Question

If $p,q,r$ are in $A.P$ and are positive, than the roots of the quadratic equations $p{x}^2+qx+r=0$ will be 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 $|\frac{r}{p}-7|\ge 4\sqrt{3}$

Given $p,q,r$ are in A.P. then $q=\frac{p+r}{2}$.....(1).

Now $p{x}^2+qx+r=0$ will have real root then $q^2-4pr\ge 0$.

or, $\frac{(p+r{)}^2}{4}-4pr\ge 0$

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

or, $r^2-14rp+49p^2\ge 48p^2$

or, $(r-7p{)}^2\ge (4\sqrt{3}p{)}^2$

or, ${(\frac{r}{p}-7)}^2\ge (4\sqrt{3}{)}^2$ [ Since $p\ne 0$ for the given equation to be quadratic]

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