given equation is $5x^2 - 6\sqrt{5}x + 9 = 0$

Comparing it with ax² + bx + c = 0
a= 5, b = -6√5, c = 9

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-6\sqrt{5})^2 - 4 \times 5 \times 9 \\[1em] = 180 - 180 \\[1em] = 0$

Since, Discriminant = 0, hence equation has two equal and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-6\sqrt{5}) ± \sqrt{(-6\sqrt{5})^2 - 4 \times 5 \times 9}}{2 \times 5} \\[1em] \Rightarrow \dfrac{6\sqrt{5} ± \sqrt{180 - 180}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5} + \sqrt{0}}{10} \text{ or } \dfrac{6\sqrt{5} - \sqrt{0}}{10} \\[1em] \Rightarrow \dfrac{6\sqrt{5}}{10} \text{ or } \dfrac{6\sqrt{5}}{10} \\[1em] \Rightarrow \dfrac{3\sqrt{5}}{5} \text{ or } \dfrac{3\sqrt{5}}{5} \\[1em] \dfrac{3}{\sqrt{5}} \text{ or } \dfrac{3}{\sqrt{5}}$

Hence, roots of the given equation are $\dfrac{3}{\sqrt{5}}, \dfrac{3}{\sqrt{5}}$.

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

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

Comparing it with ax² + bx + c = 0
a= $\sqrt{3}$, b = -2, c = -$\sqrt{3}$

$\therefore \text{Discriminant } = b^2 - 4ac \\[1em] = (-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3} \\[1em] = 4 + 12 \\[1em] = 16$

Since, Discriminant > 0 , hence equation has two distinct and real roots.

By using the formula , x = $\dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ , we obtain

$\Rightarrow \dfrac{-(-2) ± \sqrt{(-2)^2 - 4 \times \sqrt{3} \times -\sqrt{3}}}{2 \times \sqrt{3}} \\[1em] \Rightarrow \dfrac{2 ± \sqrt{4 + 12}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + \sqrt{16}}{2\sqrt{3}} \text{ or } \dfrac{2 - \sqrt{16}}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{2 + 4}{2\sqrt{3}} \text{ or } \dfrac{2- 4}{2\sqrt{3}} \\[1em] \Rightarrow \dfrac{6}{2\sqrt{3}} \text{ or } -\dfrac{2}{2\sqrt{3}} \\[1em] \sqrt{3} \text{ or } -\dfrac{1}{\sqrt{3}}$

Hence, roots of the given equation are $\sqrt{3}, -\dfrac{1}{\sqrt{3}}$.

Find the values of k so that the quadratic equation (4 - k)x² + 2(k + 2)x + (8k + 1) = 0 has equal roots.

Answer

The given equation is (4 - k)x² + 2(k + 2)x + (8k + 1) = 0

Comparing it with ax² + bx + c = 0
a= (4 - k), b = (2k + 4), c = (8k + 1)

Given,

Equation has real and equal roots

∴ b² - 4ac = 0

$\Rightarrow (2k + 4)^2 - 4 \times (4 - k) \times (8k + 1) = 0 \\[1em] \Rightarrow (4k^2 + 16 + 16k) - 4(32k + 4 - 8k^2 - k) = 0 \\[1em] \Rightarrow 4k^2 + 16 + 16k - 128k - 16 + 32k^2 + 4k = 0 \\[1em] \Rightarrow 4k^2 + 32k^2 + 16k - 128k + 4k + 16 - 16 = 0 \\[1em] \Rightarrow 36k^2 - 108k = 0 \\[1em] \Rightarrow 36k(k - 3) = 0 \\[1em] \Rightarrow 36k = 0 \text{ or } k - 3 = 0 \\[1em] k = 0 \text{ or } k = 3$

Hence , the value of k are 0, 3.

Find the values of m so that the quadratic equation 3x² - 5x - 2m = 0 has two distinct real roots.

