Answer:
there are no extraneous solutions to the equation
Step-by-step explanation:
Given
[tex]\sqrt{8x+9}[/tex] = x + 2 ( square both sides )
8x + 9 = (x + 2)² ← expand
8x + 9 = x² + 4x + 4 ( subtract 8x + 9 from both sides )
0 = x² - 4x - 5 ← in standard form
0 = (x - 5)(x + 1) ← 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 the solutions.
x = 5 : [tex]\sqrt{8(5)+9}[/tex] = [tex]\sqrt{49}[/tex] = 7
right side = 5 + 2 = 7
left side = right side ⇒ x = 5 is a solution
x = - 1 : [tex]\sqrt{8(-1)+9}[/tex] = [tex]\sqrt{1}[/tex] = 1
right side = - 1 + 2 = 1
left side = right side ⇒ x = - 1 is a solution
There are no extraneous solutions
