Question

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

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

Answer 1

The correct option is A $|\frac{r}{p}-7|\ge 4\sqrt{3}$

$2q=p+r$

for roots to be real.

$q^2-4pr⩾0$

$(\frac{p+r}{2}{)}^2-4pr⩾0$

$(p^2+r^2+2pr-16pr)⩾0$

$p^2+r^2-14pr⩾0$

$(\frac{p}{r}{)}^2+1-14\frac{p}{r}\ge 0$

$(\frac{p}{r}-7{)}^2⩾48$

$|\frac{p}{r}-7|⩾4\sqrt{3}$