If A = 5 + 2√6, then find the value of √A + \(\frac{1}{\sqrt{A}}\)

1.	If A = 5 + 2√6, then find the value of √A + \(\frac{1}{\sqrt{A}}\).
Answer» \(\sqrt{A}=\sqrt{5+2\sqrt6} = \sqrt{3+2+2\sqrt{3\times2}}\) = \(\sqrt{(\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3\sqrt2}\) = \(\sqrt{(\sqrt3) + (\sqrt2)^2} = \sqrt3+\sqrt2\) ∴ \(\sqrt{A} + \frac{1}{\sqrt{A}}=(\sqrt3+\sqrt2) + \frac{1}{(\sqrt3+\sqrt2)}\) = \((\sqrt3+\sqrt2) + \frac{\sqrt3-\sqrt2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\) = (\(\sqrt3+\sqrt2\)) + \(\frac{\sqrt3-\sqrt2}{3-2} = \sqrt3+\sqrt2+\sqrt3-\sqrt2= 2\sqrt3\).

If A = 5 + 2√6, then find the value of √A + \(\frac{1}{\sqrt{A}}\).

Answer»

\(\sqrt{A}=\sqrt{5+2\sqrt6} = \sqrt{3+2+2\sqrt{3\times2}}\)

= \(\sqrt{(\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3\sqrt2}\)

= \(\sqrt{(\sqrt3) + (\sqrt2)^2} = \sqrt3+\sqrt2\)

∴ \(\sqrt{A} + \frac{1}{\sqrt{A}}=(\sqrt3+\sqrt2) + \frac{1}{(\sqrt3+\sqrt2)}\)

= \((\sqrt3+\sqrt2) + \frac{\sqrt3-\sqrt2}{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}\)

= (\(\sqrt3+\sqrt2\)) + \(\frac{\sqrt3-\sqrt2}{3-2} = \sqrt3+\sqrt2+\sqrt3-\sqrt2= 2\sqrt3\).