• # question_answer $\alpha ,\ \beta$ are the roots of the equation ${{x}^{2}}-3x+a=0$ and $\gamma ,\ \delta$ are the roots of the equation ${{x}^{2}}-12x+b=0$. If $\alpha ,\ \beta ,\ \gamma ,\ \delta$ form an increasing G.P., then $(a,\ b)=$  [DCE 2000] A) (3, 12) B) (12, 3) C) (2, 32) D) (4, 16)

Since $\alpha ,\ \beta ,\ \gamma ,\ \delta$ form an increasing G.P., so $\alpha \delta =\beta \gamma$where$\alpha <\beta <\gamma <\delta$. On solving ${{x}^{2}}-3x+a=0$, we get $x=\frac{1}{2}(3\pm \sqrt{9-4a})$. Also $\alpha <\beta$. Hence  $\alpha =\frac{1}{2}(3-\sqrt{9-4a}),\ \beta =\frac{1}{2}(3+\sqrt{9-4a})$ Similarly from ${{x}^{2}}-12x+b=0$, we get $\gamma =\frac{1}{2}(12-\sqrt{144-4b}),\ \delta =\frac{1}{2}(12+\sqrt{144-4b})$ Substituting these values of $\alpha ,\ \beta ,\ \gamma ,\ \delta$ in $\alpha \delta =\beta \gamma$ and simplifying, we get$(a,\ b)=(2,\ 32)$. Trick: Check the alternates; only (c) satisfies the condition.