Question

If $p{x}^2$ + qx + r = 0 has no real roots and p, q, r are real such that p + r > 0 then

• A. p - q + r < 0
• B. p - q + r > 0
• C. p + r = q
• D. All of these

Answer 1

The correct option is B

p - q + r > 0

Let $\alpha +i\beta ,\alpha -i\beta$ be the roots. Then ${\alpha }^2+{\beta }^2=\frac{r}{p}>0.$ So p, r are of the same sign. Also p + r > 0. So p, r are both positive.
If q < 0, p - q + r > 0
If q > 0, $(p+r{)}^2-(p-r{)}^2=4pr\ge q^2(\because$ roots are nonreal)
$\therefore (p+r{)}^2\ge q^2+(p-r{)}^2\ge$
$q^2\therefore$ p + r > q