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 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

The sum of roots and product of roots is follows :
$\alpha +\beta =1,\alpha \beta =p\gamma +\delta =4,\gamma \delta =q$
Let r be the common ratio of the $GP\alpha ,\beta ,\gamma ,\delta$
$\begin{array}{cccc}Then& \alpha +\beta =1& and& \gamma +\delta =4\\ & \alpha +\alpha r=1& and& \alpha r^2+\alpha r^3=4\\ \Rightarrow & \alpha (1+r)=1& and& \alpha r^2(1+r)=4\\ So& (1+r)=\frac{1}{a}& and& a{r}^2\left(\frac{1}{\alpha }\right)=4\\ \Rightarrow & r=\pm 2& & \end{array}$
When r = 2, then $\alpha =\frac{1}{3}$ which is inadmissible as all given options are integral values.
Hence, r = - 2 is to be considered, then $\alpha =-1$
$Thus\alpha \beta =p=-2and\gamma \delta =q=-32$