Question

Let $\alpha$ and $\beta$ be the roots of the equation $p{x}^2+qx+r=0$. If $p,q,r$ are in AP and $\alpha +\beta =4$, then $\alpha \beta$ is equal to

• A. $-9$
• B. $9$
• C. $-5$
• D. $5$
• E. $-4$

Answer 1

The correct option is A $-9$

Given, $(\alpha ,\beta )$ are the roots of the equation

$p{x}^2+qx+r=0$

$\therefore \alpha +\beta =-\frac{q}{p}$ .... (i)

and $\alpha \beta =\frac{r}{p}$ .... (ii)

Also given, $\alpha +\beta =4$ .... (iii)

From Eqs. (i) and (iii), we get

$-\frac{q}{p}=4\Rightarrow q=-4p$ .... (iv)

Also, $p,q$ and $r$ are in AP.

Therefore, $p+r=2q$

$\Rightarrow p+r=-8p$ [from Eq. (iv)]

$\Rightarrow 9p+r=0$

$\Rightarrow \frac{r}{p}=-9....\left(v\right)$

From Eqs. (ii) and (v), we have
$\alpha \beta =-9$