Indian Institute of Foreign Trade 2008 Solved Paper

The given equation is $\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}=0$
Roots of the given quadratic equation are $-\mathit{b}\pm \frac{\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}}{2\mathit{a}}$
We know that one of the 2 roots is double the other.
$-\mathit{b}+\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}-\mathit{b}-\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}$
Therefore, $2\mathit{a}=2\ast 2\mathit{a}$
$\Rightarrow -\mathit{b}+\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}=-2\mathit{b}-2\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}$
$\Rightarrow \mathit{b}=-3\sqrt{{\mathit{b}}^2-4\mathit{a}\mathit{c}}$
Squaring on both sides, we get,
${\mathit{b}}^2=9\ast \left({\mathit{b}}^2-4\mathit{a}\mathit{c}\right)$
$8{\mathit{b}}^2=36\mathit{a}\mathit{c}$
$2{\mathit{b}}^2=9\mathit{a}\mathit{c}$
Therefore, option A is the right answer.
Alternately
Suppose two roots are x, 2x
Sum of the roots$=3\mathit{x}=-\frac{\mathit{b}}{\mathit{a}}$
Product of the roots$=2{\mathit{x}}^2=\frac{\mathit{c}}{\mathit{a}}$
Putting the value of x from the first eqn.
We get $2{\mathit{b}}^2=9\mathit{a}\mathit{c}$