Question

If p,q and r are positive and are in AP, the root of quadratic equation px²+qx+r = 0 are real for ?

Answer 1

$$\begin{gathered} Givenquadraticequationisp{x}^2+qx+r=0 \\ TherootsofthegivenquadraticequationisrealifandonlyifD\ge 0 \\ Hence{q}^2-4pr\ge 0 \\ Sincep,q,rareinAPso2q=p+r \\ Replacingqby\frac{p+r}{2}in{q}^2-4pr\ge 0,weget \\ (\frac{p+r}{2}{)}^2-4pr\ge 0 \\ \Rightarrow p^2+r^2-14pr\ge 0 \\ \Rightarrow \frac{p^2}{r^2}-14\frac{p}{r}+1\ge 0 \\ \Rightarrow (\frac{p}{r}-7{)}^2-48\ge 0 \\ \Rightarrow (\frac{p}{r}-7{)}^2\ge 48 \\ \Rightarrow \left|\frac{p}{r}-7\right|\ge 4\sqrt{3} \end{gathered}$$