Quadratic Equations — Exercise 5(E)

If x⁴ - 5x² + 4 = 0; the values of x are :

1. 1 or 2

2. ± 1 or ± 2

3. -1 and 2

4. -1 and -2

Answer

Given,

⇒ x⁴ - 5x² + 4 = 0

⇒ x⁴ - 4x² - x² + 4 = 0

⇒ x²(x² - 4) - 1(x² - 4) = 0

⇒ (x² - 1)(x² - 4) = 0

⇒ x² - 1 = 0 or x² - 4 = 0

⇒ x² = 1 or x² = 4

⇒ x = $\sqrt{1}$ or x = $\sqrt{4}$

⇒ x = ± 1 or x = ± 2.

Hence, Option 2 is the correct option.

For equation $\dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10}$; one value of x is :

1. $-\dfrac{5}{3}$

2. 10

3. -10

4. 5

Answer

Given,

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x - 5} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{x - 5 + x}{x(x - 5)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{2x - 5}{x^2 - 5x} = \dfrac{3}{10} \\[1em] \Rightarrow 10(2x - 5) = 3(x^2 - 5x) \\[1em] \Rightarrow 20x - 50 = 3x^2 - 15x \\[1em] \Rightarrow 3x^2 - 15x - 20x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 35x + 50 = 0 \\[1em] \Rightarrow 3x^2 - 30x - 5x + 50 = 0 \\[1em] \Rightarrow 3x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (3x - 5)(x - 10) = 0 \\[1em] \Rightarrow 3x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow 3x = 5 \text{ or } x = 10 \\[1em] \Rightarrow x = \dfrac{5}{3} \text{ or } x = 10.$

Hence, Option 2 is the correct option.

Which of the following is correct for the equation $\dfrac{1}{x - 3} - \dfrac{1}{x + 5}$ = 1 ?

1. x ≠ 3 and x = -5

2. x = 3 and x ≠ -5

3. x ≠ 3 and x ≠ -5

4. x > 3 and x < 5

Answer

Given,

$\dfrac{1}{x - 3} - \dfrac{1}{x + 5}$ = 1

So, in above equation.

x - 3 and x + 5 cannot be equal to zero.

⇒ x - 3 ≠ 0

⇒ x ≠ 3

⇒ x + 5 ≠ 0

⇒ x ≠ -5.

Hence, Option 3 is the correct option.

Solve :

2x⁴ - 5x² + 3 = 0

Answer

Let x² = y,

⇒ 2x⁴ - 5x² + 3 = 0

⇒ 2y² - 5y + 3 = 0

⇒ 2y² - 2y - 3y + 3 = 0

⇒ 2y(y - 1) - 3(y - 1) = 0

⇒ (2y - 3)(y - 1) = 0

⇒ 2y - 3 = 0 or y - 1 = 0      [Zero product rule]

⇒ 2y = 3 or y = 1

⇒ y = $\dfrac{3}{2}$ or y = 1.

∴ x² = $\dfrac{3}{2}$ or x² = 1

⇒ x = $\pm \sqrt{\dfrac{3}{2}} = \pm 1.22$ or x = $\sqrt{1} = \pm 1$

Hence, x = +1.22, -1.22, +1, -1.

Solve :

x⁴ - 2x² - 3 = 0

Answer

Let x² = y,

⇒ x⁴ - 2x² - 3 = 0

⇒ y² - 2y - 3 = 0

⇒ y² - 3y + y - 3 = 0

⇒ y(y - 3) + 1(y - 3) = 0

⇒ (y + 1)(y - 3) = 0

⇒ y + 1 = 0 or y - 3 = 0      [Zero product rule]

⇒ y = -1 or y = 3

∴ x² = -1 or x² = 3

Since, square of a number cannot be negative,

∴ x² = 3

⇒ x = $\sqrt{3} = \pm 1.73$

Hence, x = +1.73, -1.73.

Solve :

(x² - x)² + 5(x² - x) + 4 = 0

Answer

Let x² - x = a

Substituting value in (x² - x)² + 5(x² - x) + 4 = 0 we get,

⇒ a² + 5a + 4 = 0

⇒ a² + 4a + a + 4 = 0

⇒ a(a + 4) + 1(a + 4) = 0

⇒ (a + 1)(a + 4) = 0

⇒ a + 1 = 0 or a + 4 = 0      [Zero product rule]

⇒ a = -1 or a = -4

∴ x² - x = -1 and x² - x = -4

Solving, x² - x = -1

⇒ x² - x = -1

⇒ x² - x + 1 = 0

Comparing above equation with ax² + bx + x = 0 we get,

a = 1, b = -1, c = 1

Discriminant = D = b² - 4ac = (-1)² - 4(1)(1) = 1 - 4 = -3.

