Solve 1 p + 1 q + 1 x = 1 p + q + x .

Given 1 p + 1 q + 1 x = 1 p + q + x We have to find the solution of given quadratic equation. ∵ 1 p + 1 q + 1 x = 1 p + q + x ⇒ 1 p + q + x - 1 x = 1 p + 1 q ⇒ x - p - q - x x ( p + q + x ) = p + q p q ⇒ - ( p + q ) x ( p + q + x ) = p + q p q ⇒ x 2 + ( p + q ) x = - p q ⇒ x 2 + p x + q x + p q = 0 ⇒ x ( x + p ) + q ( x + p ) = 0 ⇒ ( x + p ) ( x + q ) = 0 Now if x + p = 0 then x = - p and if x + q = 0 then x = - q ∴ The required solution is x = - p , - q

Solve 1 p + 1 q + 1 x = 1 p + q + x .

Given 1 p + 1 q + 1 x = 1 p + q + x We have to find the solution of given quadratic equation. ∵ 1 p + 1 q + 1 x = 1 p + q + x ⇒ 1 p + q + x - 1 x = 1 p + 1 q ⇒ x - p - q - x x ( p + q + x ) = p + q p q ⇒ - ( p + q ) x ( p + q + x ) = p + q p q ⇒ x 2 + ( p + q ) x = - p q ⇒ x 2 + p x + q x + p q = 0 ⇒ x ( x + p ) + q ( x + p ) = 0 ⇒ ( x + p ) ( x + q ) = 0 Now if x + p = 0 then x = - p and if x + q = 0 then x = - q ∴ The required solution is x = - p , - q

Solve x a + a x = x b + b x .

Given x a + a x = x b + b x We have to find the solution of given quadratic equation. x a + a x = x b + b x ⇒ x 2 + a 2 a x = x 2 + b 2 b x ⇒ b x 2 + a 2 b = a x 2 + a b 2 , x ≠ 0 ⇒ ( a - b ) x 2 = a b ( a - b ) , x ≠ 0 ⇒ x 2 = a b ( a - b ) ( a - b ) , x ≠ 0 ⇒ x = ± a b , x ≠ 0 ∴ The required solution is x = ± a b

Solve x 3 - 1 ( x - 1 ) 2 - x 3 + 1 ( x + 1 ) 2 = 6

Given x 3 - 1 ( x - 1 ) 2 - x 3 + 1 ( x + 1 ) 2 = 6 We have to find the solution of given quadratic equation. ∵ x 3 - 1 ( x - 1 ) 2 - x 3 + 1 ( x + 1 ) 2 = 6 ⇒ ( x - 1 ) ( x 2 + x + 1 ) ( x - 1 ) 2 - ( x + 1 ) ( x 2 - x + 1 ) ( x + 1 ) 2 = 6 [ ∵ m 3 + n 3 = ( m + n ) ( m 2 + n 2 - m n ) ] ⇒ ( x 2 + x + 1 ) x - 1 - ( x 2 - x + 1 ) x + 1 = 6 ⇒ ( x 2 + x + 1 ) ( x + 1 ) - ( x 2 - x + 1 ) ( x - 1 ) = 6 ( x + 1 ) ( x - 1 ) , x ≠ 1 , - 1 ⇒ ( x 3 + x 2 + x + x 2 + x + 1 ) - ( x 3 - x 2 + x - x 2 + x - 1 ) = 6 ( x 2 - 1 ) ⇒ 4 x 2 + 2 = 6 x 2 - 6 ⇒ 2 x 2 = 8 ⇒ x 2 = 8 2 = 4 ⇒ x = ± 4 = ± 2 ∴ The required solution is x = ± 2

Solve 1 ( x - 1 ) ( x - 2 ) + 1 ( x - 2 ) ( x - 3 ) + 1 ( x - 3 ) ( x - 4 ) = 1 6

Given 1 ( x - 1 ) ( x - 2 ) + 1 ( x - 2 ) ( x - 3 ) + 1 ( x - 3 ) ( x - 4 ) = 1 6 We have to find the solution of given quadratic equation. ∵ 1 ( x - 1 ) ( x - 2 ) + 1 ( x - 2 ) ( x - 3 ) + 1 ( x - 3 ) ( x - 4 ) = 1 6 ⇒ ( x - 1 ) - ( x - 2 ) ( x - 1 ) ( x - 2 ) + ( x - 2 ) - ( x - 3 ) ( x - 2 ) ( x - 3 ) + ( x - 3 ) - ( x - 4 ) ( x - 3 ) ( x - 4 ) = 1 6 ⇒ 1 ( x - 2 ) - 1 ( x - 1 ) + 1 ( x - 3 ) - 1 ( x - 2 ) + 1 ( x - 4 ) - 1 ( x - 3 ) = 1 6 ⇒ 1 x - 4 - 1 x - 1 = 1 6 ⇒ x - 1 - x + 4 ( x - 4 ) ( x - 1 ) = 1 6 ⇒ ( x - 4 ) ( x - 1 ) = 18 ⇒ x 2 - 5 x + 4 - 18 = 0 ⇒ x 2 - 5 x - 14 = 0 ⇒ x 2 - 7 x + 2 x - 14 = 0 ⇒ x ( x - 7 ) + 2 ( x - 7 ) = 0 ⇒ ( x - 7 ) ( x + 2 ) = 0 Now if x - 7 = 0 then x = 7 and if x + 2 = 0 then x = - 2 ∴ The required solution is x = - 2 , 7