Given \(\sqrt{x} + \dfrac{1}{\sqrt{x}} = 4\), then what is \(x + \dfrac{1}{x}\)?
14
Let \(u = \sqrt{x} + \dfrac{1}{\sqrt{x}} = 4\).
Square both sides: \(u^2 = x + 2 + \dfrac{1}{x} = 16\)
Therefore \(x + \dfrac{1}{x} = 16 - 2 = \mathbf{14}\)
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