-3 < 0

∴ No real solution.

Solving, x² - x = -4

⇒ x² - x + 4 = 0

Comparing above equation with ax² + bx + x = 0 we get,

a = 1, b = -1, c = 4

Discriminant = D = b² - 4ac = (-1)² - 4(1)(4) = 1 - 16 = -15.

-15 < 0

∴ No real solution.

Hence, there is no real solution.

Solve :

(x² - 3x)² - 16(x² - 3x) - 36 = 0

Answer

Let x² - 3x = a

Substituting value in (x² - 3x)² - 16(x² - 3x) - 36 = 0 we get,

⇒ a² - 16a - 36 = 0

⇒ a² - 18a + 2a - 36 = 0

⇒ a(a - 18) + 2(a - 18) = 0

⇒ (a + 2)(a - 18) = 0

⇒ a + 2 = 0 or a - 18 = 0      [Zero product rule]

⇒ a = -2 or a = 18

∴ x² - 3x = -2 and x² - 3x = 18

Solving, x² - 3x = -2

⇒ x² - 3x = -2

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - 1(x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x - 1 = 0 or x - 2 = 0      [Zero product rule]

⇒ x = 1 or x = 2.

Solving, x² - 3x = 18

⇒ x² - 3x = 18

⇒ x² - 3x - 18 = 0

⇒ x² - 6x + 3x - 18 = 0

⇒ x(x - 6) + 3(x - 6) = 0

⇒ (x + 3)(x - 6) = 0

⇒ x + 3 = 0 or x - 6 = 0      [Zero product rule]

⇒ x = -3 or x = 6.

Hence, x = 1, 2, -3, 6.

Solve :

$\sqrt{\dfrac{x}{x - 3}} + \sqrt{\dfrac{x - 3}{x}} = \dfrac{5}{2}$

Answer

Let $\sqrt{\dfrac{x}{x - 3}} =$ a .......(i)

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

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = \dfrac{1}{2}$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = \dfrac{1}{4} \\[1em] \Rightarrow 4x = x - 3 \\[1em] \Rightarrow 4x - x = -3 \\[1em] \Rightarrow 3x = -3 \\[1em] \Rightarrow x = -1.$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \sqrt{\dfrac{x}{x - 3}} = 2$

Squaring both sides we get,

$\Rightarrow \dfrac{x}{x - 3} = 4 \\[1em] \Rightarrow x = 4(x - 3) \\[1em] \Rightarrow x = 4x - 12 \\[1em] \Rightarrow 4x - x = 12 \\[1em] \Rightarrow 3x = 12 \\[1em] \Rightarrow x = 4.$

Hence, x = -1, 4.

Solve :

$\Big(\dfrac{2x - 3}{x - 1}\Big) - 4\Big(\dfrac{x - 1}{2x -3}\Big) = 3$

Answer

Let $\dfrac{2x - 3}{x - 1} =$ a .......(i)

$\Rightarrow a - \dfrac{4}{a} = 3 \\[1em] \Rightarrow \dfrac{a^2 - 4}{a} = 3 \\[1em] \Rightarrow a^2 - 4 = 3a \\[1em] \Rightarrow a^2 - 3a - 4 = 0 \\[1em] \Rightarrow a^2 - 4a + a - 4 = 0 \\[1em] \Rightarrow a(a - 4) + 1(a - 4) = 0 \\[1em] \Rightarrow (a + 1)(a - 4) = 0 \\[1em] \Rightarrow a = -1 \text{ or } a = 4.$

Substituting value of a = -1 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = -1 \\[1em] \Rightarrow 2x - 3 = -1(x - 1) \\[1em] \Rightarrow 2x - 3 = -x + 1 \\[1em] \Rightarrow 2x + x = 1 + 3 \\[1em] \Rightarrow 3x = 4 \\[1em] \Rightarrow x = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}$

Substituting value of a = 4 in (i) we get,

$\Rightarrow \dfrac{2x - 3}{x - 1} = 4 \\[1em] \Rightarrow 2x - 3 = 4(x - 1) \\[1em] \Rightarrow 2x - 3 = 4x - 4 \\[1em] \Rightarrow 4x - 2x = -3 + 4 \\[1em] \Rightarrow 2x = 1 \\[1em] \Rightarrow x = \dfrac{1}{2}.$

Hence, x = $1\dfrac{1}{3}, \dfrac{1}{2}$.

