Q.If \(x = 3 + 2\sqrt{2}\), then find the value of \(\left(\sqrt{x} + \cfrac{1}{\sqrt{x}}\right)\).

Given: \(x = 3 + 2\sqrt{2}\) \(= 2 + 1 + 2\sqrt{2}\) \(= (\sqrt{2} + 1)^2\) \(\therefore \sqrt{x} = \sqrt{2} + 1\) And, \(\cfrac{1}{\sqrt{x}} = \cfrac{1}{\sqrt{2} + 1}\) \(= \cfrac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)}\) \(= \cfrac{\sqrt{2} - 1}{2 - 1}\) \(= \sqrt{2} - 1\) \(\therefore \left(\sqrt{x} + \cfrac{1}{\sqrt{x}}\right) = \sqrt{2} + 1 + \sqrt{2} - 1 = 2\sqrt{2}\)
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