Question

Let $\alpha, \beta$ be the roots of $x^2-x+p=0$ and $\gamma$, $\delta$ be the roots of $x^2-4 x+q=0$. 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 $
$\text { and } q=\gamma \delta=\left(\alpha r^2\right)\left(\alpha r^3\right)=\alpha^2 r^5=-32$
$\text { Hence, } p=-2, q=-32 .$