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If root{3-2x} = root(2x) +1, then 4x^2 =

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If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 05 Apr 2011, 19:34
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-root-3-2x-root-2x-1-then-4x-135539.html

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 05 Apr 2011, 20:26
=>(3-2x)^1/2=(2x)^1/2+1

Square both sides

=>(3-2x)=(2x)+1+2*(2x)^1/2

=>(2-4x)=2*(2x)^1/2

=>(1-2x)=(2x)^1/2

Square both sides again

1 - 4x + 4x^2 = 2x
4x^2=6x-1
So E.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 05 Apr 2011, 20:28
GMATD11 wrote:
16.) \sqrt{3-2x}=\sqrt{2x}+1 then 4x^2 =

a) 1
b) 4
c) 2-2x
d) 4x-2
e) 6x-1

Guys i solved and got 1

but OA is different .

I want to know the reason nd different approach that i need to follow.



Alternately, to verify your ans 4x^2 = 1 means x = 1/2
Substitute the value 1/2 in eq, it doesn't make LHS=RHS
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 05 Apr 2011, 21:34
root(3-2x) - root(2x) = 1

Squaring both sides :


3 - 2x + 2x - 2 * root(6x - 4x^2) = 1

=> -2 * root(6x - 4x^2) = -2

=> root(6x - 4x^2) = 1

Squaring again :

=> 6x - 4x^2 = 1

=> 4x^2 = 6x - 1


Answer - E
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 06 Apr 2011, 05:17
Squaring on both sides we have
2-4x = 2*sqrt(2x)
Squaring on both sides again we have
4x^2 = 6x-1

Answer E

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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 21 Jun 2014, 00:33
Before proceeding with the solving part , give a look at the options provided. Since none of the options has under root involved in the answer choices think to eliminate that from the target answer.

If u keep the squareroot value on one side w/o making any equation will help us to solve.
squaring both sides and bringing the non squareroot values on the one side and keeping squareroot 2 on the other side .. squaring the terms on both sides ... will give answer ....

let me know in case of any query ...
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]

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New post 21 Jun 2014, 02:59
GMATD11 wrote:
If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1



If \(\sqrt{3-2x} = \sqrt{2x} +1\), then \(4x^2\) =

(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides: \((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) --> \(3-2x=2x+2*\sqrt{2x}+1\) --> rearrange so that to have root at one side: \(2-4x=2*\sqrt{2x}\) --> reduce by 2: \(1-2x=\sqrt{2x}\) --> square again: \((1-2x)^2=(\sqrt{2x})^2\) --> \(1-4x+4x^2=2x\) --> rearrange again: \(4x^2=6x-1\).

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-root-3-2x-root-2x-1-then-4x-135539.html
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 21 Jun 2014, 02:59
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