Q.Check whether \(\cfrac{5}{6}\) and \(\cfrac{4}{3}\) are roots of the equation \(x + \cfrac{1}{x} = \cfrac{13}{6}\).

Given the equation \(x + \cfrac{1}{x} = \cfrac{13}{6}\), Or, \(\cfrac{x^2 + 1}{x} = \cfrac{13}{6}\) Multiplying both sides by 6x: \[ 6x^2 + 6 = 13x \] \[ 6x^2 - 13x + 6 = 0 \] Now, substituting \(x = \cfrac{5}{6}\) into the equation: \[ 6\left(\cfrac{5}{6}\right)^2 - 13\left(\cfrac{5}{6}\right) + 6 = 6 \cdot \cfrac{25}{36} - \cfrac{65}{6} + 6 \] \[ = \cfrac{25}{6} - \cfrac{65}{6} + 6 = \cfrac{25 - 65 + 36}{6} = \cfrac{-4}{6} = -\cfrac{2}{3} \] [≠ 0; ∓ the equation is not satisfied by \(\cfrac{5}{6}\)] Now, substituting \(x = \cfrac{4}{3}\) into the equation: \[ 6\left(\cfrac{4}{3}\right)^2 - 13\left(\cfrac{4}{3}\right) + 6 = 6 \cdot \cfrac{16}{9} - \cfrac{52}{3} + 6 \] \[ = \cfrac{32}{3} - \cfrac{52}{3} + 6 = \cfrac{32 - 52 + 18}{3} = \cfrac{-2}{3} \] [≠ 0; ∓ the equation is not satisfied by \(\cfrac{4}{3}\)] Therefore, \(\cfrac{5}{6}\) and \(\cfrac{4}{3}\) cannot be roots of the quadratic equation \(x + \cfrac{1}{x} = \cfrac{13}{6}\).
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