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

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

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New post 04 Jul 2020, 04:44
1
Bunuel wrote:
SOLUTION

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.


Thank you for the overview.

When I tackled this question the first time, I wondered whether I need to put everything what was under the radical as absolute number as I need to limit it to positive numbers only.

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> square both sides:
\((\sqrt{3-2x})^2 =(\sqrt{2x} +1)^2\) --> \(|3-2x|=|2x|+2*\sqrt{2x}+1\) --

Since the math afterwards did not look funny, I dumped the question. Could someone please explain what is wrong in my logic?
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Re: If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =  [#permalink]

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New post 25 Jul 2020, 08:21
How do we know that we should use equation/algebra to solve this question? by looking at answer choices to realize that not all choices are number?
Bunuel wrote:
SOLUTION

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.
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Re: If (3 - 2x)^(1/2) = (2x)^(1/2) + 1, then 4x^2 =   [#permalink] 25 Jul 2020, 08:21

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