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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
$\left| {\,\frac{r}{p} - 7\;} \right|\; \ge 4\sqrt 3 $
$\left| {\;\frac{p}{r} - 7\;} \right|\; < 4\sqrt 3 $
All $p$ and $r$
No $p$ and $r$
Solution
(a) $p,\;q,\;r$ are positive and are in $A.P.$
$\therefore \;q = \frac{{p + r}}{2}$ ……(i)
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 $.
