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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

\(\sqrt{3-2x} = \sqrt{2x} +1\) --> 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: \(4x^2=6x-1\).

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

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09 Jul 2012, 21:48

Hi narangvaibhav,

Yes definitely, we can use the conventional method too, \(\sqrt{3-2x} = \sqrt{2x} +1\) Squaring both the sides, \(3-2x=2x+1+2 \sqrt{2x}\) or \(1-2x=\sqrt{2x}\) Again, squaring both the sides, \(1+4x^2-4x=2x\) or \(4x^2=6x-1\)

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

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

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15 Dec 2013, 11:46

Hi guys,

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

can someone please explain why (root(2x) +1)^2 is not 2x +1 but 2x+1+2 root(2x). I thought I would just take every term to the secound power, such as root(2x)^2 + 1^2...

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

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31 Jul 2014, 04:31

Hi,

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

I have one question. I got to the equation 1 = 2x +\sqrt{2x} but then I squared 1- \sqrt{2x} = 2x to get to 4x^2 easily but I got stuck with the left term. So my question is how can I get the solution with this approach and additionally why did you guys square the term the other way round? It makes much more sense to me to square 2x in order to arrive at 4x^2 instead of squaring 1-2x.

Next, if you square \(1-\sqrt{2x}=2x\) you'll get \(1 - 2\sqrt{2x}+2x=4x^2\). As you can see we don't have this answer among the options, so we should have done the other way around.

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

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29 Oct 2014, 23:47

Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

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

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30 Oct 2014, 00:05

ColdSushi wrote:

Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

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

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30 Oct 2014, 04:02

PareshGmat wrote:

ColdSushi wrote:

Hi guys,

I'm new to this community!

Wondering why Sqrt(3-2x)^2 got expanded to [Sqrt(3-2x)][Sqrt(3+2x)] but with (1-2x)^2 it got expanded to (1-2x)(1-2x)? I.e. the postive and negative signs. I thought whenever we get something in the form of (x-y)^2, the expansion will always be (x-y)(x+y) not (x-y)(x-y)?

Thanks a mil!

\((\sqrt{3-2x})^2 = 3-2x\)

\((1-2x)^2 = (1-2x)(1-2x)\)

\((x-y)^2 = x^2 - 2xy + y^2\)

\(x^2 - y^2 = (x+y)(x-y)\)

Hope this clarifies

Wow - I totally mixed up the rules - thanks for clarifying.

Thus, only possible option which lies in the range is E.

Answer (E),

Regards,

Hi, Could you please explain how you got into this 0<x<1 from the LHS and RHS? Also, the calculations of highlighted portion in the answer choice E. TIA

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

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08 Feb 2015, 06:41

1

This post was BOOKMARKED

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.

sorry but a noob question here, i thought we aren't allowed to square both sides like this?

but instead all we can do is just to multiply both sides or divide both sides by the same number/objects etc.

cause when you square both sides, both sides is multiplying with different things right???

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