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If \sqrt of 3-4x=4-sqrt3x
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02 Aug 2021, 13:46
02 Aug 2021, 13:46
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Question Stats:
58% (03:33) correct
41% (03:29) wrong based on 31 sessions
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If \(\sqrt{3-4x}=4-\sqrt{3x}\), then \(49x^2=\)
(A) 10x + 144
(B) 10x − 169
(C) 64x − 144
(D) 64x + 169
(E) 118x − 192
(A) 10x + 144
(B) 10x − 169
(C) 64x − 144
(D) 64x + 169
(E) 118x − 192
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
If \sqrt of 3-4x=4-sqrt3x
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02 Aug 2021, 17:06
02 Aug 2021, 17:06
3
Squaring both sides results in one square root expression
\(3-4x=16-8\sqrt{3x}+3x\)
\(8\sqrt{3x}=16+3x-3+4x\)
\(64*3x=(13+7x)^2=169+26*7x+49x^2\)
\(49x^2=64*3x-169-26*7x=10x-169\) and it's B
\(3-4x=16-8\sqrt{3x}+3x\)
\(8\sqrt{3x}=16+3x-3+4x\)
\(64*3x=(13+7x)^2=169+26*7x+49x^2\)
\(49x^2=64*3x-169-26*7x=10x-169\) and it's B
Carcass wrote:
If \(\sqrt{3-4x}=4-\sqrt{3x}\), then \(49x^2=\)
(A) 10x + 144
(B) 10x − 169
(C) 64x − 144
(D) 64x + 169
(E) 118x − 192
(A) 10x + 144
(B) 10x − 169
(C) 64x − 144
(D) 64x + 169
(E) 118x − 192
Re: If \sqrt of 3-4x=4-sqrt3x
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02 Sep 2024, 09:42
02 Sep 2024, 09:42
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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