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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
09 Jul 2012, 02:23

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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.

Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
09 Jul 2012, 20:48

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

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
13 Jul 2012, 01:42

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

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.

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
15 Dec 2013, 10:46

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...

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
16 Dec 2013, 00:18

1

This post received KUDOS

Expert's post

steilbergauf wrote:

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...

If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
31 Jul 2014, 03:31

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.

Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink]
31 Jul 2014, 06:23

Expert's post

pipe19 wrote:

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.

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.