Question

Let $\alpha ,\beta$ be 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

$\because$ From the given equations $\alpha +\beta =1$ and $\alpha \beta =p$

Similarly, $\gamma +\delta =4$ and $\gamma \delta =q;(p,q\in Z)$

$\because \alpha ,\beta ,\gamma$ and $\delta$ are in G.P with common ratio ${}^{\prime }{r}^{\prime }$

$\therefore \alpha +\alpha r=1\Rightarrow \alpha (1+r)=1\cdots \cdots \left(1\right)$

And $\alpha r^2+\alpha r^3=4\Rightarrow \alpha r^2(1+r)=4\cdots \cdots \left(2\right)$

On solving eqn (1) and (2)
$r^2=4\Rightarrow r=\pm 2$

If $r$=2, $\alpha +2\alpha =1\Rightarrow \alpha =\frac{1}{3}$

If $r$=-2, $\alpha -2\alpha =1\Rightarrow \alpha =-1$

But $p=\alpha \beta \in Z\therefore r=-2$ and $\alpha =-1$

$\therefore p=-2,q={\alpha }^2{r}^5=1{(-2)}^5=-32$