SChand CLASS 10 Chapter 5 Quadratic Equations Exercise 5B

Exercise 5B

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Q1 | Ex-5B | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 1

$x^4 + 5x^2 – 36 = 0$

Sol :

Given that $x^4 + 5x^2 – 36 = 0$

Put $y = x^2$.

⇒ $y^2 + 5y – 36 = 0$

⇒ $y^2 + 9y – 4y – 36 = 0$

⇒ y(y + 9) – 4(y + 9) = 0

⇒ (y + 9)(y – 4) = 0

⇒ (y + 9) = 0 or (y – 4) = 0

⇒ y = -9 or 4

If y = -9, then $x^2 = -9$ ⇒ x is NOT real.

If y = 4, then $x^2 = 4$ ⇒ x = ±2.

Thus, x = 2 or -2

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Q2 | Ex-5B | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 2

$x^4 – 25x^2 + 144 = 0$

Sol :

Given that $x^4 – 25x^2 + 144 = 0$.

Put $y = x^2$.

⇒ $y^2 – 25y + 144 = 0$

⇒ $y^2 – 16y – 9y + 144 = 0$

⇒ y(y – 16) – 9(y – 16) = 0

⇒ (y – 9)(y – 16) = 0

⇒ (y – 9) = 0 or (y – 16) = 0

⇒ y = 9 or 16

If y = 9, then $x^2 = 9$⇒ x = ±3.

If y = 16, then $x^2 = 16$ ⇒ x = ±4.

Thus, x = -4, -3, 3, 4.

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Q3 | Ex-5B | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 3

$(x^2 + x)^2 – (x^2 + x) – 2 = 0$

Sol :

Given that $(x^2 + x)^2 – (x^2 + x) – 2 = 0$

Put $y = (x^2 + x)$.

⇒ $y^2 – y – 2 = 0$

⇒ $y^2 – 2y + y – 2 = 0$

⇒ y(y – 2) + 1(y – 2) = 0

⇒ (y + 1)(y – 2) = 0

⇒ (y + 1) = 0 or (y – 2) = 0

⇒ y = -1 or 2

If y = -1, then $x^2 + x = -1$

⇒ $x^2 + x + 1 = 0$

Since its discriminant D = 1 – 4 = -3 < 0, there is NO real values of x.

If y = 2, then

⇒ $x^2 + x = 2$

⇒ $x^2 + x – 2 = 0$

⇒ $x^2 + 2x – x – 2 = 0$

⇒ x(x + 2) – 1(x + 2) = 0

⇒ (x – 1)(x + 2) = 0

⇒ (x – 1) = 0 or (x + 2) = 0

⇒ x = 1 or -2

Thus, x = 1 or -2

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Question 4

$\left[\frac{(x – 2)}{(x + 2)}\right]^2 – 4\left[\frac{(x – 2)}{(x + 2)}\right] + 3 = 0$, x ≠ 2

Sol :

Given that $\left[\frac{(x – 2)}{(x + 2)}\right]^2 – 4\left[\frac{(x – 2)}{(x + 2)}\right] + 3 = 0$.

Put $y = \left[\frac{(x – 2)}{(x + 2)}\right]$.

⇒ $y^2 – 4y + 3 = 0$

⇒ $y^2 – 3y – y + 3 = 0$

⇒ y(y – 3) – 1(y – 3) = 0

⇒ (y – 1)(y – 3) = 0

⇒ (y – 1) = 0 or (y – 3) = 0

⇒ y = 1 or 3

If y = 1, then $\left[\frac{(x – 2)}{(x + 2)}\right] = 1$

⇒ x – 2 = x + 2

⇒ – 2 = 2

⇒ There is no value of x.

If y = 3, then $\left[\frac{(x – 2)}{(x + 2)}\right] = 3$.

⇒ x – 2 = 3(x + 2)

⇒ x – 2 = 3x + 6

⇒ x – 3x = 2 + 6

⇒ -2x = 8

⇒ x = -4

Thus, x = -4

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Q5 | Ex-5B | Class 8 | S.Chand | Mathematics | Chapter 5 | Quadratic Equations | myhelper

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Question 5

$4\left[\frac{(7x – 1)}{x}\right]^2 – 8\left[\frac{(7x – 1)}{x}\right] + 3 = 0$

Sol :

Given that $4\left[\frac{(7x – 1)}{x}\right]^2 – 8\left[\frac{(7x – 1)}{x}\right] + 3 = 0$.

Put $y = \left[\frac{(7x – 1)}{x}\right]$.

