Question

If $p{x}^2+qx+r=0$ has no real roots for real values of $p,q,r$ and $4p+2q+r>0$, then

• A. $r>0$
• B. $p+q+r>0$
• C. $p+q+r<0$
• D. $r\ge 0$

Answer 1

Answer: (A), (B)

The correct options are
A $r>0$
B $p+q+r>0$

Given that $4p+2q+r>0$ ...(1)

$f\left(x\right)=p{x}^2+qx+r$

Since, the equation $f\left(x\right)=0$ has no real roots.

Therefore, $f\left(x_1\right)f\left(x_2\right)>0$

$f\left(0\right)f\left(2\right)=r(4p+2q+r)>0$

Using (1), we get $r>0$

$f\left(1\right)f\left(2\right)=(p+q+r)(4p+2q+r)>0$

Using (1), we get $p+q+r>0$

Ans: A,B