Question

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

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

Answer 1

Answer: (A)

We have
$\alpha +\beta =1,\alpha \beta =p,$
$\gamma +\delta =4,\gamma \delta =q$
Let $r$ be the common ratio of the GP $\alpha ,\beta ,\gamma ,\delta$. Then
$\alpha +\beta =1\Rightarrow \alpha +\alpha r=1\Rightarrow \alpha (1+r)=1$
$\gamma +\delta =4\Rightarrow \alpha r^2+\alpha r^3=4$
$\Rightarrow \alpha r^2(1+r)=4$
$\text{ Thus, }\frac{\alpha r^2(1+r)}{\alpha (1+r)}=4\Rightarrow r^2=4\Rightarrow r=\pm 2$
$\text{ When }\alpha (1+r)r=2$
$\text{ we get }\alpha (1+r)=1\Rightarrow \alpha =1/3$
$\text{ In this case }p=\alpha \beta =\alpha (\alpha r)={\alpha }^{2}r$
$=\frac{1}{9}(2)=\frac{2}{9}$
which is not an integer.
Thus, $r=-2$. In this case, $\alpha (1+r)=1\Rightarrow \alpha =-1$.
$\therefore p={\alpha }^{2}r=(-1{)}^2(-2)=-2$
$\text{ and }q=\gamma \delta =\left(\alpha r^2\right)\left(\alpha r^3\right)={\alpha }^2{r}^5=-32$
$\text{ Hence, }p=-2,q=-32.$