Question

p, q, r are real numbers such that $4pr>q^2$ and $p+r>q$. which of the following is/are true?

• A. 4p + r > 2q
• B. p + q + r > 0
• C. p > 0
• D. r > 0

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A 4p + r > 2q
B p + q + r > 0
C p > 0
D r > 0

Let $f\left(x\right)=p{x}^2+qx+r$
Here $D=q^2-4pr<0\Rightarrow$ roots are complex
and $f(-1)=p-z+r=0.\Rightarrow p>0$
$\therefore f\left(0\right),f\left(1\right),f(-2)>0$.