OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Ex 5(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 5 Quadratic Equations Ex 5(b)

Solve the following equations by reducing them to quadratic equations :

Question 1.
x⁴ + 5x² – 36 = 0
Solution:
x⁴ + 5x² – 36 = 0
Let x² = y, then
(x²)² + 5(x²) – 36 = 0
⇒ y² + 5y – 36 = 0
⇒ y² + 9y – 4y – 36 = 0

⇒ y (y + 9) – 4 (y + 9) = 0
⇒ (y + 9)(y – 4) = 0
Either y + 9 = 0, then y = – 9
or y – 4 = 0, then y = 4

(i) If y = – 9, then x² = – 9 x = ±\(\sqrt{-9}\)
which is not real solution
or y = 4, then x² = 4
⇒ x² – 4 = 0
⇒ (x)² – (2)² = 0
⇒ (x + 2) (x – 2) = 0
Either x + 2 = 0, then x = – 2
or x – 2 = 0. then x = 2
∴ x = 2, – 2

Question 2.
x⁴ – 25x² + 144 = 0
Solution:
x⁴ – 25x² + 144 = 0
⇒ (x²)² – 25 (x)² + 144 = 0
Let x² = y, then
y² – 25y + 144 = 0
⇒ y² – 16y – 9y + 144 = 0

⇒ y (y – 16) – 9 (y – 16) = 0
⇒ (y – 16) (y – 9) = 0
Either y – 16 = 0, then y = 16
or y – 9 = 0, then y = 9
If y = 16, then x² = 16 ⇒ x² – 16 = 0
⇒ x² – (4)² = 0
⇒ (x + 4) (x – 4) = 0
Either x + 4 = 0, then x = – 4
or x – 4 = 0, then x = 4
If y = 9, then x² = 9 ⇒ x² – 9 = 0
⇒ x² – (3)² = 0
⇒ (x + 3) (x – 3) = 0
Either x + 3 = 0, then x = – 3
or x – 3 = 0, then x = 3
∴ x = 4, – 4, 3, – 3

Question 3.
(x² + x)² – (x² + x) – 2 = 0
Solution:
(x² + x)² – (x² + x) – 2 = 0
Let x² + x = y, then y² – y – 2 = 0
⇒ y² – 2y + y – 2 = 0

⇒ y(y – 2) + 1 (y – 2) = 0
⇒ (y – 2) (y + 1) = 0
Either y – 2 = 0, then y = 2
or y + 1 = 0, then y = – 1
If y = 2, then
x² + x = 2 ⇒ x² + x – 2 = 0
⇒ x² + 2x – x – 2 = 0
⇒ x (x + 2) – 1 (x + 2) = 0
⇒ (x + 2) (x – 1) = 0
Either x + 2 = 0, then x = – 2
or x – 1 = 0, then x = 1
If x = – 1, then
x² + x = – 1
⇒ x² + x + 1 = 0
But it has no real solution
∴ x = – 2, 1

Question 4.
\(\left[\frac{x-2}{x+2}\right]^2-4\left[\frac{x-2}{x+2}\right]\) + 3 = 0, x ≠ 2
Solution:
\(\left[\frac{x-2}{x+2}\right]^2-4\left[\frac{x-2}{x+2}\right]\) = 0
Let \(\frac{x-2}{x+2}\) = y, then
y² – 4y + 3 = 0
⇒ y² – y – 3y + 3 = 0

⇒ y(y – 1) – 3(y – 1) = 0
⇒ (y -1)(y – 3) = 0
Either y – 1 = 0, then y = 1
or y – 3 = 0, then y = 3
If y = 1, then
\(\frac{x-2}{x+2}\) = 1 ⇒ x – 2 = x + 2 = 0 = 4
Which is not possible
If y = 3, then
\(\frac{x-2}{x+2}\) = \(\frac { 3 }{ 1 }\) ⇒ 3x + 6 = x – 2
⇒ 3x – x = – 2 – 6 ⇒ 2x = – 8
⇒ x = – 4
∴ x = – 4

Question 5.
4\(\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right)\) + 3 = 0
Solution:
Let 4\(\left(\frac{7 x-1}{x}\right)^2-8\left(\frac{7 x-1}{x}\right)\) + 3 = 0, then
Let \(\frac { 7x -1 }{ x }\) = y, then
4y² – 8y + 3 = 0
⇒ 4y² – 6y – 2y + 3 = 0

