Question

Let $x_1, x_2$ be the roots of $x^2 - 3x + a = 0$ and $x_3, x_4$ be the roots of $x_2 - 12x + b = 0$. If $x_1 < x_2 < x_3 < x_4$ and $x_ 1 , x_2, x_3, x_4$ are in G.P. then ab equals

• A. $\frac{24}{5}$
• B. 64
• C. 16
• D. 8

Answer 1

Answer: (B)

$x_{1} + x_{2} = 3; x_{1}.x_{2} = a$
$x_{3} + x_{4} = 12; x_{3}.x_{4} = b$
Let r be the common ratio of G.P., then
$\frac{x_{1}\left(1+r\right)}{x_{1}r^{2}\left(1+r\right)} = \frac{3}{12}\,\Rightarrow r = \pm2$
Take $r = 2$ ($\because$ G.P. is increasing)
$\because x_{1} +x_{2} = 3 \quad\Rightarrow x_{1}\left(1+r\right) = 3\quad\Rightarrow x_{1} = 1$
$\therefore ab = x_{1} x_{2} x_{3} x_{4} = 1.2.4.8 = 64$