Solve 2 x + 3 + 6 x + 6 = 1

Given 2 x + 3 + 6 x + 6 = 1 We have to find the solution of given quadratic equation. ∵ 2 x + 3 + 6 x + 6 = 1 ⇒ 2 ( x + 6 ) + 6 ( x + 3 ) ( x + 3 ) ( x + 6 ) = 1 ⇒ 2 ( x + 6 ) + 6 ( x + 3 ) = ( x + 3 ) ( x + 6 ) ⇒ 2 x + 12 + 6 x + 18 = x 2 + 6 x + 3 x + 18 ⇒ x 2 + x - 12 = 0 ⇒ x 2 + 4 x - 3 x - 12 = 0 [ ∵ 12 = 3 × 4 ] ⇒ x ( x + 4 ) - 3 ( x + 4 ) = 0 ⇒ ( x - 3 ) ( x + 4 ) = 0 Now if x - 3 = 0 then x = 3 and if x + 4 = 0 then x = - 4 ∴ The required solution is x = - 4 , 3 .

Solve 2 x + 3 + 6 x + 6 = 1

Given 2 x + 3 + 6 x + 6 = 1 We have to find the solution of given quadratic equation. ∵ 2 x + 3 + 6 x + 6 = 1 ⇒ 2 ( x + 6 ) + 6 ( x + 3 ) ( x + 3 ) ( x + 6 ) = 1 ⇒ 2 ( x + 6 ) + 6 ( x + 3 ) = ( x + 3 ) ( x + 6 ) ⇒ 2 x + 12 + 6 x + 18 = x 2 + 6 x + 3 x + 18 ⇒ x 2 + x - 12 = 0 ⇒ x 2 + 4 x - 3 x - 12 = 0 [ ∵ 12 = 3 × 4 ] ⇒ x ( x + 4 ) - 3 ( x + 4 ) = 0 ⇒ ( x - 3 ) ( x + 4 ) = 0 Now if x - 3 = 0 then x = 3 and if x + 4 = 0 then x = - 4 ∴ The required solution is x = - 4 , 3 .

Solve 1 x + 1 + 2 x + 5 = 1 2

Given 1 x + 1 + 2 x + 5 = 1 2 We have to find the solution of given quadratic equation. ∵ 1 x + 1 + 2 x + 5 = 1 2 ⇒ 1 ( x + 5 ) + 2 ( x + 1 ) ( x + 1 ) ( x + 5 ) = 1 2 ⇒ 2 [ ( x + 5 ) + 2 ( x + 1 ) ] = ( x + 1 ) ( x + 5 ) ⇒ 2 x + 10 + 4 x + 4 = x 2 + 5 x + x + 5 ⇒ x 2 = 9 ⇒ x = ± 9 = ± 3 ∴ The required solution is x = - 3 , 3 .

Solve x 3 + 3 x = 3 1 3

Given x 3 + 3 x = 3 1 3 We have to find the solution of given quadratic equation. ∵ x 3 + 3 x = 3 1 3 ⇒ x 2 + 9 3 x = 10 3 ⇒ 3 ( x 2 + 9 ) = 10 × 3 x ⇒ x 2 - 10 x + 9 = 0 ⇒ x 2 - 9 x - x + 9 = 0 [ ∵ 9 = ( - 1 ) × ( - 9 ) ] ⇒ x ( x - 9 ) - 1 ( x - 9 ) = 0 ⇒ ( x - 9 ) ( x - 1 ) = 0 Now if x - 9 = 0 then x = 9 and if x - 1 = 0 then x = 1 ∴ The required solution is x = 9 , 1 .

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

Given x - 2 x + 2 + 6 ( x - 2 x - 6 ) = 1 We have to find the solution of given quadratic equation. ∵ x - 2 x + 2 + 6 ( x - 2 x - 6 ) = 1 ⇒ ( x - 2 ) ( x - 6 ) + 6 ( x - 2 ) ( x + 2 ) ( x + 2 ) ( x - 6 ) = 1 ⇒ ( x - 2 ) ( x - 6 ) + 6 ( x - 2 ) ( x + 2 ) = ( x + 2 ) ( x - 6 ) ⇒ x 2 - 2 x - 6 x + 12 + 6 x 2 - 24 = x 2 + 2 x - 6 x - 12 ⇒ 6 x 2 - 4 x = 0 ⇒ x ( 6 x + 4 ) = 0 Now either x = 0 or 6 x + 4 = 0 ⇒ x = - 4 6 = - 2 3 ∴ The required solution is x = 0 , - 2 3 .

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

Given ( 2 x + 1 ) + 3 2 x + 1 = 4 We have to find the solution of given quadratic equation. ∵ ( 2 x + 1 ) + 3 2 x + 1 = 4 ⇒ ( 2 x + 1 ) 2 + 3 ( 2 x + 1 ) = 4 ⇒ 4 x 2 + 4 x + 1 + 3 = 4 ( 2 x + 1 ) ⇒ 4 x 2 - 4 x = 0 ⇒ x 2 - x = 0 ⇒ x ( x - 1 ) = 0 Now either x = 0 or x - 1 = 0 ⇒ x = 1 ∴ The required solution is x = 0 , 1 .

Solve x x + 1 + x + 1 x = 2 1 6 .

Given x x + 1 + x + 1 x = 2 1 6 We have to find the solution of given quadratic equation. ∵ x x + 1 + x + 1 x = 2 1 6 ⇒ x 2 + ( x + 1 ) 2 x ( x + 1 ) = 13 6 ⇒ 6 x 2 + 6 ( x + 1 ) 2 = 13 x ( x + 1 ) ⇒ 6 x 2 + 6 x 2 + 12 x + 6 = 13 x 2 + 13 x ⇒ x 2 + x - 6 = 0 ⇒ x 2 + 3 x - 2 x - 6 = 0 ⇒ x ( x + 3 ) - 2 ( x + 3 ) = 0 ⇒ ( x - 2 ) ( x + 3 ) = 0 Now if ( x - 2 ) = 0 then x = 2 and if ( x + 3 ) = 0 then x = - 3 ∴ The required solution is x = 2 , - 3 .

Solve 3 x + 2 + 3 3 x + 2 = - 4

Given 3 x + 2 + 3 3 x + 2 = - 4 We have to find the solution of given quadratic equation. ∵ 3 x + 2 + 3 3 x + 2 = - 4 ⇒ ( 3 x + 2 ) 2 + 3 3 x + 2 = - 4 ⇒ 9 x 2 + 12 x + 4 + 3 = - 12 x - 8 ⇒ 9 x 2 + 24 x + 15 = 0 ⇒ 3 x 2 + 8 x + 5 = 0 ⇒ 3 x 2 + 3 x + 5 x + 5 = 0 ⇒ 3 x ( x + 1 ) + 5 ( x + 1 ) = 0 ⇒ ( 3 x + 5 ) ( x + 1 ) = 0 Now if ( 3 x + 5 ) = 0 then x = - 5 3 and if ( x + 1 ) = 0 then x = - 1 ∴ The required solution is x = - 5 3 , - 1 .