$I f x_{1} + x_{2} = 3, x_{3} + x_{4} = 12 a n d x_{1}, x_{2}, x_{3}, x_{4} a r e i n i n c r e a \sin g G . P ., t h e n t h e e q u a t i o n h a v i n g x_{1,} x_{2} a s r o o t s i s$

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$x^{2} - 3 x + 2 = 0$

$x^{2} - 9 x + 8 = 0$

$x^{2} - 12 x + 32 = 0$

$x^{2} - 6 x + 8 = 0$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t x_{1} = a, x_{2} = a r, x_{3} = a r^{2}, x_{4} = a r^{3} \\ G i v e n x_{1} + x_{2} = 3 \Rightarrow a + a r = 3 \Rightarrow a (a + r) = 3 - - (1) \\ a n d x_{3} + x_{4} = 12 \Rightarrow a r^{2} + a r^{3} = 12 \Rightarrow a r^{2} (1 + r) = 12 - - (2) \\ \frac{(2)}{(1)} \Rightarrow r^{2} = 4 \Rightarrow r = 2 \\ f r o m (1) a (1 + 2) = 3 \Rightarrow a = 1 \\ ∴ x_{1} = 1, x_{2} = 2 \\ Re q u i r e d E q u a t i o n i s x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2} = 0 \\ \Rightarrow x^{2} - 3 x + 2 = 0 \end{array}$