Question

Eloise started to solve a radical equation in this way:

square root of negative 2x plus 1 − 3 = x
square root of negative 2x plus 1 − 3 + 3 = x + 3
square root of negative 2x plus 1 = x + 3
square root of negative 2x plus 1 − 1 = x + 3 − 1
square root of negative 2 x = x + 2
(square root of negative 2 x)2 = (x − 4)2
−2x = x2 − 8x + 16
−2x + 2x = x2 + 8x + 16 + 2x
0 = x2 + 10x + 16
0 = (x + 2)(x + 8)

x + 2 = 0 x + 8 = 0
x + 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?

She added 2x after squaring both sides.
She subtracted 1 before squaring both sides.
She factored x2 + 10x + 16 incorrectly.
She did not check for extraneous solutions.

Answer 1

If I understand this correctly, the original equation is
sqrt(2x+1) - 3 = x
then it became sqrt(2x+1) = x+3
so the problem is that sqrt(2x+1) is not the same as sqrt(2x) + 1 so you can't subtract 1 without squaring both sides first

Answer 2

thanks i appreciate it :)