Answer

The given equation is 3x² - 5x - 2m = 0

Comparing it with ax² + bx + c = 0
a= 3, b = -5, c = -2m

Given,

Equation has two real and distinct roots

∴ b² - 4ac > 0

$\Rightarrow (-5)^2 - 4 \times 3 \times -2m \gt 0 \\[1em] \Rightarrow 25 + 24m \gt 0 \\[1em] \Rightarrow 24m \gt -25 \\[1em] m \gt -\dfrac{25}{24}$

Hence, the required value is m > $-\dfrac{25}{24}$

Find the value(s) of k for which each of the following quadratic equation has equal roots:

(i) 3kx² = 4(kx - 1)

(ii) (k + 4)x² + (k + 1)x + 1 = 0

Also, find the roots for that value(s) of k in each case.

Answer

(i) Given,

$3kx^2 = 4(kx - 1) \\[0.5em] \Rightarrow 3kx^2 - 4(kx - 1) = 0 \\[0.5em] \Rightarrow 3kx^2 - 4kx + 4 = 0 \\[0.5em]$

Comparing it with ax² + bx + c = 0
a= 3k, b = -4k, c = 4

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (-4k)^2 - 4 \times 3k \times 4 = 0 \\[0.5em] \Rightarrow 16k^2 - 48k = 0 \\[0.5em] \Rightarrow 16k(k - 3) = 0 \\[0.5em] \Rightarrow 16k = 0 \text{ or } k - 3 = 0 \\[0.5em] k = 0 \text{ or } k = 3$

But k cannot be equal to 0 as that will make a = 3k = 0 , which will make roots equal to infinity.

∴ k = 3

Hence equation is 9x² - 12x + 4 = 0

$\Rightarrow 9x^2 - 6x - 6x + 4 = 0 \\[0.5em] \Rightarrow 3x(3x - 2) - 2(3x - 2) = 0 \\[0.5em] \Rightarrow (3x - 2)(3x - 2) = 0 \\[0.5em] \Rightarrow 3x - 2 = 0 \text{ or } 3x - 2 = 0 \\[0.5em] x = \dfrac{2}{3} \text{ or } x = \dfrac{2}{3}$

Hence the value of k is 3 and the roots are $\dfrac{2}{3}, \dfrac{2}{3}.$

(ii) The given equation is (k + 4)x² +(k + 1)x + 1 = 0

Comparing it with ax² + bx + c = 0
a = (k + 4), b = (k + 1), c = 1

Given,

Equation has equal roots

∴ b² - 4ac = 0

$\Rightarrow (k + 1)^2 - 4 \times (k + 4) \times 1 = 0 \\[1em] \Rightarrow (k^2 + 1 + 2k) - 4(k + 4) = 0 \\[1em] \Rightarrow k^2 + 1 + 2k - 4k - 16 = 0 \\[1em] \Rightarrow k^2 - 2k - 15 = 0 \\[1em] \Rightarrow k^2 - 5k + 3k - 15 = 0 \\[1em] \Rightarrow k(k - 5) + 3(k - 5) = 0 \\[1em] \Rightarrow (k + 3)(k - 5) = 0 \\[1em] \Rightarrow k + 3 = 0 \text{ or } k - 5 = 0 \\[1em] k = -3 \text{ or } k = 5$

∴ When k = -3 , equation is x² - 2x + 1 = 0

$\Rightarrow x^2 - x - x + 1 = 0 \\[1em] \Rightarrow x(x - 1) - 1(x - 1) = 0 \\[1em] \Rightarrow (x - 1)(x - 1) = 0 \\[1em] \Rightarrow x - 1 = 0 \text{ or } x - 1 = 0 \\[1em] x = 1 \text{ or } x = 1$

∴ When k = 5 , equation is 9x² + 6x + 1 = 0

$\Rightarrow 9x^2 + 3x + 3x + 1 = 0 \\[1em] \Rightarrow 3x(3x + 1) + 1(3x + 1) = 0 \\[1em] \Rightarrow (3x + 1)(3x + 1) = 0 \\[1em] \Rightarrow 3x + 1 = 0 \text{ or } 3x + 1 = 0 \\[1em] x = -\dfrac{1}{3} \text{ or } -\dfrac{1}{3}$

k = -3, 5
When k = -3 , roots are 1, 1
When k = 5 , roots are $-\dfrac{1}{3}, -\dfrac{1}{3}$

Find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer

Let first number be x

Since difference between two numbers is 3 hence, the other number is (x + 3).