⇒ $4y^2 – 8y + 3 = 0$

⇒ $4y^2 – 2y – 6y + 3 = 0$

⇒ 2y(2y – 1) – 3(2y – 1) = 0

⇒ (2y – 3)(2y – 1) = 0

⇒ (2y – 3) = 0 or (2y – 1) = 0

⇒ $y = \frac{3}{2}$ or $\frac{1}{2}$

If $y =\frac{3}{2}$, then $\frac{(7x – 1)}{x}= \frac{3}{2}$

⇒ 2(7x – 1) = 3x

⇒ 14x – 2 = 3x

⇒ 14x – 3x = 2

⇒ 11x = 2

⇒ $x = \frac{2}{11}$

If $y =\frac{1}{2}$, then $\frac{(7x – 1)}{x} = \frac{1}{2}$

⇒ 2(7x – 1) = x

⇒ 14x – 2 = x

⇒ 14x – x = 2

⇒ $x = \frac{2}{13}$

Thus, $x = \frac{2}{11}$ or $\frac{2}{13}$

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Question 6

$4^x – 5.2^x + 4 = 0$

Sol :

Given that $4^x – 5.2^x + 4 = 0$

$⇒ (2^x)^2 – 5.2^x + 4 = 0$

Put $y = 2^x$

⇒ $y^2 = 4^x$

⇒ $y^2 – 5y + 4 = 0$

⇒ $y^2 – 4y – y + 4 = 0$

⇒ y(y – 4) – 1(y – 4) = 0

⇒ (y – 1)(y – 4) = 0

⇒ (y – 1) = 0 or (y – 4) = 0

⇒ y = 1 or 4

If y = 1, then $2^x = 1$ ⇒ $2^x = 2^0$⇒ x = 0.

If y = 4, then $2^x = 4$ ⇒$2^x = 2^2$⇒ x = 2.

Thus, x = 0 or 2.

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Question 7

$16.4^x+2 – 16.2^x+1 + 1 = 0$

Sol :

Given that $16.4^x+2 – 16.2^x+1 + 1 = 0$

⇒ $16.4^x.4^2 – 16.2^x.2 + 1 = 0$

⇒ $256(2^x)^2 – 32(2^x) + 1 = 0$

Put $y = 2^x$

⇒ 256y^2 – 32y + 1 = 0$⇒$(16y – 1)^2 = 0$⇒ 16y – 1 = 0 ⇒$y = \frac{1}{16}$If$y = \frac{1}{16}$, then$2^x = 2^{-4}$ ⇒ x = -4.

Thus, x = -4.

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Question 8

$3^{4x+1}– 2.3^{2x+2} – 81 = 0$

Sol :

Given that $3^{4x+1}– 2.3^{2x+2} – 81 = 0$

⇒ $3^{4x}.3^1 – 2.3^{2x}.3^{2} – 81 = 0$

⇒ $3(3^{2x})^{2} – 18(3^{2x}) – 81 = 0$

⇒ (3^{2x})^2 – 6(3^{2x}) – 27 = 0

Put $y = 3^{2x}$

⇒ $y^2 – 6y – 27 = 0$

⇒ $y^2 – 9y + 3y – 27 = 0$

⇒ y(y – 9) + 3(y – 9) = 0

⇒ (y + 3)(y – 9) = 0

⇒ (y + 3) = 0 or (y – 9) = 0

⇒ y = -3 or 9

If y = -3, then $3^{2x} = -3$ ⇒ There is NO real values of x as $a^x > 0$.

If y = 9, then $3^{2x} = 3^2$ ⇒ 2x = 2

Thus, x = 1.

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Question 9

$\left[\frac{(2x – 3)}{(x – 1)}\right] – 4\left[\frac{(x – 1)}{(2x – 3)}\right] = 3$, x ≠ 1 and 3/2

Sol :

Given that $\left[\frac{(2x – 3)}{(x – 1)}\right] – 4\left[\frac{(x – 1)}{(2x – 3)}\right] = 3$.

Put $y =\frac{ (2x – 3)}{(x – 1)}$

⇒ $y – \frac{4}{y} = 3$

⇒ $\frac{(y^2 – 4)}{y} = 3$

⇒$(y^2 – 4) = 3y$

⇒ $y^2 – 3y – 4 = 0$

⇒ $y^2 – 4y + y – 4 = 0$

⇒ y(y – 4) + 1(y – 4) = 0

⇒ (y + 1)(y – 4) = 0

⇒ (y + 1) = 0 or (y – 4) = 0

⇒ y = -1 or 4

If y = -1, then $\frac{(2x – 3)}{(x – 1) }= -1$

⇒ 2x – 3 = -(x – 1)

⇒ 2x – 3 = -x + 1

⇒ 2x + x = 3 + 1

⇒ 3x = 4

⇒ $x = \frac{4}{3}$

If y = 4, then $\frac{(2x – 3)}{(x – 1) }= 4$

⇒ 2x – 3 = 4(x – 1)

⇒ 2x – 3 = 4x – 4

⇒ 2x – 4x = 3 – 4

⇒ -2x = -1

⇒ $x =\frac{1}{2}$

Thus, $x = \frac{1}{2}$ or $\frac{4}{3}$.

