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 $p-q+r>0$

Since it has no real roots,
we can compare it with $x^2+x+1=0$
Hence
$p=1,q=1$ and $r=1$
Therefore
$p-q+r$
$=1-1+1$
$=1$
$>0$
Hence only B is true.