OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Ex 10(b)

The availability of step-by-step OP Malhotra Maths Class 11 Solutions Chapter 10 Quadratic Equations Ex 10(b) can make challenging problems more manageable.

S Chand Class 11 ICSE Maths Solutions Chapter 10 Quadratic Equations Ex 10(b)

Solve the following equations :

Question 1.
x⁴ – 5x² + 6 = 0.
Solution:
Given eqn. be, x⁴ – 5x² + 6 = 0 … (1)
put x² = t in eqn. (1) ; we have
t² – 5t + 6 = 0
⇒ (t – 2) (t – 3) = 0
⇒ t – 2 = 0 or t – 3 = 0
⇒ x² = 2 or x² = 3
⇒ x = ± \(\sqrt{2}\) or x = ± \(\sqrt{3}\)
∴ x = ± \(\sqrt{2}\), ± \(\sqrt{3}\)

Question 2.
x⁵ + 242 = \(\frac{243}{x^5}\).
Solution:
Given eqn. be,
x⁵ + 242 = \(\frac{243}{x^5}\) … (1)
putting x⁵ = y in eqn. (1) ; we have
y + 242 = \(\frac { 243 }{ y }\) ⇒ y² + 242y – 243 = 0
⇒ y² – y + 243y – 243 = 0
⇒ y(y – 1) + 243 (y – 1) = 0
⇒ (y – 1) (y + 243) = 0
⇒ y = 1 = 0 or y + 243 = 0
⇒ y = 1 or y = – 243
⇒ x⁵ = 1 or x⁵ = (- 3)⁵
⇒ x = 1 or x = – 3
Thus, s = – 1, – 3

Question 3.
10x^-2 – 9 – x^-4 = 0.
Solution:
Given eqn. be, 10x^-2 – 9 – x^-4 = 0 … (1)
putting x^-2 = t in eqn. (1); we have
10t – 9 – t² = 0
⇒ t² – 10t + 9 = 0
⇒ t² – t – 9t + 9 = 0
⇒ t (t – 1) – 9 (t – 1) = 0
⇒ (t – 9)(t – 1) = 0
either t = 9 = 0 or t – 1 = 0
⇒ x^-2 = 9 or x^-2 – 1 = 0
⇒ \(\frac{1}{x^2}\) = 9 or \(\frac{1}{x^2}\) = 1
⇒ x = ± \(\frac { 1 }{ 3 }\) or x = ± 1
Thus, x = ± \(\frac { 1 }{ 3 }\), ± 1

Question 4.
3^2x – 10 x 3^x + 9 = 0.
Solution:
Given eqn. be,
3^2x – 10 x 3^x + 9 = 0 …(1)
putting 3^x = t in eqn. (1) ; we have
t² – 10t + 9 = 0
⇒ t² – 9t – t + 9 = 0
⇒ t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (t – 1) = 0
either t – 9 = 0 or t – 1 = 0
⇒ t = 9 or t = 1
⇒ 3^x = 3^x or 3^x = 3°
⇒ x = 2 or x = 0
Thus, x = 0, 2

Question 5.
2^2x-1 – 9 x 2^x-2 + 1 = 0.
Solution:
Given equation be,
2^2x-1 – 9 x 2^x-2 + 1 = 0
⇒ (2^x)² x 2^-1 – 9 x 2^x x 2^-2 + 1 = 0 …(1)
putting 2^x = t in eqn. (1); we have
t² x \(\frac { 1 }{ 2 }\) – 9 x t x \(\frac { 1 }{ 4 }\) + 1 = 0
⇒ 2t² – 9t + 4 = 0
⇒ 2t² – 8t – t + 4 = 0
⇒ 2t (t – 4) – 1 (t – 4) = 0
⇒ (2t – 1) (t – 4) = 0
either 2t – 1 = 0 or t – 4 = 0
⇒ t = \(\frac { 1 }{ 2 }\) or t = 4
⇒ 2^x = \(\frac { 1 }{ 2 }\) = 2^-1 or 2^x = 2²
⇒ x = – 1 or x = 2
∴ x = – 1 or 2

Question 6.
3^2x+1 + 3² = 3^x+3 + 3^x.
Solution:
Given eqn. be,
3^2x+1 + 3² = 3^x+3 + 3^x
⇒ (3^x)² x 3 + 9 = 3^x (3³ + 1) …(1)
putting 3^x = t in eqn. (1) ; we have
3t² + 9 = 28t ⇒ 3t² – 28t + 9 = 0
⇒ 3t² – 27t – t + 9 = 0
⇒ 3t(t – 9) – 1 (t – 9) = 0
⇒ (t – 9) (3t – 1) = 0
either t – 9 = 0 or 3t – 1 = 0
⇒ t = 9 or t = \(\frac { 1 }{ 3 }\)
⇒ 3^x = 3² or 3^x = 3^-1
⇒ x = 2 or x = – 1
Thus, x = 2, – 1

