Question

If the sum of the roots of $p{x}^2+qx+r=0$ is equal to the sum of their squares then $2rp=$

• A. $q(p-q)$
• B. $p(p+q)$
• C. $p(p-q)$
• D. $q(p+q)$

Answer 1

The correct option is D $q(p+q)$

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

$sum=\frac{-q}{p}=\alpha +\beta$

Sum of squares of roots $={\alpha }^2+{\beta }^2={(\alpha +\beta )}^2-2\alpha \beta =\frac{q^2}{p^2}-\frac{2r}{p}$

$\alpha +\beta ={\alpha }^2+{\beta }^2$

$\frac{-q}{p}=\frac{q^2}{p^2}-\frac{2r}{p}$

$q^2-2rp+qp=0$

$2rp=q(q+p)$