Mathematics
Mathematics, 01.12.2021 07:20, meganwintergirl

Eloise started to solve a radical equation in this way: square root of negative 2x plus 1 βˆ’ 3 = x
square root of negative 2x plus 1 βˆ’ 3 + 3 = x + 3
square root of negative 2x plus 1 = x + 3
square root of negative 2x plus 1 βˆ’ 1 = x + 3 βˆ’ 1
square root of negative 2 x = x + 2
(square root of negative 2 x)2 = (x βˆ’ 4)2
βˆ’2x = x2 βˆ’ 8x + 16
βˆ’2x + 2x = x2 + 8x + 16 + 2x
0 = x2 + 10x + 16
0 = (x + 2)(x + 8)

x + 2 = 0 x + 8 = 0
x + 2 βˆ’ 2 = 0 βˆ’ 2 x + 8 βˆ’ 8 = 0 βˆ’ 8
x = βˆ’2 x = βˆ’8

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

What error did Eloise make?

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.

answer
Answers: 3

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