Solve :

$\Big(\dfrac{3x + 1}{x + 1}\Big) + \Big(\dfrac{x + 1}{3x + 1}\Big) = \dfrac{5}{2}$

Answer

Let $\Big(\dfrac{3x + 1}{x + 1}\Big) = a$ .......(i)

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

Substituting value of a = $\dfrac{1}{2}$ in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = \dfrac{1}{2} \\[1em] \Rightarrow 2(3x + 1) = x + 1 \\[1em] \Rightarrow 6x + 2 = x + 1 \\[1em] \Rightarrow 6x - x = 1 - 2 \\[1em] \Rightarrow 5x = -1 \\[1em] \Rightarrow x = -\dfrac{1}{5}$

Substituting value of a = 2 in (i) we get,

$\Rightarrow \Big(\dfrac{3x + 1}{x + 1}\Big) = 2 \\[1em] \Rightarrow 3x + 1 = 2(x + 1) \\[1em] \Rightarrow 3x + 1 = 2x + 2 \\[1em] \Rightarrow 3x - 2x = 2 - 1 \\[1em] \Rightarrow x = 1.$

Hence, x = 1, $-\dfrac{1}{5}$.

Solve :

$9(x^2 + \dfrac{1}{x^2}) - 9(x + \dfrac{1}{x}) - 52 = 0$

Answer

Let $x + \dfrac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} + 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 - 2 .......(ii)$

Substituting the values from equations (i) and (ii) we get,

⇒ 9(a² - 2) - 9a - 52 = 0

⇒ 9a² - 18 - 9a - 52 = 0

⇒ 9a² - 9a - 70 = 0

⇒ 9a² - 30a + 21a - 70 = 0

⇒ 3a(3a - 10) + 7(3a - 10) = 0

⇒ (3a + 7)(3a - 10) = 0

⇒ (3a + 7) = 0 or 3a - 10 = 0      [Zero product rule]

⇒ 3a = -7 or 3a = 10

⇒ $a = -\dfrac{7}{3} \text{ or } a = \dfrac{10}{3}$

Considering a = $\dfrac{10}{3}$ we get,

$\Rightarrow x + \dfrac{1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = \dfrac{10}{3} \\[1em] \Rightarrow 3(x^2 + 1) = 10x \\[1em] \Rightarrow 3x^2 + 3 = 10x \\[1em] \Rightarrow 3x^2 - 10x + 3 = 0 \\[1em] \Rightarrow 3x^2 - 9x - x + 3 = 0 \\[1em] \Rightarrow 3x(x - 3) - 1(x - 3) = 0 \\[1em] \Rightarrow (3x - 1)(x - 3) = 0 \\[1em] \Rightarrow 3x - 1 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow 3x = 1 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{1}{3} \text{ or } x = 3.$

Considering a = $-\dfrac{7}{3}$ we get,

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

Comparing 3x² + 7x + 3 = 0 with ax² + bx + c = 0 we get,

a = 3, b = 7 and c = 3.

We know that,

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

$\Rightarrow x = \dfrac{-7 \pm \sqrt{(7)^2 - 4(3)(3)}}{2(3)} \\[1em] = \dfrac{-7 \pm \sqrt{49 - 36}}{6} \\[1em] = \dfrac{-7 \pm \sqrt{13}}{6}.$

Hence, x = 3, $\dfrac{1}{3}, \dfrac{-7 \pm \sqrt{13}}{6}.$

Solve :

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

Answer

Let $x - \dfrac{1}{x} = a$ ........(i)

Squaring, both sides we get,

$\Rightarrow x^2 + \dfrac{1}{x^2} - 2 = a^2 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = a^2 + 2 .......(ii)$

Substituting the values from equations (i) and (ii) we get,

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

⇒ a² + 2 - 3a - 2 = 0

⇒ a² - 3a = 0

⇒ a(a - 3) = 0

⇒ a = 0 or a - 3 = 0

⇒ a = 0 or a = 3.

