Question

Let $\alpha ,\beta$ be the roots of $x^2-x+p=0and\gamma ,\delta$ be the roots of $x^2-4x+q=0$. If $\alpha ,\beta ,\gamma ,\delta$ are in GP, then the integral values of $p$ and $q$, respectively, are:

• A. $-2,-32$
• B. $-2,3$
• C. $-6,3$
• D. $-6,-32$

Answer 1

The correct option is A $-2,-32$

$\alpha ,\beta$ are the roots of $x^2-x+p=0$

$\therefore \alpha +\beta =1$ ...(1)

and $\alpha \beta =p$ ...(2)

$\gamma ,\delta$ are the roots of $x^2-4x+q=0$

$\therefore \gamma +\delta =4$ ...(3)

and $\gamma \delta =q$ ...(4)

$\alpha ,\beta ,\gamma ,\delta$ are in G.P.

Let $r$ be the common ratio

So, $\beta =\alpha r,\gamma =\alpha r^2$, $\delta =\alpha r^3$ ...(5)

$\frac{\alpha +\beta }{\gamma +\delta }=\frac{1}{4}$ ...(from 1 and 3)

$\therefore \frac{\alpha +\alpha r}{\alpha r^2+\alpha r^3}=\frac{1}{4}$ ...(from 5)

$\therefore \frac{1}{r^2}=\frac{1}{4}$

$\therefore r^2=4$

$\therefore r=\pm 2$

Case(1) $r=2$

Using equation (1) and (5), we get

$\therefore \alpha =\frac{1}{3}$

Again using equation (5), we get

$\beta =\frac{2}{3},\gamma =\frac{4}{3},\delta =\frac{8}{3}$

$\therefore p=\alpha \beta =\frac{2}{9}$

and $q=\gamma \delta =\frac{32}{9}$

Case(2) $r=-2$

Using equation (1) and (5), we get

$\therefore \alpha =-1$

Again using equation (5), we get

$\beta =2,\gamma =-4,\delta =8$

$\therefore p=\alpha \beta =-2$

and $q=\gamma \delta =-32$

So, the integral values of $p$ and $q$ are $-2$ and $-32$ respectively.