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)