Question

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

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

Answer 1

The correct option is B $qr=p^2+\frac{c}{a}$

$a{(p+q)}^2+2bqp+c=0\Rightarrow a{(p+r)}^2+2bpr+c=0$

$\therefore$ $q$ and $r$ satisfy the following equation

$a{(p+x)}^2+2bpx+c=0$

so they are the roots of equation.

$\Rightarrow a{x}^2+2(ap+bp)x+c+a{p}^2=0$

Product of roots$=qr$

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