⇒ 2y(2y – 3) – 1 (2y – 3) = 0
⇒ (2y – 3) (2y – 1) = 0
Either 2y – 3 = 0, then 2y = 3 ⇒ y = \(\frac { 3 }{ 2 }\)
or 2y – 1 = 0, then 2y = 1 ⇒ y = \(\frac { 1 }{ 2 }\)
If y = \(\frac { 3 }{ 2 }\) , then
\(\frac{7 x-1}{x}=\frac{3}{2}\)
⇒ 14x – 2 = 3x ⇒ 14x – 3x = 2
⇒ 11x = 2 ⇒ x = \(\frac { 2 }{ 11 }\)
y = \(\frac { 1 }{ 2 }\), then
\(\frac{7 x-1}{x}=\frac{1}{2}\) ⇒ 14x – 2 = x
⇒ 14x – x = 2 ⇒ 13x = 2
⇒ x = \(\frac { 2 }{ 13 }\)
∴ x = \(\frac { 2 }{ 11 }\), \(\frac { 2 }{ 13 }\)

Question 6.
4^x – 5.2^x + 4 = 0
Solution:
4^x – 5.2^x + 4 = 0
⇒ [(2)²]^x – 5.(2)^x + 4 = 0
⇒ (2^x)² – 5.2^x + 4 = 0
Let 2^x = y, then
y² – 5y + 4 = 0
⇒ y² – y – 4y + 4 = 0

⇒ y(y – 1) – 4(y – 1) = 0
⇒ (y – 1) (y – 4) = 0
Either y – 1 = 0, then y = 1
or y – 4 = 0, then y = 4
If y = 1, then
2^x = 1 ⇒ 2x = 2° (∵ 2° = 1)
Comparing,
∴ x = 0
If y = 4, then
2^x = 4 ⇒ 2^x = 2²
Comparing,
∴ x = 2
∴ x = 0, 2

Question 7.
16.4^x+2 – 16.2^x+1 + 1 = 0
Solution:
16.4^x+2 – 16.2^x+1 + 1 = 0
⇒ 16.(2²)^x+2 – 16(2)^x+1 + 1 = 0
⇒ 16.2^x+2– 16.2^x+1+ 1 = 0
⇒ 16.2^x.2⁴ – 16.2^x.2¹ + 1 = 0
⇒ 16 x 16.2^2x – 16 x 2.2^x + 1 = 0
⇒ 256.2^2x – 32.2^x + 1 = 0
Let 2^x = y, then
256.y² – 32y + 1 = 0
256y² – 16y – 16y + 1 = 0

16y(16y – 1) – 1 (16y – 1) = 0
⇒ (16y – 1)(16y – 1) = 0
Either 16y – 1 = 0, then 16y = 1 ⇒ y = \(\frac { 1 }{ 16 }\)
or 16y – 1 = 0, then 16y = 1 ⇒ y = \(\frac { 1 }{ 16 }\)
When y = \(\frac { 1 }{ 16 }\), then
2^x = \(\frac { 1 }{ 16 }\) 2^x = \(\frac{1}{(2)^4} \Rightarrow 2^x=2^{-4}\)
Comparing
x = – 4

Question 8.
3^4x+1 – 2 x 3^2x+1 – 81 = 0
Solution:
3^4x+1 – 2 x 3^2x+1 – 81 = 0
3^4x.3¹ – 2 x 3^2x x 3² – 81 = 0
⇒ (3^2x)² x 3 – 2 x 9 x 3^2x – 81 = 0
⇒ 3 (32x)² – 18 (3^2x) – 81 = 0
Let 3^2x = y, then
3y² – 18y -81 = 0 ⇒ y² – 6y – 27 = 0
⇒ y² – 9y + 3y – 27 = 0

⇒ y(y – 9) + 3(y – 9) = 0
⇒ (y – 9) (y + 3) = 0
Either y – 9 = 0, then y = 9
or y + 3 = 0, then y = – 3
If y = 9, then
3^2x = 9 = 3²
Comparing
2x – 2 ⇒ x = 1
If y = – 3, then
3^3x = – 3 which has no real values
∴ x = 1

