Answer:
x = - 1 is extraneous
Step-by-step explanation:
[tex]\sqrt{4x+5}[/tex] = x ( square both sides )
4x + 5 = x² ( subtract 4x + 5 from both sides )
0 = x² - 4x - 5 ← in standard form
0 = (x - 5)(x + 1) = 0 ← in factored form
Equate each factor to zero and solve for x
x - 5 = 0 ⇒ x = 5
x + 1 = 0 ⇒ x = - 1
As a check
Substitute these values into the equation and if both sides are equal then they are solutions.
x = 5
left side = [tex]\sqrt{4(5)+5}[/tex] = [tex]\sqrt{20+5}[/tex] = [tex]\sqrt{25}[/tex] = 5
right side = x = 5
Then x = 5 is a solution of the equation
x = - 1
left side = [tex]\sqrt{4(-1)+5}[/tex] = [tex]\sqrt{-4+5}[/tex] = [tex]\sqrt{1}[/tex] = 1
right side = x = - 1
Then x = - 1 is an extraneous solution
