First slide

Theory of equations

Question

Let $x_{1}, x_{2}$ be the roots of the equation $x^{2} - 3 x + p = 0$ and let $x_{3}, x_{4}$ be the roots of the equation $x^{2} - 12 x + q = 0$ .If the number $x_{1}, x_{2}, x_{3}, x_{4}$ (in order) form an increasing G.P., then

Moderate

A

$p = 2, q = 16$
B

$p = 2, q = 32$
C

$p = 4, q = 16$
D

$p = 4, q = 32$

Solution

It is given that
$x_{1}, x_{2}$ are roots of $x^{2} - 3 x + p = 0$
$\Rightarrow x_{1} + x_{2} = 3, x_{1} x_{2} = p$
$x_{3}, x_{4}$ are roots of $x^{2} - 12 x + q = 0$
$\Rightarrow x_{3} + x_{4} = 12$ and $x_{3} x_{4} = q$
It is given that $x_{1}, x_{2}, x_{3}, x_{4}$ form an increasing G.P. Therefore,
$x_{1} = a, x_{2} = a r, x_{3} = a r^{2}, x_{4} = a r^{3},$ where $r > 1$
Now ,
$\begin{array}{cc} x_{1} + x_{2} & = 3 & \Rightarrow & a (1 + r) & = 3 \\ x_{3} + x_{4} & = 12 & \Rightarrow & a r^{2} (1 + r) & = 12 \end{array}\} \Rightarrow r = 2$ and $a = 1$
$∴ x_{1} = 1, x_{2} = 2, x_{3} = 4, x_{4} = 8$
Thus , $p = x_{1} x_{2} = 2$ and $q = x_{3} x_{4} = 32 .$

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