Question 9.
\(\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)\) = 3 where x ≠ 1 and x ≠ \(\frac { 3 }{ 2}\)
Solution:
Let \(\frac{2 x-3}{x-1}\) = y, then \(\frac{x-1}{2 x-3}=\frac{1}{y}\)
∴ y – \(\frac { 4 }{ y }\) = 3
⇒ y² – 4 = 3y ⇒ y² – 3y – 4 = 0
⇒ y² – 4y + y – 4 = 0

⇒ y(y – 4) + 1 (y – 4) = 0
⇒ (y – 4)(y + 1) = 0
Either y – 4 = 0, then y = 4
or y + 1 = 0, then y = – 1
If y = 4, then
\(\frac { 2x-3 }{ x-1 }\) = 4 ⇒ 4x – 4 = 2x – 3
⇒ 4x – 2x = – 3 + 4 ⇒ 2x = 1
⇒ x = \(\frac { 1 }{ 2 }\)
If y = – 1, then
\(\frac { 2x-3 }{ x-1 }\) = – 1 ⇒ – x + 1 = 2x – 3
⇒ 2x + x = 1 + 3 ⇒ 3x = 4
⇒ x = \(\frac { 4 }{ 3 }\)
∴ x = \(\frac { 1 }{ 2 }\), \(\frac { 4 }{ 3 }\)

Question 10.
\(\left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right)\)
Solution:
Let x – \(\frac { 1 }{ x }\) = y, then
\(\left(x+\frac{1}{x}\right)^2\) = x² + \(\frac { 1 }{ x² }\) + 2 = x² + \(\frac { 1 }{ x² }\) – 2 + 4
= (x – \(\frac { 1 }{ x }\))² + 4 = y² + 4
∴ \(\left(x+\frac{1}{x}\right)^2=4+\frac{3}{2}\left(x-\frac{1}{x}\right)\)
⇒ y² + 4 = 4 + \(\frac { 3 }{ 2 }\)y
⇒ y² – \(\frac { 3 }{ 2 }\)y + 4 – 4 = 0 ⇒ y² – \(\frac { 3 }{ 2 }\)y = 0
⇒ y(y – \(\frac { 3 }{ 2 }\)) = 0
or y – \(\frac { 3 }{ 2 }\) = 0, then y = \(\frac { 3 }{ 2 }\)
∴ y = 0, \(\frac { 3 }{ 2 }\)
If y = 0, then
x – \(\frac { 1 }{ x }\) = 0
⇒ x² – 1 = 0 ⇒ (x + 1) (x – 1) = 0
Either x + 1 = 0, then x = – 1
or x – 1 = 0, then x = 1
If x = \(\frac { 3 }{ 2 }\), then
x – \(\frac { 1 }{ x }\) = \(\frac { 3 }{ 2 }\) ⇒ 2x² – 2 = 3x
⇒ 2x² – 3x – 2 = 0
⇒ 2x² – 4x + x – 2 = 0

⇒ 2x (x – 2) + 1 (x – 2) = 0
⇒ (x – 2) (2x + 1) = 0
Either x – 2 = 0, then x = 2
or 2x + 1 – 0, then 2x = – 1 ⇒ x = \(\frac { -1 }{ 2 }\)
x = 1. -1, 2, \(\frac { -i }{ 2 }\)

Question 11.
\(\sqrt{2 x+7}\) = x + 2
Solution:
\(\sqrt{2 x+7}\) = x + 2
Squaring both sides,
2x + 7 = (x + 2)²
⇒ 2x + 7 = x² + 4x + 4
⇒ x² + 4x + 4 – 2x – 7 = 0
⇒ x² + 2x – 3 = 0
⇒ x² + 3x – x – 3 = 0

⇒ x (x + 3) – 1 (x + 3) = 0
⇒ (x + 3) (x – 1) = 0
Either x + 3 = 0, then x = – 3
or x – 1 = 0, then x = 1
∴ x = 1, – 3
Check : If x = 1, then
L.H.S. = \(\sqrt{2 x+7}\) = \(\sqrt{2 + 7}\) = \(\sqrt{9}\) = 3
R.H.S. = x + 2 = 1 + 2 = 3
x = 1 is its solution
If x = – 3, then
L.H.S. = \(\sqrt{2 \times(-3)+7}+\sqrt{-6+7}=\sqrt{1}\) = 1
R.H.S. = – 3 + 2 = – 1
∵ L.H.S. = R.H.S.
∴ x = – 3 is not its solution
Hence x = 1

