If x = \(\sqrt {1 + \frac{\sqrt {3}}{2}}\) - \(\sqrt {1 - \frac{\sqrt {3}}{2}}\),
then find value of \(\frac{\sqrt {2}\;-\;x}{\sqrt {2}\;+\;x}\).
Answer & explanation
Correct answer: option 2
x = \(\sqrt {1+\frac{\sqrt {3}}{2}}\) - \(\sqrt {1-\frac{\sqrt {3}}{2}}\)
Multiply and divide by 2 for making perfect square
x = \(\sqrt {1+\frac{2\sqrt {3}}{4}}\) - \(\sqrt {1-\frac{2\sqrt {3}}{4}}\)
x = \(\sqrt {\frac{{(\sqrt {3}+1)}^{2}}{4}}\) - \(\sqrt {\frac{{(\sqrt {3}-1)}^{2}}{4}}\)
x = \(\frac{\sqrt {3}+1}{2}\) - \(\frac{\sqrt {3}+1}{2}\)
x = 1
Put & find
⇒ \(\frac{\sqrt {2}\;-\;x}{\sqrt {2}\;+\;x}\) = \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;+\;1}\) = \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;+\;1}\) × \(\frac{\sqrt {2}\;-\;1}{\sqrt {2}\;-\;1}\) = (\(\sqrt {2}\) - 1)2 = 3 - 2\(\sqrt {2}\)