If a = \(\frac{\sqrt {5}}{3}\) then find the value of \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {5}}{3}\)?
Answer & explanation
Correct answer: option 3
a = \(\frac{\sqrt {5}}{3}\)
1 + a = 1 + \(\frac{\sqrt {5}}{3}\) = \(\frac{3 + \sqrt {5}}{3}\)
= \(\frac{6 + 2\sqrt {3}}{6}\) (multiply & divide by 2)
= \(\frac{(\sqrt {5} + 1)^2}{6}\)
Similarly,
1 - a = \(\frac{(\sqrt {5} - 1)^2}{6}\)
Now,
⇒ \(\sqrt {1 + a}\) + \(\sqrt {1 - a}\) + \(\frac{\sqrt {5}}{3}\) = \(\sqrt {(\frac{\sqrt {3} + 1)^2}{4}}\) + \(\sqrt {(\frac{\sqrt {3} - 1)^2}{4}}\) + \(\frac{\sqrt {3}}{2}\)
= \(\frac{\sqrt {5}\;+\;1}{\sqrt{6}}\) + \(\frac{\sqrt {5}\;-\;1}{\sqrt{6}}\) + \(\frac{\sqrt {5}}{3}\)
= \(\frac{\sqrt {5}\;+\;1\;+\sqrt {3}\;-\;1}{\sqrt{6}}\) + \(\frac{\sqrt {5}}{3}\)
= \(\frac{2\sqrt{5}}{\sqrt{6}}\) + \(\frac{\sqrt {5}}{3}\)
=\(\sqrt{5}\) \(\frac{6 \;+ \; \sqrt{6}}{3\sqrt{6}}\)