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

Answer: E.
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 09 Jul 2012, 05:16
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
0 < 4x^2 < 4

Lets check the options now,
(A) 1 (0 < 4x^2 < 4)
(B) 4 (0 < 4x^2 < 4)
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but 0 < 4x^2 < 4)
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but 0 < 4x^2 < 4)
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and 0 < 4x^2 < 4)

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

Answer (E),

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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 09 Jul 2012, 06:26
cyberjadugar wrote:
Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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

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

Hi,

Put x=1,
LHS= 1,
RHS=2.414
or LHS < RHS

Put x=0,
LHS= 1.732
RHS=1
LHS>RHS,

Thus, 0< x < 1, for equality to hold true.
or 0 < 2x < 2, squaring,
0 < 4x^2 < 4

Lets check the options now,
(A) 1 (0 < 4x^2 < 4)
(B) 4 (0 < 4x^2 < 4)
(C)2 − 2x (0< x < 1, thus, 0< 2 − 2x < 2, but 0 < 4x^2 < 4)
(D) 4x − 2 (0< x < 1, thus, -2< 4x-2 < 2, but 0 < 4x^2 < 4)
(E) 6x − 1 (0< x < 1, thus, -1< 6x-1 < 5, and 0 < 4x^2 < 4)

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

Answer (E),

Regards,

Hi

Can we solve equations and get the answer?
Like taking \sqrt{2x} to right and squaring the both sides
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 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

Answer (E).

Regards,

PS: But that isn't much fun. :twisted:
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 13 Jul 2012, 01:42
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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.

Answer: E.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 29 Apr 2013, 05:40
Bunuel wrote:
If \sqrt{3-2x} = \sqrt{2x} +1, then 4x^2 =

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



Can someone help me solve this?

\sqrt{(3-2x)} = \sqrt{2x} + 1
\sqrt{(3-2x)} + 1 = \sqrt{2x}
If I square both sides,
2 - \sqrt{(3-2x)} = 2x
So, 4x^2 = 4 - (3-2x) - 2\sqrt{(3-2x)}

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

Please help
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 29 Apr 2013, 05:56
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sdas wrote:
Bunuel wrote:
If \sqrt{3-2x} = \sqrt{2x} +1, then 4x^2 =

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



Can someone help me solve this?

\sqrt{(3-2x)} = \sqrt{2x} + 1
\sqrt{(3-2x)} + 1 = \sqrt{2x}
If I square both sides,
2 - \sqrt{(3-2x)} = 2x
So, 4x^2 = 4 - (3-2x) - 2\sqrt{(3-2x)}

I am unable to reach to E

However, if I squared the main problem without shifting 1, then I am reaching to E

Please help


You could solve this way too:

\sqrt{3-2x} = \sqrt{2x} +1

Re-arrange: \sqrt{3-2x}-1 = \sqrt{2x};

Square: (3-2x)-2\sqrt{3-2x}+1=2x;

Re-arrange and reduce by 2: \sqrt{3-2x}=2-2x;

Square: 3-2x=4-8x+4x^2

4x^2=6x-1.

Hope it's clear.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 29 Apr 2013, 07:15
Thanks Bunuel. I missed the last 2 steps....
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 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...

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

Thanks in advance!


Must know property: (a+b)^2=a^2+2ab+b^2.

Check the following link for more: algebra-101576.html

Hope this helps.
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 17 Apr 2014, 02:20
Bunuel wrote:
If \sqrt{3-2x} = \sqrt{2x} +1, then 4x^2 =

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

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Question: 16
Page: 22
Difficulty: 600


\sqrt{3-2x} = \sqrt{2x} +1

\sqrt{3-2x} - \sqrt{2x} = 1

Squaring both sides

3 - 2x + 2 \sqrt{(3-2x) . 2x} + 2x = 1

3 + \sqrt{24x - 16x^2} = 1

24x - 16x^2 = 4

16x^2 = 24x - 4

4x^2 = 6x - 1

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

I hope you understand what I'm asking. :wink:

Thanks for any help!
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Re: If root{3-2x} = root(2x) +1, then 4x^2 = [#permalink] New post 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.

I hope you understand what I'm asking. :wink:

Thanks for any help!


First of all please read this:

rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum)

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.

There are two ways of solving this given above:
if-root-3-2x-root-2x-1-then-4x-135539.html#p1104029
if-root-3-2x-root-2x-1-then-4x-135539.html#p1218265
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Re: If root{3-2x} = root(2x) +1, then 4x^2 =   [#permalink] 31 Jul 2014, 06:23
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