Question

Solve the following equation using quadratic formula:

`(x + 3)/(2x + 3) = (x + 1)/(3x + 2)`

Answer 1

⇒ `(x + 3)/(2x + 3) = (x + 1)/(3x + 2)`

⇒ (x + 3)(3x + 2) = (x + 1)(2x + 3)

⇒ (3x² + 2x + 9x + 6) = (2x² + 3x + 2x + 3)

⇒ 3x² + 11x + 6 = 2x² + 5x + 3

⇒ 3x² – 2x² + 11x – 5x + 6 – 3 = 0

⇒ 3x² + 6x + 3 = 0

Comparing equation x² + 6x + 3 = 0 with ax² + bx + c = 0, we get :

a = 1, b = 6 and c = 3

By formula,

`x = (-b ± sqrt(b^2 - 4ac))/(2a)`

Substituting values we get:

⇒ `x = (-(6) ± sqrt((6)^2 - 4 xx (1) xx (3)))/(2 xx 1)`

= `(-6 ± sqrt(36 - 12))/2`

= `(-6 ± sqrt(24))/2`

= `(-6 ± sqrt(6 xx 4))/2`

= `(-6 ± 2sqrt(6))/2`

= `(2(-3 ± sqrt(6)))/2`

= `-3 ± sqrt(6)`

= `-3 + sqrt(6)` or `-3 - sqrt(6)`

Hence, `x = {(-3 + sqrt(6)),(-3 - sqrt(6))}`.