Question

If $a(p+q{)}^2+2bpq+c=0$ and $a(p+r{)}^2+2bpr+c=0$, then $p^2+\frac{c}{a}$ is

• A. $pr$
• B. $rq$
• C. $r^2$
• D. $q^2$

Answer 1

The correct option is B $rq$

$a(p+q{)}^2+2bpq+c=0$ and $a(p+r{)}^2+2bpr+c=0$
The given equation can be written as
$\Rightarrow a{q}^2+2(a+b)pq+(c+a{p}^2)=0....\left(1\right)$
$\Rightarrow a{r}^2+2(a+b)pr+(c+a{p}^2)=0....\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we can say that $q$ and $r$ satisfy the quadratic equation,
$\Rightarrow a{x}^2+2(a+b)px+(c+a{p}^2)=0$

Now,
$p^2+\frac{c}{a}=\frac{a{p}^2+c}{a}$
Which is product of the roots,

Hence, $p^2+\frac{c}{a}=qr$