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

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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 ....

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: \(4x^2=6x-1\).

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