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., the integral values of p,q are respectively

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

Answer 1

The correct option is A -2, -32

Let $r$ be the common ratio of the G.P. $\alpha ,\beta ,\gamma ,\delta$ then $\beta =\alpha r,\gamma =\alpha r^{2}and\delta =\alpha r^3$
$\therefore \alpha +\beta =1\Rightarrow \alpha +\alpha r=1\Rightarrow \alpha (1+r)=\mathrm{1......}\left(i\right)$
$\alpha \beta =p\Rightarrow \alpha \left(\alpha r\right)=p\Rightarrow {\alpha }^{2}r=p......\left(ii\right)$
$\gamma +\delta =4\Rightarrow \alpha r^2+\alpha r^3=4\Rightarrow \alpha r^2(1+r)=\mathrm{4.......}\left(iii\right)$
and $\gamma \delta =q\Rightarrow \alpha r^2.\alpha r^3=q\Rightarrow {\alpha }^2{r}^5=q.....\left(iv\right)$
Dividing (iii) by (i), we get $r^2=4\Rightarrow r=\pm 2$
If we take $r=2$, then $\alpha$ is not integral, and thus p and q will also be not integral, so we take $r=-2$,
Substituting $r=-2$ in (i), we get $\alpha =-1$
Now, form (ii), we have $p={\alpha }^{2}r=(-1{)}^2(-2)=-2$
and from (iv), we have $q={\alpha }^2{r}^5=(-1{)}^2(-2{)}^5=-32$
$\Rightarrow (p,q)=(-2,-32).$