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If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =
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20 Jul 2020, 05:18
20 Jul 2020, 05:18
Expert Reply
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Question Stats:
46% (01:54) correct
53% (02:22) wrong based on 26 sessions
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If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
Re: If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =
[#permalink]
20 Jul 2020, 05:19
20 Jul 2020, 05:19
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more GRE Quant Theory and Practice, see
GRE Prep Club - MATH Book for the GRE Exam
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more GRE Quant Theory and Practice, see
GRE Prep Club - MATH Book for the GRE Exam
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Re: If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =
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20 Jul 2020, 10:01
20 Jul 2020, 10:01
1
Carcass wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1
GIVEN: \(\sqrt{3-2x} = \sqrt{2x} +1\)
Square both sides to get: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\)
In other words: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)(\sqrt{2x} +1)\)
Use FOIL to expand right side: \((\sqrt{3-2x})^2 =(\sqrt{2x})^2 + 1\sqrt{2x} + 1\sqrt{2x} + 1^2\)
Simplify right side to get: \(3-2x=2x+2\sqrt{2x}+1\)
Subtract 1 from both sides to get: \(2-2x=2x+2\sqrt{2x}\)
Subtract 2x from both sides to get: \(2-4x=2\sqrt{2x}\)
Divide both sides by 2 to get: \(1-2x=\sqrt{2x}\)
Square both sides to get: \((1-2x)^2=(\sqrt{2x})^2\)
Expand and simplify to get: \(1-4x+4x^2=2x\)
Add 4x to both sides: \(1 + 4x^2=6x\)
Subtract 1 from both sides to get: \(4x^2=6x-1\)
Answer: E
Cheers,
Brent
