Question

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

• A. $qr=p^2$
• B. $qr=p^2+\frac{c}{a}$
• C. $qr=-p^2$
• D. none of these

Answer 1

Answer: (B)

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

As $a(p+q{)}^2+2bpq+c=0$ and $a(p+r{)}^2+2bpr+c=0$
Then $q,r$ are roots of $a(p+x{)}^2+2bpx+c=0$
$\Rightarrow a{p}^2+a{x}^2+2pax+2bpx+c=0\Rightarrow a{x}^2+(2pa+2pb)x+c+a{p}^2=0$
Hence, product of roots is
$qr=\frac{c+a{p}^2}{a}\Rightarrow qr=p^2+\frac{c}{a}$