Question

If $p,q,r$ are positive and are in A.P., the roots of quadratic equation $p{x}^2+qx+r=0$ are all real for

• A. $|\frac{r}{p}-7|\ge 4\sqrt{3}$
• B. $|\frac{p}{r}-7|\ge 4\sqrt{3}$
• C. All $p$ and $r$
• D. No $p$ and $r$

Answer 1

Answer: (B)

$|\frac{p}{r}-7|\ge 4\sqrt{3}$

Since, $p,q,r$ are in A.P.

Therefore, $2q=p+r$

$p{x}^2+qx+r=0$ has real roots

$\Rightarrow D=q^2-4pr\ge 0$

$\Rightarrow {\left(\frac{p+r}{2}\right)}^2-4pr\ge 0$

$\Rightarrow \left(\frac{p^2+r^2+2pr-16pr}{2}\right)\ge 0$

$\Rightarrow p^2+r^2-14pr\ge 0$

$\Rightarrow {\left(\frac{p}{r}\right)}^2+1-14\frac{p}{r}+49\ge 49$

$\Rightarrow {(\frac{p}{r}-7)}^2\ge 48$

$\Rightarrow |\frac{p}{r}-7|\ge 4\sqrt{3}$

Ans: B