Question

Let $x_1.x_2$ be the roots of the equation $x_2-3x+p=0$ and let $x_3{x}_4$ be the roots of the equation $x^2-12x+q=0$. If the numbers $x_1,x_2,x_3,x_4$ (In order) form an increasing $GP.,$ then

• A. $p=2,q=16$
• B. $p=2,q=32$
• C. $p=4,q=16$
• D. $p=4,q=32$

Answer 1

The correct option is A $p=2,q=16$

$x_1+x_2=3,x_1{x}_2=p$

$x_3+x_4=12,x_3{x}_4=q$

let $x_1=a,x_2=ar,x_3=a{r}^2,x_4=a{r}^3$

$a(1+k)=3,a^2{r}^{2}p$

$a{r}^2(1+x)=12,a^2{x}^5=q$

$\Rightarrow r^2=\frac{12}{3}=4$

$\begin{array}{cc}\text{r}=\text{2}& \text{and}a(1+2)=3\\ & \Rightarrow a=1\end{array}$

$p=a^{2}r=2$

$q=a^2{r}^5=2^5=32$

option $B$ is correct.