Question

Solve: 2x − 3 = `sqrt(2x^2-2x+21)`

Answer 1

Given:

2x − 3 = `sqrt(2x^2-2x+21)`

To eliminate the square root, square both sides:

`(2x-3)^2 = (sqrt(2x^2 - 2x+21))^2`

(2x − 3)² = 2x² − 2x + 21

(2x − 3)² = 4x² − 12x + 9

4x² − 12x + 9 = 2x² − 2x + 21

Subtract the right side from both sides:

(4x² − 12x + 9) − (2x² − 2x + 21) = 0

4x² − 12x + 9 − 2x² + 2x − 21 = 0

⇒ 2x² − 10x − 12 = 0

Solve the quadratic

2x² − 10x − 12 = 0

x² − 5x − 6 = 0

x² − 5x − 6 = (x − 6) (x + 1) = 0

x = 6 or x = −1

Because we squared both sides, we must check both roots in the original equation:

LHS:

2(6) − 3 = 12 − 3 = 9

RHS:

`sqrt(2(6)^2 - 2(6) + 21)`

`= sqrt(72-12+21)`

= 9

LHS: 

2(−1) −3

= −2 −3

= −5

RHS:

`sqrt(2(-1)^2 - 2(-1) + 21)`

`= sqrt(2+2+21)`

`=sqrt25`

= 5

LHS = −5, 

LHS = −5, RHS = 5 ⇒ Not equal

x = 6​