Question

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

Answer 1

$p_1{2}_{1}r\to A.P$

$p{x}^2+2x+r=02=p+r$

$x=\frac{-2\pm \sqrt{2^2-4}pr}{2p}$

$\frac{-2\pm \sqrt{\frac{p^2+r^2+2}{4}}pr-4pr}{2p}$

$=\frac{-2\pm \frac{1}{2}\sqrt{p^2+r^2+2pr-8pr}}{2p}$

$=\frac{-2\pm \frac{1}{2}\sqrt{p^2+r^2-6pr}}{2p}$

for Real roots

$2^2-4pr\le 0$

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

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

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

${(\frac{p}{r}-7)}^2-48\ge 0$

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

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