Q.If \(2x = \sqrt{5} + 1\), then show that \(x^2 - x - 1 = 0\).

\(2x = \sqrt{5} + 1\) i.e., \(x = \cfrac{\sqrt{5}}{2} + \cfrac{1}{2}\) \(\therefore x^2 - x - 1\) \(= \left(\cfrac{\sqrt{5}}{2} + \cfrac{1}{2}\right)^2 - \left(\cfrac{\sqrt{5}}{2} + \cfrac{1}{2}\right) - 1\) \(= \cfrac{5}{4} + \cfrac{1}{4} + \cfrac{2 \times \sqrt{5} \times 1}{2 \times 2} - \cfrac{\sqrt{5}}{2} - \cfrac{1}{2} - 1\) \(= \cfrac{6}{4} + \cfrac{\sqrt{5}}{2} - \cfrac{\sqrt{5}}{2} - \cfrac{3}{2}\) \(= \cfrac{3}{2} - \cfrac{3}{2} = 0\) (Proved)
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