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Mathematics, 26.11.2019 19:31 alex43079

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eloise started to solve a radical equation in this way:

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

x + 2 = 0 x + 8 = 0
x + 2 โˆ’ 2 = 0 โˆ’ 2 x + 8 โˆ’ 8 = 0 โˆ’ 8
x = โˆ’2 x = โˆ’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.

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Answers: 1

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