Question 12.
2\(\sqrt{2 x+1}\) – 2x = 1
Soluion:
2\(\sqrt{2 x+1}\) – 2x = 1
⇒ 2\(\sqrt{2 x+1}\) = 1 + 2x
Squaring both sides,
4(2x + 1) = (1 + 2x)²
⇒ 8x + 4 = 1 + 4x + 4x²
⇒ 4x² + 4x + 1 – 8x – 4 = 0
⇒ 4x² – 4x – 3 = 0
⇒ 4x² – 6x + 2x – 3 = 0

⇒ 2x (2x – 3) + 1 (2x – 3) = 0
⇒ (2x – 3) (2x + 1) = 0
Either 2x – 3 = 0, then 2x = 3 ⇒ x = \(\frac { 3 }{ 2 }\)
or 2x + 1 = 0 then 2x = -1 ⇒ x = \(\frac { -1 }{ 2 }\)
Check:
if x = \(\frac { 3 }{ 2 }\), then
L.H.S. = 2\(\sqrt{2 x+1}\) – 2x
= 2\(\sqrt{2 \times \frac{3}{2}+1}-2 \times \frac{3}{2}=2 \sqrt{3+1}-3\)
= 2\(\sqrt{4}\) – 3 = 2 x 2 – 3 = 4 – 3 = 1 = R.H.S.
∴ x = \(\frac { 3 }{ 2 }\) is a solution -1
If x = \(\frac { -1 }{ 2 }\), then
L.H.S. = \(2 \sqrt{2 x+1}-2 x=2 \sqrt{2\left(\frac{-1}{2}\right)+1}-2\) x (\(\frac { -1 }{ 2 }\))
= 2\(\sqrt{-1 + 1}\) + 1 ⇒ 2 x \(\sqrt{0}\) + 1 ⇒ 0 + 1 = 1 = R.H.S.
∴ x = \(\frac { -1 }{ 2 }\) is also a solution
∴ x = \(\frac { 3 }{ 2 }\), \(\frac { -1 }{ 2 }\)

Question 13.
\(\sqrt{4 x-3}+\sqrt{2 x+3}\) = 6
Solution:
\(\sqrt{4 x-3}+\sqrt{2 x+3}\) = 6
⇒ \(\sqrt{4 x-3}=6-\sqrt{2 x+3}\) = 6
Squaring both sides
4x – 3 = 36 + (2x + 3) – 12 72^+3 4x-3 = 36 + 2x + 3-12 72x + 3 ⇒ 4x – 3 – 36 – 2x – 3 = -12 72x+”3 ⇒ 2x- 42 = -1272×71
Dividing by 2,
x – 21 = -6 72x + 3
Again squaring,
x² -42x + 441 = 36 (2x + 3)
⇒ x² – 42x + 441 = 72x + 108
⇒ x² – 42x + 441 – 72x – 108 = 0
⇒ x² – 114x + 333 = 0
⇒ x² – 111x – 3x + 333 = 0

⇒ x(x- 111) – 3 (x – 111) = 0
⇒ (x – 111) (x – 3) = 0
Either x – 111 = 0, then x = 111
or x – 3 =0, then x = 3
If x = 111, then
L.H.S. = \(\sqrt{4 \times 111-3}+\sqrt{2 \times 111+3}\)
= \(\sqrt{444-3}+\sqrt{222+3}=\sqrt{441}+\sqrt{225}\)
= 21 + 15 = 36 ≠ R.H.S.
∴ x = 111 is not its Solution
If x = 3, then
L.H.S. = \(\sqrt{4 \times 3-3}+\sqrt{2 \times 3+3}\) = \(\sqrt{12-3}\) + \(\sqrt{6+3}\)
= \(\sqrt{9}\) + \(\sqrt{9}\) = 3 + 3 = 6 = R.H.S
∴ x = 3 is its solution
Hence x = 3