• # question_answer If $p,\ q,\ r$ are in A.P. and are positive, the roots of the quadratic equation $p{{x}^{2}}+qx+r=0$ are all real for [IIT 1995] A) $\left| \,\frac{r}{p}-7\ \right|\ \ge 4\sqrt{3}$ B) $\left| \ \frac{p}{r}-7\ \right|\ <4\sqrt{3}$ C) All $p$and $r$ D) No $p$ and $r$

$p,\ q,\ r$ are positive and are in A.P. $\therefore \ q=\frac{p+r}{2}$                    ......(i) $\because$ The roots of $p{{x}^{2}}+qx+r=0$ are real $\Rightarrow$ ${{q}^{2}}\ge 4pr$$\Rightarrow$${{\left[ \frac{p+r}{2} \right]}^{2}}\ge 4pr$    [using (i)] $\Rightarrow$ ${{p}^{2}}+{{r}^{2}}-14pr\ge 0$ $\Rightarrow$ ${{\left( \frac{r}{p} \right)}^{2}}-14\left( \frac{r}{p} \right)+1\ge 0$ $(\because \ p>0\ \text{and}\ p\ne 0)$ $\Rightarrow$ ${{\left( \frac{r}{p}-7 \right)}^{2}}-48\ge 0$$\Rightarrow$${{\left( \frac{r}{p}-7 \right)}^{2}}-{{(4\sqrt{3})}^{2}}\ge 0$ $\Rightarrow$ $\left| \ \frac{r}{p}-7\ \right|\ \ge 4\sqrt{3}$.