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15 Aug 2023, 03:00
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Difficulty:
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Question Stats:
82% (01:44) correct
18%
(02:17)
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Which of the following is equal to \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)?
A. \(2+\sqrt{3}\)
B. \(2-\sqrt{3}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\frac{\sqrt{6} + \sqrt{2}}{4}\)
E. \(\frac{7 - 2\sqrt{6}}{5}\)
A. \(2+\sqrt{3}\)
B. \(2-\sqrt{3}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\frac{\sqrt{6} + \sqrt{2}}{4}\)
E. \(\frac{7 - 2\sqrt{6}}{5}\)
15 Aug 2023, 03:09
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Bunuel
\(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)
Multiply the numerator and denominator by (\(\sqrt{6} - \sqrt{2}\))
\(\frac{(\sqrt{6} - \sqrt{2}) (\sqrt{6}) - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}\)
\(\frac{(\sqrt{6} - \sqrt{2})^2}{6-2}\)
\(\frac{6+2-2\sqrt{12}}{6-2}\)
\(\frac{8-4\sqrt{3}}{4}\)
\(\frac{4(2-\sqrt{3})}{4}\)
\(2-\sqrt{3}\)
Option B
15 Aug 2023, 05:37
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For anyone wondering how gmatophobia knew to multiply the numerator and denominator by \(\sqrt{6} - \sqrt{2}\), a sign that that approach will work is that all of the answer choices have integer denominators.
So, it makes sense to see what happens if you multiply \(\sqrt{6} + \sqrt{2}\) in the denominator by \(\sqrt{6} - \sqrt{2}\) to get \(\sqrt{6}^2 - \sqrt{2}^2 = 6 - 2 = 4\), which is an integer value, in the denominator.
So, it makes sense to see what happens if you multiply \(\sqrt{6} + \sqrt{2}\) in the denominator by \(\sqrt{6} - \sqrt{2}\) to get \(\sqrt{6}^2 - \sqrt{2}^2 = 6 - 2 = 4\), which is an integer value, in the denominator.
