Problem
ALG-B1-M06-P008 Reciprocals
#8
★★☆☆☆ Level 2 of 5
Let \(p+q+r=0\), \(p^2+q^2+r^2=18\), \(pqr=6\). Find \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\).
First find \(pq+qr+rp\).
From \((p+q+r)^2=0\) we get \(18+2(pq+qr+rp)=0\), so \(pq+qr+rp=-9\).
Therefore \(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{pq+qr+rp}{pqr}=\frac{-9}{6}=-\frac{3}{2}\).
The problem connects zero sum with a symmetric fraction.