If x 1 , x 2 are the roots of the quadratic equation a x 2 + b x + c = 0 and c ≠ 0 , find the value of a x 1 + b - 2 + a x 2 + b - 2 in terms of a , b , c .

Given x 1 , x 2 are the roots of quadratic equation a x 2 + b x + c = 0 . Sum of roots is x 1 + x 2 = - b a Product of roots is x 1 · x 2 = c a Now, a x 1 2 + b x 1 + c = 0 ⇒ x 1 a x 1 + b = - c ⇒ a x 1 + b = - c x 1 Similarly, a x 2 2 + b x 2 + c = 0 ⇒ x 2 a x 2 + b = - c ⇒ a x 2 + b = - c x 2 Now, a x 1 + b - 2 + a x 2 + b - 2 = x 1 2 + x 2 2 c 2 = 1 c 2 x 1 + x 2 2 - 2 x 1 x 2 = 1 c 2 b 2 - 2 a c a 2 ∴ a x 1 + b - 2 + a x 2 + b - 2 = b 2 - 2 a c a 2 c 2

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