Question

If $p,q,r$ are positive and are in AP, then the roots of the quadratic equation $p{x}^2+qx+r=0$ are complex for

• A. $\left|\left(\frac{r}{p}\right)–7\right|\ge 4\sqrt{3}$
• B. $\left|\left(\frac{p}{r}\right)–7\right|<4\sqrt{3}$
• C. All $p\text{and}r$
• D. No $p\text{and}r$

Answer 1

The correct option is A

$\left|\left(\frac{r}{p}\right)–7\right|\ge 4\sqrt{3}$

Explanation for the correct option

Step 1: Determination of the value of $q$

$\begin{array}{rcl}\therefore 2q& =& p+r\\ & \Rightarrow & q=\frac{p+r}{2}\end{array}$

Step 2: Formation of the equation to find the required roots

The quadratic equation of the form $a{x}^2+bx+c=0$ then the real roots will be when the discriminant is greater than or equal to zero.

$\mathit{D}\mathbf{=}{\mathit{q}}^{\mathbf{2}}\mathbf{–}\mathbf{4}\mathit{p}\mathit{r}\mathbf{\ge }\mathbf{0}$, where $q=b,p=a,\text{and}r=c$

Put the value $\begin{array}{rcl}& & \left(q=\frac{p+r}{2}\right)\end{array}$ in the above equation

$\begin{array}{rcl}\therefore {\left(\frac{p+r}{2}\right)}^2–4pr& \ge & 0\\ & \Rightarrow & \frac{{\left(p+r\right)}^2-16pr}{4}\ge 0\\ & \Rightarrow & p^2+r^2+2pr-16pr\ge 0\\ & \Rightarrow & p^2+r^2-14pr\ge 0\\ \Rightarrow p^2+r^2& \ge & 14pr\end{array}$

Divide by $pr$ on both the sides, then

$\begin{array}{rcl}\therefore \left(\frac{p}{r}\right)+\left(\frac{r}{p}\right)& \ge & 14\end{array}$

Now, put $t=\frac{r}{p}$ the above equation will become,

$\begin{array}{rcl}\therefore \left(\frac{1}{t}\right)+t& \ge & 14\\ & \Rightarrow & 1+t^2\ge 14t\\ & \Rightarrow & 1+t^2-14t\ge 0\end{array}$

Step 3: Solve the inequality by the perfect square method.

Add and subtract $49$ from the LHS,

$\begin{array}{rcl}\therefore 1+t^2-14t+49-49& \ge & 0\\ & \Rightarrow & t^2-14t+7^2-48\ge 0\\ & \Rightarrow & t^2-14t+7^2\ge 48\\ {\Rightarrow \left(t–7\right)}^2& \ge & {\left(4\sqrt{3}\right)}^2\end{array}$

Take the square root on both the sides,

$\begin{array}{rcl}\therefore |t–7|& \ge & 4\sqrt{3}\\ & \Rightarrow & \left|\left(\frac{r}{p}\right)–7\right|\ge 4\sqrt{3}\left(\because \mathrm{t}=\frac{\mathrm{r}}{\mathrm{p}}\right)\end{array}$

Hence, option (A) is correct.