Bunuel
What is the value of \((\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) ?
A. -4
B. -2
C. -1
D. 1
E. 2
Fresh GMAT Club Tests' Question Solution: Let us assume \(a=(\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\)
Now, we need the value of a
If you notice, we have a square root over \(4-\sqrt{15}\) which if removed can be helpful. So, let us square both sides
\(⇒a^2=(\sqrt{4-\sqrt{15}})^2\times (4 + \sqrt{15})^2\times (\sqrt{6} - \sqrt{10})^2\)
\(⇒a^2=(4-\sqrt{15})\times(4+\sqrt{15})\times(4+\sqrt{15})\times(\sqrt{6}-\sqrt{10})^2\)
\(⇒a^2=(4^2-(\sqrt{15})^2)\times (4+\sqrt{15})\times (6+10-2\sqrt{60})\) (couple of identities \((a+b)(a-b)=a^2-b^2\) and \((a-b)^2=a^2+b^2-2ab)\)
\(⇒a^2=(16-15)\times (4+\sqrt{15})\times (16-2\sqrt{60})\)
\(⇒a^2=(4+\sqrt{15})\times (16-4\sqrt{15})\)
\(⇒a^2=(4+\sqrt{15})\times 4(4-\sqrt{15})\)
\(⇒a^2=4\times (4+\sqrt{15})(4-\sqrt{15})\)
\(⇒a^2=4(4^2-(\sqrt{15})^2)\)
\(⇒a^2=4\)
\(⇒a=-2\)
Note: we are ignoring \(a=2\) possibility because if you observe \(a=(\sqrt{4-\sqrt{15}})(4 + \sqrt{15})(\sqrt{6} - \sqrt{10})\) and the term \((\sqrt{6} - \sqrt{10})\) is negative and other two terms \((\sqrt{4-\sqrt{15}})\) and \((4 + \sqrt{15})\) are positive, so the ultimate result will be negative
Hence the right answer is
Option B