Question

If \( p,q,r \) are positive and are in A.P., the roots of the equation \( px^2+qx+r=0 \) are all real for

• A. \( \left|\dfrac{r}{p}-7\right|\ge 4\sqrt{3} \)
• B. \( \left|\dfrac{p}{r}-7\right|\ge 4\sqrt{3} \)
• C. all \( p \) and \( q \)
• D. No \( p \) and \( r \)

Hint:

When coefficients of a quadratic are in A.P., first express the middle term using the extremes, then apply the discriminant condition for real roots.

Answer 1

The Correct Option is B

Since \( p,q,r \) are in A.P., \[ q=\frac{p+r}{2}. \] Step 1: For the quadratic equation \[ px^2+qx+r=0 \] to have all real roots, its discriminant must be non-negative: \[ q^2-4pr \ge 0. \]
Step 2: Substituting \( q=\frac{p+r}{2} \), \[ \left(\frac{p+r}{2}\right)^2 - 4pr \ge 0. \] \[ \Rightarrow \frac{p^2+2pr+r^2-16pr}{4} \ge 0. \] \[ \Rightarrow p^2-14pr+r^2 \ge 0. \] 
Step 3: Dividing throughout by \( pr>0 \), \[ \frac{p}{r}+\frac{r}{p} \ge 14. \] Let \( x=\frac{p}{r}>0 \). Then, \[ x+\frac{1}{x} \ge 14. \] 
Step 4: Solving, \[ x^2-14x+1 \ge 0 \Rightarrow x \ge 7+4\sqrt{3} \quad \text{or} \quad x \le 7-4\sqrt{3}. \] Hence, \[ \left|\frac{p}{r}-7\right|\ge 4\sqrt{3}. \]