If p,q,r are positive and are in A.P the roots of quadratic equations $px^2+qx+r=0$ are real for

Question

The answer is $|\frac rp -7|\ge 4\sqrt 3$

Since they are in AP $$2q=p+r$$ For x to be real $$q^2-4pr\ge 0$$ Then $$(\frac{p+r}{2})^2-4pr\ge 0$$ $$p^2+r^2-14pr\ge 0$$ I don’t know that to do next. Please help.

Thanks!

Answer 1

Next divide through by $p^2$, complete the square and take square roots:

$$p^2+r^2-14pr\ge 0$$ $$\left(\frac rp\right)^2 - 14 \frac rp \ge -1$$ $$\left(\frac rp\right)^2 - 14 \frac rp +49 \ge 48$$ $$\left(\frac rp-7\right)^2 \ge 48$$ $$\left|\frac rp-7\right| \ge \sqrt{48}$$ $$\left|\frac rp-7\right| \ge 4\sqrt{3}$$

Answer 2

Hint Divide both sides of the equation by $p^2$ to produce a polynomial in $\frac{r}{p}$ and then complete the square.