Question 7.
\(\sqrt{x^2-3 x}=4 x^2-12 x-3\)
Solution:
Given equation be,
\(\sqrt{x^2-3 x}=4 x^2-12 x-3\) – 3 …(1)
putting x² – 3x = t in eqn. (1); we have
\(\sqrt{t}\) = 4t – 3 ;
on squaring both sides ; we have
⇒ t = (4t – 3)² ⇒ t = 16t² – 24t + 9 ⇒ 16t² – 25t + 9 = 0

Question 8.
\(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}}\) = 5.
Solution:
Given equation be
\(\sqrt{\frac{x^2+2}{x^2-2}}+6 \sqrt{\frac{x^2-2}{x^2+2}}\) = 5
putting \(\sqrt{\frac{x^2+2}{x^2-2}}\) = y in eqn. (1) ; we have
y + \(\frac { 6 }{ y }\) = 5
⇒ y² – 5y + 6 = 0
⇒ (y – 2)(y – 3) = 0
⇒ y – 2 = 0 or y – 3 = 0
⇒ y = 2 or y = 3
Case-I: When y = 2 ⇒ \(\sqrt{\frac{x^2+2}{x^2-2}}\) = 2
On squaring both sides ; we have
\(\frac{x^2+2}{x^2-2}\) = 4
⇒ x² + 2 = 4x² – 8
⇒ 3x² = 10
⇒ x² = \(\frac { 10 }{ 3 }\)
⇒ x = ± \(\sqrt{\frac{10}{3}}\)

Case-II: When y = 3 ⇒ \(\sqrt{\frac{x^2+2}{x^2-2}}\) = 3
On squaring both sides ; we have
\(\frac{x^2+2}{x^2-2}\) = 9 ⇒ x² + 2 = 9x² – 18
⇒ 8x² = 20
⇒ x² = \(\frac { 5 }{ 2 }\)
⇒ x = ± \(\sqrt{\frac{5}{2}}\)
Hence x = ± \(\sqrt{\frac{10}{3}}, \pm \sqrt{\frac{5}{2}}\)

Question 9.
\(\sqrt{\frac{2 x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2 x^2+1}}\) = 5.
Solution:
Given equation be
\(\sqrt{\frac{2 x^2+1}{x^2-1}}+6 \sqrt{\frac{x^2-1}{2 x^2+1}}\) = 5 … (1)
putting \(\sqrt{\frac{2 x^2+1}{x^2-1}}\) = t in eqn. (1); we have
t + \(\frac { 6 }{ t }\) = 5
⇒ t² – 5t + 6 = 0
⇒ (t – 2)(t – 3) = 0
either t – 2 = 0 or t – 3 = 0
⇒ t = 2 or t = 3
Case-I: When t = 2
⇒ \(\sqrt{\frac{2 x^2+1}{x^2-1}}\) = 2
On squaring both sides ; we have 2*2 + 1
\(\frac{2 x^2+1}{x^2-1}\) = 4
⇒ 2x² + 1 = 4x² – 4
⇒ 2x² = 5
⇒ x = \(\pm \sqrt{\frac{5}{2}}\)

Case-II: When t = 3

Question 10.
x (x – 1) (x + 2) (x- 3) + 8 = 0.
Solution:
Given equation be
x (x – 1) (x + 2) (x – 3) + 8 = 0
⇒ (x² – x) [x² – x – 6] + 8 = 0 … (1)
putting x² – x = t in eqn. (1) ; we have
t(t – 6) + 8 = 0 ⇒ t² – 6t + 8 = 0
⇒ t² – 4t – 2t + 8 = 0
⇒ t(t – 4) – 2(t – 4) = 0
⇒ (t – 4)(t – 2) = 0
either t – 4 = 0 or t – 2 = 0
⇒ t = 4 or t = 2
Case-I: When t – 4
⇒ x² – x – 4 = 0
⇒ x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
⇒ x = \(\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 1 \times(-4)}}{2}\)
⇒ x = \(\frac{1 \pm \sqrt{17}}{2}\)

