Answer:
x = 3 is extraneous
Step-by-step explanation:
Given
[tex]\sqrt{4x-8}[/tex] = x - 5 ( square both sides to clear the radical )
4x - 8 = (x - 5)² ← expand using FOIL
4x - 8 = x² - 10x + 25 ( subtract 4x - 8 from both sides )
0 = x² - 14x + 33 ← in standard form
0 = (x - 11)(x - 3) ← in factored form
Equate each factor to zero and solve for x
x - 11 = 0 ⇒ x = 11
x - 3 = 0 ⇒ x = 3
As a check
Substitute these values into the equation and if both sides are equal then they are solutions
x = 11
left side = [tex]\sqrt{44-8}[/tex] = [tex]\sqrt{36}[/tex] = 6
right side = 11 - 5 = 6
Thus x = 11 is a solution
x = 3
left side = [tex]\sqrt{12-8}[/tex] = [tex]\sqrt{4}[/tex] = 2
right side = 3 - 5 = - 2 ≠ 2
Thus x = 3 is an extraneous solution