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(root2*root5)/[(root7-root2-root5)(root7+root2+root5)]

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(root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 09 Dec 2015, 08:48
11
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

65% (01:58) correct 35% (01:49) wrong based on 338 sessions

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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 27 Jan 2016, 23:15
4
3
Bunuel wrote:
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum



The question tests our use of algebraic identities. There are a few things in the expression that give us hints on how to proceed.

Note the \(\sqrt{2}*\sqrt{5}\) in the numerator. This reminds us of \(2ab\) terms in \((a+b)^2\) and \((a-b)^2\). There are three terms in each factor of the denominator but one term has only positive signs while the other has negative signs too. This reminds us of \(a^2 - b^2\) expansion.

\((\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5}) =[\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})]\)

Here \(a = \sqrt{7}\)

\(b = (\sqrt{2}+\sqrt{5})\)

\((a + b)(a - b) = a^2 - b^2\)

\([\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2\)

Now use \((x + y)^2 = x^2 + y^2 + 2xy\) on \((\sqrt{2}+\sqrt{5})^2\)

\((\sqrt{2}+\sqrt{5})^2 = \sqrt{2}^2 + \sqrt{5}^2 + 2*\sqrt{2}*\sqrt{5}= 7 + 2*\sqrt{2}*\sqrt{5}\)

Plugging this value back in the denominator:

\([\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = 7 - ( 7 + 2*\sqrt{2}*\sqrt{5})\)

\([\sqrt{7}-(\sqrt{2}+\sqrt{5})][\sqrt{7}+(\sqrt{2}+\sqrt{5})] = - 2*\sqrt{2}*\sqrt{5}\)

Plugging this value of denominator back in the fraction:

\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= \frac{\sqrt{2}*\sqrt{5}}{-2*\sqrt{2}*\sqrt{5}} = \frac{-1}{2}\)

Answer (C)
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 09 Dec 2015, 09:23
1
Bunuel wrote:
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum



\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)...
lets solve the denominator...
\({(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2\)..
this is equal to \(-2*\sqrt{10}\)...
therefore the fraction becomes \(\sqrt{10}/2*\sqrt{10}\)= -1/2
C
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 09 Dec 2015, 09:59
chetan2u wrote:
Bunuel wrote:
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum



\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)...
lets solve the denominator...
\({(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2\)..
this is equal to \(-2*\sqrt{10}\)...
therefore the fraction becomes \(\sqrt{10}/2*\sqrt{10}\)= -1/2
C




for some crazy reason i am getting -20 as my answer. Is there a different sign in between? Coz, i also got (-2 Sqroot 10) when i simplified the parenthesis.
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 09 Dec 2015, 22:44
rohan89 wrote:
chetan2u wrote:
Bunuel wrote:
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum



\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)...
lets solve the denominator...
\({(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}= (\sqrt{7})^2 - (\sqrt{2}+\sqrt{5})^2\)..
this is equal to \(-2*\sqrt{10}\)...
therefore the fraction becomes \(\sqrt{10}/2*\sqrt{10}\)= -1/2
C




for some crazy reason i am getting -20 as my answer. Is there a different sign in between? Coz, i also got (-2 Sqroot 10) when i simplified the parenthesis.


if you too have got the denominator as \(-2\sqrt{10}\), the numerator is\(\sqrt{2}*\sqrt{5}\)=\(\sqrt{10}\)...
strike off the common term \(\sqrt{10}\)..
what is left is -\(\frac{1}{2}\)
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 16 Jan 2016, 09:33
(√7 - √2 - √5)*( √7+√2+√5) = - (√2+√5 - √7) *( √7+√2+√5) = - [(√2+√5)^2 – 7]
=> -(2 + 5 + 2√10 – 7) = -2√10
√2*√5/ -2√10 = -1/2
Answer C.
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 27 Jan 2016, 20:43
Bunuel wrote:
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)

A. -2
B. -1
C. -1//2
D. 1/2
E. 1

Source: GMATQuantum



Bunuel...Do we have any formula to deal with the denominator?

I solved by multiplying each term....If there is any quick way, please let us know

Thanks,
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]  [#permalink]

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New post 24 Dec 2018, 07:32
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Re: (root2*root5)/[(root7-root2-root5)(root7+root2+root5)]   [#permalink] 24 Dec 2018, 07:32
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