Case-II: When t = 2 ⇒ x² – x – 2 = 0
⇒ x² – 2x + x – 2 = 0
⇒ x (x – 2) + 1 (x – 2) = 0
⇒ (x + 1) (x – 2) = 0
⇒ x = – 1, 2
Hence x = – 1, 2, \(\frac{1 \pm \sqrt{17}}{2}\)

Question 11.
(x – 7) (x – 3) (x + 1) (x + 5) = 1680.
Solution:
Given eqn. be
(x – 7) (x – 3) (x + 1)(x + 5) = 1680
⇒ {(x – 7) (x + 5)} {(x-3) (x+ 1)} = 1680
⇒ {x² – 2x – 35} {x² – 2x – 3} = 1680 … (1)
putting x² – 2x = t in eqn. (1); we get
(t – 35) (t – 3) = 1680
⇒ t² – 38t + 105 – 1680 = 0
⇒ t² – 38t – 1575 = 0
∴ t = \(\frac{38 \pm \sqrt{1444+6300}}{2}=\frac{38 \pm 88}{2}\)
⇒ t = 63, – 25

Case-I: When t = 63 ⇒ x² – 2x – 63 = 0

Question 12.
(2x – 7) (x² – 9) (2x + 5) = 91.
Solution:
Given equation be,
(2x – 7) (x² – 9) (2x + 5) = 91
⇒ {(2x – 7) (x + 3)} {(x – 3)(2x + 5)} = 91
⇒ (2x² – x – 21)(2x² – x – 15) = 91 …(1)
putting 2x² – x = t in eqn. (1) ; we have
(t – 21) (t – 15) = 91
⇒ t² – 36t + 224 = 0
⇒ t² – 8t – 28t + 224 = 0
⇒ t (t – 8) – 28 (t – 8) = 0
⇒ (t – 8) (t – 28) = 0
either t – 8 = 0 or t – 28 = 0
⇒ t = 8 or t = 28
Case-I: When t = 8 ⇒ 2x² – x – 8 = 0
⇒ x = \(\frac{-(-1) \pm \sqrt{(-1)^2-4 \times 2 \times(-8)}}{2 \times 2}\)
⇒ x = \(\frac{1 \pm \sqrt{65}}{4}\)

Case-II: When t = 28
⇒ 2x² – x – 28 = 0
⇒ 2x² – 8x + 7x – 28 = 0
⇒ 2x (x – 4) + 7 (x – 4) = 0
⇒ (x – 4) (2x + 7) = 0
either x – 4 = 0 or 2x + 7 = 0
⇒ x = 4 or x = – \(\frac { 7 }{ 2 }\)
Hence, x = + 4, – \(\frac { 7 }{ 2 }\), \(\frac{1 \pm \sqrt{65}}{4}\)

Question 13.
By substituting y = 2^x, or otherwise, solve the equation
2^2x + 2^x+2 – 4 x 2³ = 0.
Solution:
Given eqn. be,
2^2x + 2^x+2 – 4 x 2³ = 0 …(1)
putting 2^2x = y in eqn. (1) ; we have
y² + y . 2² – 32 = 0
⇒ y² + 4y – 32 = 0
⇒ y² + 8y – 4y – 32 = 0
⇒ y(y + 8) – 4(y + 8) = 0
⇒ (y + 8)(y – 4) = 0
either y + 8 = 0 or y – 4 = 0
⇒ y = – 8 or y = 4
⇒ 2^x = – 8
it has no real solution (∵ 2^x > 0)
or 2^x = 4 = 2² ∴ x = 2
Thus x = 2

Question 14.
2^x² : 2^x = 8 : 1.
Solution:

Question 15.
2^2x+3 + 2^x+3 = 1 + 2^x
Solution:
Given eqn. be 2^2x+3 + 2^x+3 = 1 + 2^x
⇒ (2^x)² x 2³ + 2^x x 2³ = 1 + 2^x …(1)
putting 2^x = t in eqn. (1) ;
we have 8t² + 8t = 1 + t
⇒ 8t² + 7t – 1 = 0
⇒ 8t² + 8t – t – 1 = 0
⇒ 8t (t + 1) – 1 (t + 1) = 0
⇒ (t + 1) (8t – 1) = 0
either t + 1 = 0 or 8t – 1 = 0
⇒ t = – 1 or t = \(\frac { 1 }{ 8 }\)
⇒ 2^x = – 1, which is not possible as 2^x > 0
when t = \(\frac { 1 }{ 8 }\) ⇒ 2^x = 2^-3 ⇒ x = – 3

Question 16.
\(4^x-3^{x-\frac{1}{2}}=3^{x+\frac{1}{2}}-2^{2 x-1}\)
Solution:
Given equation be