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

The correct option is B $64$

$x^2-3x+a=0$
$x_1+x_2=3;x_1.x_2=a$

$x^2-12x+b=0$
$x_3+x_4=12;x_3.x_4=b$

Let $r$ be the common ratio of G.P., then
$\frac{x_1+x_2}{x_3+x_4}=\frac{x_1(1+r)}{x_1{r}^2(1+r)}=\frac{3}{12}$
$\Rightarrow r=2(\because x_1<x_2<x_3<x_4)$

$\because x_1+x_2=3\Rightarrow x_1(1+r)=3\Rightarrow x_1=1$
$\therefore ab=x_1{x}_2{x}_3{x}_4=1\cdot 2\cdot 4\cdot 8=64$