Question

Solve the following equation by factorisation method:

`sqrt(6x+7) = x`

Answer 1

Given:

`sqrt(6x+7) = x`

Since the square root is defined and equals x,

x ≥ 0 and 6x + 7 ≥ 0

`(sqrt(6x+7))^2 = x^2`

6x + 7 = x²

x² − 6x − 7 = 0

Two numbers whose product = −7 and sum = −6:

−7 and 1

(x − 7) (x + 1) = 0

x − 7 = 0

⇒ x = 7

x + 1 = 0

⇒ x = −1

Check against the condition

x = 7:

`sqrt(6(7)+7)`

`=sqrt49`

= 7

 x = −1:

`sqrt(6(-1)+7)`

`=sqrt1`

= 1≠ −1

x = 7