Considering a = 0 we get,

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

Considering a = 3 we get,

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

Comparing x² - 3x - 1 = 0 with ax² + bx + c = 0 we get,

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

We know that,

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

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 4}}{2} \\[1em] = \dfrac{3 \pm \sqrt{13}}{2}.$

Hence, x = 1, -1, $\dfrac{3 \pm \sqrt{13}}{2}$.

Solve :

(x² + 5x + 4)(x² + 5x + 6) = 120

Answer

Let x² + 5x = a we get,

⇒ (a + 4)(a + 6) = 120

⇒ a² + 6a + 4a + 24 = 120

⇒ a² + 10a + 24 - 120 = 0

⇒ a² + 10a - 96 = 0

⇒ a² + 16a - 6a - 96 = 0

⇒ a(a + 16) - 6(a + 16) = 0

⇒ (a - 6)(a + 16) = 0

⇒ (a - 6) = 0 or (a + 16) = 0

⇒ a = 6 or a = -16.

Considering a = 6 we get,

⇒ x² + 5x = 6

⇒ x² + 5x - 6 = 0

⇒ x² + 6x - x - 6 = 0

⇒ x(x + 6) - 1(x + 6) = 0

⇒ (x - 1)(x + 6) = 0

⇒ (x - 1) = 0 or (x + 6) = 0

⇒ x = 1 or x = -6.

Considering a = -16 we get,

⇒ x² + 5x = -16

⇒ x² + 5x + 16 = 0

In this case,

D = b² - 4ac = 5² - 4(1)(16) = 25 - 64 = -39 < 0.

It means roots are imaginary in this case.

Hence, x = 1, -6.

Solve : $3\Big(\dfrac{3x - 1}{2x + 3}\Big) - 2\Big(\dfrac{2x + 3}{3x - 1}\Big) = 5$

Answer

Solving,

$\Rightarrow 3\Big(\dfrac{3x - 1}{2x + 3}\Big) - 2\Big(\dfrac{2x + 3}{3x - 1}\Big) = 5 \\[1em] \Rightarrow \dfrac{3(3x - 1)^2 - 2(2x + 3)^2}{(2x + 3)(3x - 1)} = 5 \\[1em] \Rightarrow 3(3x - 1)^2 - 2(2x + 3)^2 = 5(2x + 3)(3x - 1) \\[1em] \Rightarrow 3(9x^2 - 6x + 1) - 2(4x^2 + 12x + 9) = 5(6x^2 - 2x + 9x - 3) \\[1em] \Rightarrow 27x^2 - 18x + 3 - 8x^2 - 24x - 18 = 5(6x^2 + 7x - 3) \\[1em] \Rightarrow 19x^2 - 42x - 15 = 30x^2 + 35x - 15 \\[1em] \Rightarrow 30x^2 - 19x^2 + 35x + 42x - 15 + 15 = 0 \\[1em] \Rightarrow 11x^2 + 77x = 0 \\[1em] \Rightarrow 11x(x + 7) = 0 \\[1em] \Rightarrow 11x = 0 \text{ or } x + 7 = 0 \\[1em] \Rightarrow x = 0 \text{ or } x = -7.$

Hence, x = 0 \text{ or } x = -7.

Solve: 5^x+1 + 5^2-x = 5³ + 1

Answer

Given,

5^x+1 + 5^2-x = 5³ + 1

⇒ 5^x.5¹ + 5².5^-x = 125 + 1

⇒ 5^x.5¹ + $\dfrac{5^2}{5^x}$ = 126

Let 5^x be y.

⇒ 5y + $\dfrac{25}{\text{y}}$ = 126

⇒ 5y² + 25 = 126y

⇒ 5y² - 126y + 25 = 0

⇒ 5y² - 125y - y + 25 = 0

⇒ 5y(y - 25) - (y - 25) = 0

⇒ (y - 25)(5y - 1) = 0

⇒ (y - 25) = 0 or (5y - 1) = 0

⇒ y = 25 or y = $\dfrac{1}{5}$

⇒ y = 5² or y = 5^-1

Substituting the value of y,

⇒ 5^x = 5² or 5^x = 5^-1

⇒ x = 2 or -1

Hence, the value of x = 2 or -1.