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Question 10

$\left(x + \frac{1}{x}\right)^2 = 4 + \left(\frac{3}{2}\right)\left(x – \farc{1}{x}\right)$

Sol :

Given that $\left(x + \frac{1}{x}\right)^2 = 4 + \left(\frac{3}{2}\right)\left(x –\frac{ 1}{x}\right)$

⇒ $\left(x – \frac{1}{x}\right)^2 + 4 = 4 + \left(\frac{3}{2}\right)\left(x – \frac{1}{x}\right)$ [since $(a + b)^2 = (a – b)^2 + 4ab$]

⇒ $\left(x – \frac{1}{x}\right)^2 = \left(\frac{3}{2}\right)\left(x – \frac{1}{x}\right)$

Put $y = \left(x –\frac{1}{x}\right)$

⇒ $y^2 = \frac{3y}{2}$

⇒ $y^2 – \frac{3y}{2}= 0$

⇒ $y\left(y – \frac{3}{2}\right) = 0$

⇒ y = 0 or $\left(y – \frac{3}{2}\right) = 0$

⇒ y = 0 or $\frac{3}{2}$

If y = 0, then $\left(x – \frac{1}{x}\right) = 0$

⇒ $x = \frac{1}{x}$ ⇒ x2 = 1 ⇒ x = ±1

If $y =\frac{ 3}{2}$, then $\left(x – \frac{1}{x}\right) = \frac{3}{2}$

⇒ $\frac{(x^2 – 1)}{x} = \frac{3}{2}$

⇒ $2(x^2 – 1) = 3x$

⇒ $2x^2 – 2 = 3x$

⇒ $2x^2 – 3x – 2 = 0$

⇒ $2x^2 – 4x + x – 2 = 0$

⇒ 2x(x – 2) + 1(x – 2) = 0

⇒ (2x + 1)(x – 2) = 0

⇒ (2x + 1) = 0 or (x – 2) = 0

⇒ $x = \frac{-1}{2}$ or 2

Thus, x = -1, 1, $\frac{-1}{2}$ or 2.

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Question 11

√(2x + 7) = x + 2

Sol :

Given that √(2x + 7) = x + 2. Then, 2x + 7 >= 0 and x + 2 >= 0.

Squaring both sides,

⇒ $(2x + 7) = (x + 2)^2$

⇒ $2x + 7 = x^2 + 4x + 4$

⇒ $x^2 + 2x – 3 = 0$

⇒ $x^2 + 3x – x – 3 = 0$

⇒ x(x + 3) – 1(x + 3) = 0

⇒ (x – 1)(x + 3) = 0

⇒ (x – 1) = 0 or (x + 3) = 0

⇒ x = 1 or -3

If x = -3, then x + 2 = -3 + 2 = – 1 < 0, so x ≠ -3.

Thus, x = 1

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Question 12

2√(2x + 1) – 2x = 1

Sol :

Given that 2√(2x + 1) – 2x = 1

⇒ 2√(2x + 1) = 1 + 2x. Then, 2x + 1 >= 0.

Squaring both sides,

⇒ $4(2x + 1) = (1 + 2x)^2$

⇒ $8x + 4 = 1 + 4x^2 + 4x$

⇒ $4x^2 – 4x – 3 = 0$

⇒ $4x^2 – 6x + 2x – 3 = 0$

⇒ 2x(x – 3) + 1(2x – 3) = 0

⇒ (2x + 1)(2x – 3) = 0

⇒ (2x + 1) = 0 or (2x – 3) = 0

⇒ $x = \frac{-1}{2}$ or $\frac{3}{2}$

Since at $x =\frac{-1}{2}$ or $\frac{3}{2}$, then value of 1 + 2x = 0, 4.

Thus, $x = \frac{-1}{2}$ or $\frac{3}{2}$

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Question 13

√(4x – 3) + √(2x + 3) = 6

Sol :

Given that √(4x – 3) + √(2x + 3) = 6

⇒ √(4x – 3) = 6 – √(2x + 3)

Squaring both sides,

⇒ $4x – 3 = [6 – √(2x + 3)]^2$

⇒ 4x – 3 = 36 + 2x + 3 – 12√(2x + 3)

⇒ 2x – 42 = -12√(2x + 3)

⇒ 6√(2x + 3) = 21 – x,

Then, 21 – x >= 0 ⇒ x =< 21.

Squaring both sides,

⇒ $36(2x + 3) = (21 – x)^2$

⇒ $72x + 108 = 441 + x^2 – 42x$

⇒ $x^2 – 114x + 333 = 0$

⇒ $x^2 – 111x – 3x + 333 = 0$

⇒ x(x – 111) – 3(x – 111) = 0

⇒ (x – 3)(x – 111) = 0

⇒ (x – 3) = 0 or (x – 111) = 0

⇒ x = 3 or 111

Since x <= 21, thus x = 3