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If x>0 then 1/[(root(2x)+root(x)]=?

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If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 14 Dec 2010, 18:08
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If x>0 then \frac{1}{\sqrt{2x}+\sqrt{x}}=?

A. \frac{1}{\sqrt{3x}}

B. \frac{1}{2\sqrt{3x}}

C. \frac{1}{x\sqrt{2}}

D. \frac{\sqrt{2}-1}{\sqrt{x}}

E. \frac{1+\sqrt{2}}{\sqrt{x}}

I have a doubt about the ans. Can someone explain this.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 16 Mar 2012, 04:30, edited 1 time in total.
Edited the question and the answer choices.
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 14 Dec 2010, 18:52
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.


Hi!

unless you're missing exponents, then the answer should be (A).

v(2x) is the same as 2vx. So, if we're adding 2vx + 1vx, we get 3vx.

Of course, v(3x) is the same as 3vx, so (A) is correct.

Are you sure you've reproduced the question correctly?
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 14 Dec 2010, 18:55
the question is correct but i guess the oa is wrong. thanks
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 14 Dec 2010, 19:00
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ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.


Square roots in denominator should remind you of difference of squares.
\frac{1}{{\sqrt{2x}+\sqrt{x}}}

Multiply and divide by {\sqrt{2x}-\sqrt{x}

You get \frac{1}{{\sqrt{2x}+\sqrt{x}}} * ({\sqrt{2x}-\sqrt{x}})/({\sqrt{2x}-\sqrt{x}})

You get ({\sqrt{2x}-\sqrt{x})/x
Taking root x common from the numerator, we get ({\sqrt{2}-1)/\sqrt{x}

Answer (D)
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 14 Dec 2010, 19:08
Oh.. if "v" is supposed to be "root of", then my response is completely wrong. Please use the math fonts for roots (or say "root").
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 15 Dec 2010, 01:43
Expert's post
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

a. 1/v(3x)
b.1/[2v(2x)]
c.1/(xv2)
d. v2-1/vx
e. 1+v2/vx

I have a doubt about the ans. Can someone explain this.


ajit257 please make sure the stem and the answer choices are not ambiguous.

As OA is given as D then I guess v means square root, so the question should be:

If x>0 then \frac{1}{\sqrt{2x}+\sqrt{x}}=?

Factor out \sqrt{x} --> \frac{1}{\sqrt{x}(\sqrt{2}+1)} --> multiply and divide by \sqrt{2}-1 --> \frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}.

Answer: D.
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 15 Dec 2010, 04:16
firstly apologies.....well it was hard to analyse the question in the first place the way it was written...I would not have got it wrong if it was a sq rt....the ans choice led me to ask the question...but thanks guys at-least i got the right question now.
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 16 Mar 2012, 04:22
VeritasPrepKarishma, thank you
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 17 Mar 2012, 06:20
Bunuel wrote:
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

Factor out \sqrt{x} --> \frac{1}{\sqrt{x}(\sqrt{2}+1)} --> multiply and divide by \sqrt{2}-1 --> \frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}.

Answer: D.



Could you please explain why you multiply and divide by \sqrt{2}-1 ...whats the concept here?

Thanks!
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Re: if x>0 then 1/[v(2x) + vx] [#permalink] New post 17 Mar 2012, 07:04
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Impenetrable wrote:
Bunuel wrote:
ajit257 wrote:
Q. if x>0 then 1/[v(2x) + vx] = ?

Factor out \sqrt{x} --> \frac{1}{\sqrt{x}(\sqrt{2}+1)} --> multiply and divide by \sqrt{2}-1 --> \frac{\sqrt{2}-1}{\sqrt{x}(\sqrt{2}^2-1^2)}=\frac{\sqrt{2}-1}{\sqrt{x}}.

Answer: D.



Could you please explain why you multiply and divide by \sqrt{2}-1 ...whats the concept here?

Thanks!


It's called rationalisation and is performed to eliminate irrational expression in the denominator. For this particular case we are doing rationalisation by applying the following rule: (a-b)(a+b)=a^2-b^2.

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

Hope it' clear.
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 20 Mar 2012, 09:30
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can you please explain last step . how did you get root x in the numerator
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 20 Mar 2012, 20:26
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can you please explain last step . how did you get root x in the numerator


\frac{1}{\sqrt{2x} + \sqrt{x}}*\frac{\sqrt{2x} - \sqrt{x}}{\sqrt{2x} - \sqrt{x}} = \frac{\sqrt{2x} - \sqrt{x}}{x}
= \frac{\sqrt{x}(\sqrt{2} - 1)}{\sqrt{x}*\sqrt{x}}
(Take \sqrt{x} common from the two terms of the numerator. Split the x in the denominator into \sqrt{x}*\sqrt{x}

Now cancel the \sqrt{x} from the numerator and denominator.

You get \frac{(\sqrt{2} - 1)}{\sqrt{x}}
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 23 Mar 2012, 03:54
thanks for the explanations. they are very helpful.
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 26 Jul 2012, 03:08
ajit257 wrote:
If x>0 then \frac{1}{\sqrt{2x}+\sqrt{x}}=?

A. \frac{1}{\sqrt{3x}}

B. \frac{1}{2\sqrt{3x}}

C. \frac{1}{x\sqrt{2}}

D. \frac{\sqrt{2}-1}{\sqrt{x}}

E. \frac{1+\sqrt{2}}{\sqrt{x}}

I have a doubt about the ans. Can someone explain this.



Just by rationalising the denominator we get D
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 01 Sep 2014, 18:31
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Re: If x>0 then 1/[(root(2x)+root(x)]=? [#permalink] New post 01 Sep 2014, 20:21
rajathpanta wrote:
ajit257 wrote:
If x>0 then \frac{1}{\sqrt{2x}+\sqrt{x}}=?

A. \frac{1}{\sqrt{3x}}

B. \frac{1}{2\sqrt{3x}}

C. \frac{1}{x\sqrt{2}}

D. \frac{\sqrt{2}-1}{\sqrt{x}}

E. \frac{1+\sqrt{2}}{\sqrt{x}}

I have a doubt about the ans. Can someone explain this.



Just by rationalising the denominator we get D



Just adding a small note as all guys above have explained.......

Rationalising would change the middle sign in the problem for the numerator

Only option D fits in...
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Re: If x>0 then 1/[(root(2x)+root(x)]=?   [#permalink] 01 Sep 2014, 20:21
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