Given , sum of the squares of number = 117

∴ x² + (x + 3)² = 117

$\Rightarrow x^2 + x^2 + 9 + 6x = 117 \\[1em] \Rightarrow 2x^2 + 6x + 9 - 117 = 0 \\[1em] \Rightarrow 2x^2 + 6x - 108 = 0 \\[1em] \Rightarrow 2(x^2 + 3x - 54) = 0 \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow x - 6 = 0 \text{ or } x + 9 = 0 \\[1em] x = 6 \text{ or } x = -9$

Since, numbers are natural hence, x ≠ -9.

∴ x = 6 , x + 3 = 9.

Hence, the required numbers are 6, 9.

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Answer

Let the larger number be x , so the smaller number is 16 - x.

Given, twice the square of the larger part exceeds the square of the smaller part by 164

$\therefore 2x^2 - (16 - x)^2 = 164 \\[1em] \Rightarrow 2x^2 - (256 + x^2 - 32x) = 164 \\[1em] \Rightarrow 2x^2 - 256 - x^2 + 32x - 164 = 0 \\[1em] \Rightarrow x^2 + 32x - 420 = 0 \\[1em] \Rightarrow x^2 + 42x - 10x - 420 = 0 \\[1em] \Rightarrow x(x + 42) - 10(x + 42) = 0 \\[1em] \Rightarrow (x - 10)(x + 42) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x + 42 = 0 \\[1em] x = 10 \text{ or } x = -42$

Since, numbers are natural hence, x ≠ -42.

∴ x = 10 , 16 - x = 6.

Hence, the required numbers are 10, 6.

Two natural numbers are in the ratio 3 : 4 . Find the numbers if the difference between their squares is 175.

Answer

Since, the numbers are in the ratio 3 : 4, hence the numbers be 3x and 4x.

Given , difference between their squares = 175

∴ (4x)² - (3x)² = 175

$\Rightarrow 16x^2 - 9x^2 = 175 \\[1em] \Rightarrow 7x^2 = 175 \\[1em] \Rightarrow x^2 = \dfrac{175}{7} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x + 5 = 0 \\[1em] x = 5 \text{ or } x = -5$

Since, numbers are natural hence, x ≠ -5.

∴ x = 5, 3x = 15 , 4x = 20.

Hence, the required numbers are 15, 20.

Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation and solve it to find the sides of the squares.

Answer

Area of a square = (side)²

∴ Area of first square = x² and Area of second square = (x + 4)²

Given, sum of areas of two squares is = 656 cm²

∴ x² + (x + 4)² = 656

$\Rightarrow x^2 + x^2 + 16 + 8x = 656 \\[1em] \Rightarrow 2x^2 + 8x + 16 - 656 = 0 \\[1em] \Rightarrow 2x^2 + 8x - 640 = 0 \\[1em] \Rightarrow 2(x^2 + 4x - 320) = 0 \\[1em] \Rightarrow x^2 + 4x - 320 = 0 \\[1em] \Rightarrow x^2 + 20x - 16x - 320 = 0 \\[1em] \Rightarrow x(x + 20) - 16(x + 20) = 0 \\[1em] \Rightarrow (x + 20)(x - 16) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 16 = 0 \\[1em] x = -20 \text{ or } x = 16$

Since, length cannot be negative hence x ≠ -20.

∴ x = 16 , x + 4 = 20.

Hence, the sides of two squares are 16 cm and 20 cm.

