Q.If \[ x = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} \quad \text{and} \quad xy = 1 \] then show that \[ \frac{x^2 + xy + y^2}{x^2 - xy + y^2} = \frac{12}{11} \]

Since \(xy = 1\), \[ \therefore y = \frac{1}{x} = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} \] \[ \therefore x + y = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}} + \frac{\sqrt{7} - \sqrt{3}}{\sqrt{7} + \sqrt{3}} \] \[ = \frac{(\sqrt{7} + \sqrt{3})^2 + (\sqrt{7} - \sqrt{3})^2}{(\sqrt{7} - \sqrt{3})(\sqrt{7} + \sqrt{3})} \] \[ = \frac{2[(\sqrt{7})^2 + (\sqrt{3})^2]}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{2(7 + 3)}{7 - 3} = \frac{20}{4} = 5 \] Now, \[ \frac{x^2 + xy + y^2}{x^2 - xy + y^2} = \frac{x^2 + y^2 + xy}{x^2 + y^2 - xy} = \frac{(x + y)^2 - 2xy + xy}{(x + y)^2 - 2xy - xy} \] \[ = \frac{(x + y)^2 - xy}{(x + y)^2 - 3xy} = \frac{5^2 - 1}{5^2 - 3 \times 1} = \frac{25 - 1}{25 - 3} = \frac{24}{22} = \frac{12}{11} \quad \text{[Proved]} \]
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