Question

Solve the radical equation.

Answer 1

there are no extraneous solutions to the equation

Explanation:

Given

`√(8x+9)` = 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 :
`√(8(5)+9)` =
`√(49)` = 7

right side = 5 + 2 = 7

left side = right side ⇒ x = 5 is a solution

x = - 1 :
`√(8(-1)+9)` =
`√(1)` = 1

right side = - 1 + 2 = 1

left side = right side ⇒ x = - 1 is a solution

There are no extraneous solutions

Answer 2

There are no extraneous solutions

Explanation:

sqrt(8x+9) = x+2

Square each side

(sqrt(8x+9))^2 = (x+2)^2

8x+9 = (x+2)^2

FOIL

8x+9 = x^2 +2x+2x+4

Combine like term

8x+9 = x^2 +4x+4

Subtract 8x from each side

8x-8x+9 = x^2 +4x-8x+4

9 = x^2 -4x+4

Subtract 9 from each side

0 = x^2 -4x-5

Factor

0=(x-5) (x+1)

Using the zero product property\

x-5 =0 x+1=0

x=5 x=-1

Check the solutions

x=5

sqrt(8*5+9) = 5+2

sqrt(40+9) = 5+2

sqrt(49) =7

7=7

x=-1

sqrt(8*(-1)+9) = -1+2

sqrt(-8+9) = -1+2

sqrt(1) =1

1=1

Both solutions are true solutions