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![(root2*root5)/[(root7-root2-root5)(root7+root2+root5)]](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "(root2*root5)/[(root7-root2-root5)(root7+root2+root5)]"){kind=icon}
09 Dec 2015, 08:48
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
70% (01:59) correct
30%
(01:58)
wrong
based on 965
sessions
History
Date
Time
Result
Not Attempted Yet
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)
A. -2
B. -1
C. -1//2
D. 1/2
E. 1
Source: GMATQuantum
A. -2
B. -1
C. -1//2
D. 1/2
E. 1
Source: GMATQuantum
27 Jan 2016, 23:15
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Bunuel
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)
General Discussion
09 Dec 2015, 09:23
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Bunuel
\(\frac{\sqrt{2}*\sqrt{5}}{(\sqrt{7}-\sqrt{2}-\sqrt{5})(\sqrt{7}+\sqrt{2}+\sqrt{5})}=\)...
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
