Solve 4 x - 1 - 3 · 2 x - 1 + 2 = 0 .

Given, 4 x - 1 - 3 · 2 x - 1 + 2 = 0 Let, 2 x - 1 = y ...(i) ⇒ y 2 - 3 y + 2 = 0 ⇒ y 2 - 2 y - y + 2 = 0 ⇒ y ( y - 2 ) - 1 ( y - 2 ) = 0 ⇒ ( y - 2 ) ( y - 1 ) = 0 ⇒ y = 1 , 2 From (i), 2 x - 1 = y ⇒ 2 x - 1 = 2 0 , 2 x - 1 = 2 1 ⇒ x - 1 = 0 , x - 1 = 1 ∴ x = 1 , 2

Solve x x - 3 + x - 3 x = 5 2 , when x ≠ 0 and x ≠ 3 .

Given, x x - 3 + x - 3 x = 5 2 Let, x x - 3 = t ⇒ t + 1 t = 5 2 ⇒ 2 t 2 - 5 t + 2 = 0 ⇒ 2 t 2 - 4 t - t + 2 = 0 ⇒ 2 t ( t - 2 ) - 1 ( t - 2 ) = 0 ⇒ ( 2 t - 1 ) ( t - 2 ) = 0 ⇒ t = 1 2 , t = 2 Now, x x - 3 = 4 or x x - 3 = 1 4 ⇒ x = 4 x - 12 or 4 x = x - 3 ⇒ 3 x = 12 or 3 x = - 3 ⇒ x = 4 or x = - 1 ∴ x = 4 , - 1

Solve 2 x + 1 x 2 - 7 x + 1 x + 5 = 0 , when x ≠ 0 .

Given, 2 x + 1 x 2 - 7 x + 1 x + 5 = 0 Let, x + 1 x = t ⇒ 2 t 2 - 7 t + 5 = 0 ⇒ 2 t 2 - 2 t - 5 t + 5 = 0 ⇒ 2 t ( t - 1 ) - 5 ( t - 1 ) = 0 ⇒ t = 5 2 , t = 1 but x + 1 x ≥ 2 ∀ x ∈ R + Then, x + 1 x = 5 2 is only possible ⇒ 2 x 2 - 5 x + 2 = 0 ⇒ ( 2 x - 1 ) ( x - 2 ) = 0 ⇒ x = 1 2 , 2 Now, if x + 1 x = 1 ⇒ x 2 - x + 1 = 0 ⇒ x = 1 ± 1 - 4 2 ∴ x = 1 ± 3 i 2

Solve x 2 + 1 x 2 - 5 x + 1 x + 6 = 0 , when x ≠ 0 .

Given, x 2 + 1 x 2 - 5 x + 1 x + 6 = 0 Let, x + 1 x = t ⇒ t 2 - 2 - 5 t + 6 = 0 ⇒ t 2 - 5 t + 4 = 0 ⇒ t 2 - 4 t - t + 4 = 0 ⇒ t ( t - 4 ) - 1 ( t - 4 ) = 0 ⇒ t - 4 t - 1 = 0 ⇒ t = 4 , t = 1 Then, x + 1 x = 4 ⇒ x 2 + 1 - 4 x = 0 ⇒ x = 4 ± 16 - 4 1 2 = 4 ± 12 2 = 4 ± 2 3 2 ⇒ x = 2 ± 3 Now, if x + 1 x = 1 ⇒ x 2 - x + 1 = 0 ⇒ x = 1 ± 1 - 4 2 ∴ x = 1 ± 3 i 2

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Intermediate Second Year Mathematics Paper 2A>Algebra>Chapter 3 - Quadratic Expressions>Exercise 3(a)

Andhra Pradesh Board Solutions for Exercise 1: Exercise 3(a)

Author:Andhra Pradesh Board

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 3(a)

Attempt the practice questions from Exercise 1: Exercise 3(a) with hints and solutions to strengthen your understanding. Intermediate Second Year Mathematics Paper 2A solutions are prepared by Experienced Embibe Experts.