Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

# Eloise started to solve a radical equation in this way: − 3 = x − 3 + 3 = x + 3 = x + 3 − 1 = x + 3 − 1 = x + 2 ( )2 = (x − 4)2 −2x = x2 − 8x + 16 −2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16

Eloise started to solve a radical equation in this way:

− 3 = x − 3 + 3 = x + 3 = x + 3− 1 = x + 3 − 1 = x + 2 ( )2 = (x − 4)2−2x = x2 − 8x + 16−2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 160 = (x + 2)(x + 8)

x + 2 = 0 x + 8 = 0x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8 x = −2 x = −8

Both solutions are extraneous because they don't satisfy the original equation.

What error did Eloise make? (2 points)

a

She added 2x after squaring both sides.

b

She subtracted 1 before squaring both sides.

c

She factored x2 + 10x + 16 incorrectly.

d

She did not check for extraneous solutions.