The length of a rectangular garden is 12m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Answer

Let the value of breadth be x metre

So, length = (x + 12) metre

Perimeter = 2(Length + Breadth) = 2(x + x + 12) = 2(2x + 12) = (4x + 24) metre

Area = Length $\times$ Breadth = x(x + 12) = (x² + 12x) metre²

Given, area is equal to 4 times the perimeter

∴ x² + 12x = 4(4x + 24)

$\Rightarrow x^2 + 12x = 16x + 96 \\[1em] \Rightarrow x^2 + 12x - 16x - 96 = 0 \\[1em] \Rightarrow x^2 - 4x - 96 = 0 \\[1em] \Rightarrow x^2 - 12x + 8x - 96 = 0 \\[1em] \Rightarrow x(x - 12) + 8(x - 12) = 0 \\[1em] \Rightarrow (x - 12)(x + 8) = 0 \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 8 = 0 \\[1em] x = 12 \text{ or } x = -8 \\[1em]$

Since, breadth cannot be negative hence, x ≠ -8

∴ x = 12 , x + 12 = 24

Hence, the length of the garden is 24m and breadth is 12m.

A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30m barbed wire , he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side fence. Find the dimensions of his garden.

Answer

Let x metres be the length of the side opposite to unfenced side, then length of each of two others sides = $\dfrac{1}{2}(30 -x).$

According to question,

$\Rightarrow x \times \dfrac{1}{2}(30 - x) = 100 \\[1em] \Rightarrow 15x - \dfrac{x^2}{2} = 100 \\[1em] \Rightarrow \dfrac{30x - x^2}{2} = 100 \text{ (On taking L.C.M.) }\\[1em] \Rightarrow 30x - x^2 = 200 \text { (On cross multiplying) } \\[1em] \Rightarrow x^2 - 30x + 200 = 0 \\[1em] \Rightarrow x^2 - 20x - 10x + 200 = 0 \\[1em] \Rightarrow x(x - 20) - 10(x - 20) = 0 \\[1em] \Rightarrow (x - 10)(x - 20) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x - 20 = 0 \\[1em] x = 10 \text{ or } x = 20$

∴ if x = 10, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 10) \\[1em] = \dfrac{1}{2} \times 20 \\[1em] = 10$

∴ if x = 20, $\dfrac{1}{2}(30 - x) :$

$= \dfrac{1}{2}(30 - 20) \\[1em] = \dfrac{1}{2} \times 10 \\[1em] = 5$

Hence, the dimensions of garden are 10m x 10m or 20m x 5m.

The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Answer

Let the shortest side be x metre

According to question,

Third side = (x + 1)m

Hypotenuse = (2x - 1)m

For right angled triangle,

(Hypotenuse)² = (Perpendicular)² + (Base)²

$\therefore (2x - 1)^2 = (x + 1)^2 + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = x^2 + 1 + 2x + x^2 \\[1em] \Rightarrow 4x^2 + 1 - 4x = 2x^2 + 2x + 1 \\[1em] \Rightarrow 4x^2 - 2x^2 - 4x - 2x + 1 - 1 = 0 \\[1em] \Rightarrow 2x^2 - 6x = 0 \\[1em] \Rightarrow 2x(x - 3) = 0 \\[1em] \Rightarrow 2x = 0 \text{ or } x - 3 = 0 \\[1em] x = 0 \text{ or } x = 3$

Since, side's length cannot be equal to 0 , hence x ≠ 0.

∴ x = 3, (x + 1) = 4, (2x - 1) = 5

Hence, the sides of triangle are
Shortest side = 3 m
Hypotenuse = 5 m
Third side = 4 m

A wire, 112 cm long, is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Answer

Sum of length of other two sides + Hypotenuse = 112

or, Sum of length of other two sides = 112 - Hypotenuse

∴ Sum of length of other two sides = 112 - 50 = 62 cm

Let the length of perpendicular = x cm

So, the length of base = (62 - x) cm

For right angled triangle,

(Perpendicular)² + (Base)² = (Hypotenuse)²

$\therefore x^2 + (62 - x)^2 = (50)^2 \\[1em] \Rightarrow x^2 + 3844 + x^2 - 124x = 2500 \\[1em] \Rightarrow 2x^2 - 124x + 3844 - 2500 = 0 \\[1em] \Rightarrow 2x^2 - 124x + 1344 = 0 \\[1em] \Rightarrow 2(x^2 - 62x + 672) = 0 \\[1em] \Rightarrow x^2 - 62x + 672 = 0 \\[1em] \Rightarrow x^2 - 48x - 14x + 672 = 0 \\[1em] \Rightarrow x(x - 48) - 14(x - 48) = 0 \\[1em] \Rightarrow (x - 48)(x - 14) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x - 14 = 0 \\[1em] x = 48 \text{ or } x = 14$

∴ The length of other two sides are 48 cm and 14 cm.

