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If sq rt (3-2x) = sq rt (2x) + 1.....

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by chendawg » Sun Mar 06, 2011 1:57 pm

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D

E

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Difficulty—

If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=

A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1

Source: OG 12

OA after some discussion.
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by GMATGuruNY » Sun Mar 06, 2011 3:30 pm
chendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=

A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1

Source: OG 12

OA after some discussion.
Since none of the answer choices contains a √, we need to perform algebra that will clear all the √ symbols:

√(3-2x) = √2x + 1

[√(3-2x)]² = (√2x + 1)²

3 - 2x = (√2x + 1)(√2x + 1)

3 - 2x = 2x + √2x + √2x + 1

2 - 2x = 2x + 2√2x

1 - x = x + √2x

1 - 2x = √2x

(1 - 2x)² = (√2x)²

(1 - 2x)(1 - 2x) = 2x

1 - 2x - 2x + 4x² = 2x

4x² = 6x - 1.

The correct answer is E.
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by anshumishra » Sun Mar 06, 2011 3:40 pm
chendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=

A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1

Source: OG 12

OA after some discussion.
√(3-2x) - √2x = 1; square both sides
=> 3-2x + 2x - 2√(3-2x)(2x) = 1
=> 2√(3-2x)(2x) = 2
=> √(3-2x)(2x) = 1; square both sides
=> (3-2x)*2x = 1
=> 6x - 4x^2 = 1
=> 4x^2 = 6x - 1, E
Thanks
Anshu

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by chendawg » Wed Mar 09, 2011 8:33 am
OA is indeed E.

I'm not sure if I would be able to figure out to do another square under test conditions, is there a clue for future problems?
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by Scott@TargetTestPrep » Mon Apr 16, 2018 3:47 pm
chendawg wrote:If sq rt (3-2x) = sq rt (2x) + 1, then 4x^2=

A. 1
B. 4
C. 2 - 2x
D. 4x - 2
E. 6x - 1
Squaring both sides of the equation, we have:

3 - 2x = 2x + 1 + 2√(2x)

2 - 4x = 2√(2x)

1 - 2x = √(2x)

Squaring both sides of the equation again, we have:

1 + 4x^2 - 4x = 2x

4x^2 = 6x - 1

Answer:E

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