Solve-for-p-q-r-p-q-r-p-2-q-2-r-2-pq-r- Tinku Tara June 8, 2024 Algebra 0 Comments FacebookTweetPin Question Number 208241 by Frix last updated on 08/Jun/24 Solveforp,q,rp+q+r=αp2+q2+r2=βpq=r Answered by mr W last updated on 08/Jun/24 p2+q2+(pq)2=β(p+q)2+(pq)2−2(pq)=β(α−pq)2+(pq)2−2(pq)=β2(pq)2−2(α+1)(pq)+α2−β=0⇒pq=α+1±1+2α−α2+2β2=r⇒p+q=α−rp,qarerootsofx2−(α−r)x+r=0⇒p,q=α−r±(α−r)2−4r2 Commented by Frix last updated on 08/Jun/24 �� Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-208242Next Next post: K-0-4-pi-ln-cosx-dx- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.