$\text{Area} = \dfrac{1}{2} \times \text{Perpendicular} \times \text{Base} \\[1em] = \dfrac{1}{2} \times 48 \times 14 \\[1em] = 336$

Hence, the area of triangle is 336 cm^2.

The speed of a boat in still water is 11 km/h. It can go 12 km upstream and return downstream to original point in 2 hours 45 minutes. Find the speed of the stream.

Answer

Let the speed of the stream be x km/h.

Speed of boat in -

Still water = 11 km/h

Upstream = (11 - x) km/h

Downstream = (11 + x) km/h

Given,

Boat can go 12 km upstream and return downstream to original point in 2 hours 45 minutes.

Converting 2 hours 45 minutes to hours:

2 hours 45 minutes = ((2 x 60) + 45) minutes = 165 minutes.

165 minutes = $\dfrac{165}{60}$ hours

$\therefore \dfrac{12}{11 - x} + \dfrac{12}{11 + x} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{12(11 + x) + 12(11 - x)}{(11 + x)(11 - x)} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{132 + 12x + 132 - 12x}{121 - x^2} = \dfrac{165}{60} \\[1em] \Rightarrow \dfrac{264}{121 - x^2} = \dfrac{11}{4}\\[1em] \Rightarrow 264 \times 4 = 11(121 - x^2) \\[1em] \Rightarrow 1056 = 1331 - 11x^2 \\[1em] \Rightarrow 11x^2 = 1331 - 1056 \\[1em] \Rightarrow 11x^2 = 275 \\[1em] \Rightarrow x^2 = \dfrac{275}{11} \\[1em] \Rightarrow x^2 = 25 \\[1em] \Rightarrow x^2 - 25 = 0 \\[1em] \Rightarrow (x - 5)(x + 5) = 0 \\[1em] \Rightarrow x = 5, -5$

Since speed cannot be negative hence, x ≠ -5.

Hence, speed of stream is 5 km/h.

A man spent ₹2800 on buying a number of plants priced at ₹x each. Because of the number involved, the supplier reduced the price of each plant by one rupee. The man finally paid ₹2730 and received 10 more plants . Find x.

Answer

In first case,

Amount spent = ₹2800

Price of each plant = ₹x

No. of plants = $\dfrac{2800}{x}$

In second case,

Amount spent = ₹2730

Price of plant is reduced by ₹1 , so new price = ₹(x - 1)

Given, in this case 10 more plants were received hence, no. of plants = $\dfrac{2800}{x} + 10$

According to question,

$\Rightarrow \big(\dfrac{2800}{x} + 10\big)(x - 1) = 2730 \\[1em] \Rightarrow \big(\dfrac{2800 + 10x}{x}\big)(x - 1) = 2730 \\[1em] \Rightarrow \dfrac{(2800 + 10x)(x - 1)}{x} = 2730 \\[1em] \Rightarrow \dfrac{2800x - 2800 + 10x^2 - 10x}{x} = 2730 \\[1em] \Rightarrow 10x^2 + 2790x - 2800 = 2730x \text{ (On cross multiplication) } \\[1em] \Rightarrow 10x^2 + 60x - 2800 = 0 \\[1em] \Rightarrow x^2 + 6x - 280 = 0 \text{ (On dividing by 10) } \\[1em] \Rightarrow x^2 + 20x - 14x - 280 = 0 \\[1em] \Rightarrow x(x + 20) - 14(x + 20) = 0 \\[1em] \Rightarrow (x - 14)(x + 20) = 0 \\[1em] x = 14 \text{ or } x = -20$

Since price cannot be negative hence , x ≠ -20

Hence, the value of x is 14.

