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

Answer: (B)

$p-q+r>0$

Let $\alpha +i\beta ,\alpha -i\beta$ be the roots.

$\Rightarrow$ $(\alpha +i\beta )(\alpha -i\beta )=\frac{r}{p}$
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 non-real).
$\therefore (p+r{)}^2\ge q^2+(p-r{)}^2\ge q^2\therefore p+r>q$.