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

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Manager
Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 118
Concentration: Finance, General Management
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If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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17 Jan 2011, 21:51
3
8
00:00

Difficulty:

75% (hard)

Question Stats:

60% (01:34) correct 40% (01:42) wrong based on 298 sessions

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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-root-3-2x-root-2x-1-then-4x-135539.html
Manager
Joined: 19 Sep 2010
Posts: 54
Location: Pune, India
Re: OG Diag #16 PS  [#permalink]

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18 Jan 2011, 02:04
3
1
tonebeeze wrote:
This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Hey,
below mentioned is my strategy to solve this problem,

Sqrt both side you will get,
3-2x= 2x+1+2sqrt{2x}
3-2x-2x-1= 2sqrt{2x}
2-4x= 2sqrt{2x}
devide by 2 both side
1-2x = sqrt{2x}
Again sqrt of both side
1+4x^2-4x= 2x
1+4x^2-6x=0
4x^2= 6x-1

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Re: OG Diag #16 PS  [#permalink]

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18 Jan 2011, 15:32
SoniaSaini wrote:
tonebeeze wrote:
This PS quadratics problem on the OG Diagnostic really is a time sapper. Any suggestions regarding the most efficient way to solve these problem types?

If $$\sqrt{3-2x} = \sqrt{2x} +1$$, then $$4x^2$$ =
(A) 1
(B) 4
(C) 2 − 2x
(D) 4x − 2
(E) 6x − 1

Hey,
below mentioned is my strategy to solve this problem,

Sqrt both side you will get,
3-2x= 2x+1+2sqrt{2x}
3-2x-2x-1= 2sqrt{2x}
2-4x= 2sqrt{2x}
devide by 2 both side
1-2x = sqrt{2x}
Again sqrt of both side
1+4x^2-4x= 2x
1+4x^2-6x=0
4x^2= 6x-1

Same way to solve it (about 2min).
Senior Manager
Joined: 08 Nov 2010
Posts: 347
Re: OG Diag #16 PS  [#permalink]

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18 Feb 2011, 15:14
can someone explain me this stage plz:

Sqrt both side you will get,
3-2x= 2x+1+2sqrt{2x}

thanks.
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Math Expert
Joined: 02 Sep 2009
Posts: 49496
Re: OG Diag #16 PS  [#permalink]

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18 Feb 2011, 16:00
6
4
144144 wrote:
can someone explain me this stage plz:

Sqrt both side you will get,
3-2x= 2x+1+2sqrt{2x}

thanks.

Apply $$(a+b)^2=a^2+2ab+b^2$$.

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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Re: OG Diag #16 PS  [#permalink]

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18 Feb 2011, 22:57
thanks. ur explanations is super clear. +1
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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25 Jun 2012, 13:05
Solving the equation, I arrive to this point:
$$2 - 2\sqrt{2x}$$$$=$$$$4x$$

However, there are more than one way to get $$4x^2$$. For example:
We multiply both sides by $$x$$:
$$2x - 2x\sqrt{2x}$$$$=$$$$4x^2$$

So, the answer not necessarily is $$6x -1$$

How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution?
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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26 Jun 2012, 03:37
2
1
metallicafan wrote:
Solving the equation, I arrive to this point:
$$2 - 2\sqrt{2x}$$$$=$$$$4x$$

How can we know how to manipulate the equation in order to arrive to one of the choices and not another valid solution?

After this point, you see that none of the options has a square root sign. You need to get rid of it to get to the answer. Hence, it is apparent that you need to square it again. You then take the square root term to one side and the rest of the terms to the other side and square both sides of the equation. This helps you get rid of the square root sign.
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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10 May 2013, 10:16
wouldnt [sqr(3-2x)]^2 give us +/- (3-2x)????
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Posts: 49496
Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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11 May 2013, 03:14
mokura wrote:
wouldnt [sqr(3-2x)]^2 give us +/- (3-2x)????

We have $$(\sqrt{3-2x})^2$$, which equals to $$3-2x$$, the same way as $$(\sqrt{x})^2=x$$.

If it were $$\sqrt{(3-2x)^2}$$, then it would equal to $$|3-2x|$$, the same way as $$\sqrt{x^2}=|x|$$.

Hope it's clear.

P.S. Open discussion of this question is here: if-root-3-2x-root-2x-1-then-4x-135539.html In case of any questions please post in this thread.
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 =  [#permalink]

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11 May 2013, 04:43
mokura wrote:
wouldnt [sqr(3-2x)]^2 give us +/- (3-2x)????

And one more thing,

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

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16 Jul 2018, 18:10
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Re: If rot{3-2x} = root(2x) +1, then 4x^2 = &nbs [#permalink] 16 Jul 2018, 18:10
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