Forty years hence, Mr. Pratap's age will be the square of what it was 32 years ago. Find his present age.

Answer

Let the present age of Mr. Pratap be x years.

After 40 years his age will be (x + 40) years

32 years before his age was (x - 32) years

According to question ,

$\Rightarrow (x - 32)^2 = (x + 40) \\[1em] \Rightarrow x^2 + 1024 - 64x = x + 40 \\[1em] \Rightarrow x^2 - 64x - x + 1024 - 40 = 0 \\[1em] \Rightarrow x^2 - 65x + 984 = 0 \\[1em] \Rightarrow x^2 - 24x - 41x + 984 = 0 \\[1em] \Rightarrow x(x - 24) - 41(x - 24) = 0 \\[1em] \Rightarrow (x - 41)(x - 24) = 0 \\[1em] \Rightarrow x - 41 = 0 \text{ or } x - 24 = 0 \\[1em] x = 41 \text{ or } x = 24$

But x ≠ 24 as it is less than 32.

∴ x = 41.

The present age of Mr. Pratap is 41 years.

The total expenses of a trip for certain number of people is ₹ 18,000. If three more people join them, then the share of each reduces by ₹ 3,000. Take x to be the original number of people, form a quadratic equation in x and solve it to find the value of x.

Answer

Let no. of people be x.

Total expense = ₹ 18,000

Expense per person = ₹ $\dfrac{18000}{x}$

Given,

If three more people join them, then the share of each reduces by ₹ 3,000.

No, of people now = x + 3

Expense per person = ₹ $\dfrac{18000}{x + 3}$

According to question,

$\Rightarrow \dfrac{18000}{x} - \dfrac{18000}{x + 3} = 3000 \\[1em] \Rightarrow \dfrac{18000(x + 3) - 18000x}{x(x + 3)} = 3000 \\[1em] \Rightarrow \dfrac{18000x + 54000 - 18000x}{x^2 + 3x} = 3000 \\[1em] \Rightarrow \dfrac{54000}{x^2 + 3x} = 3000 \\[1em] \Rightarrow x^2 + 3x = \dfrac{54000}{3000} \\[1em] \Rightarrow x^2 + 3x = 18 \\[1em] \Rightarrow x^2 + 3x - 18 = 0 \\[1em] \Rightarrow x^2 + 6x - 3x - 18 = 0 \\[1em] \Rightarrow x(x + 6) - 3(x + 6) = 0 \\[1em] \Rightarrow (x - 3)(x + 6) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 6 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -6.$

Since, no. of people cannot be negative.

∴ x = 3.

Hence, original number of people = 3.

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed. If it takes 3 hours to complete the total journey then form a quadratic equation and solve it to find its original average speed.

Answer

Given,

A car travels a distance of 72 km at a certain average speed of x km per hour and then travels a distance of 81 km at an average speed of 6 km per hour more than its original average speed.

Total time taken to complete the journey = 3 hours

$\therefore \dfrac{72}{x} + \dfrac{81}{x + 6} = 3 \\[1em] \Rightarrow \dfrac{72(x + 6) + 81x}{x(x + 6)} = 3 \\[1em] \Rightarrow 72(x + 6) + 81x = 3x(x + 6) \\[1em] \Rightarrow 72x + 432 + 81x = 3x^2 + 18x \\[1em] \Rightarrow 3x^2 + 18x - 72x - 81x - 432 = 0 \\[1em] \Rightarrow 3x^2 - 135x - 432 = 0 \\[1em] \Rightarrow 3(x^2 - 45x - 144) = 0 \\[1em] \Rightarrow x^2 - 45x - 144 = 0 \\[1em] \Rightarrow x^2 - 48x + 3x - 144 = 0 \\[1em] \Rightarrow x(x - 48) + 3(x - 48) = 0 \\[1em] \Rightarrow (x + 3)(x - 48) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 48 = 0 \\[1em] \Rightarrow x = -3 \text{ or } x = 48.$