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Question Number 208241 by Frix last updated on 08/Jun/24

Solve for p, q, r p+q+r=α p^2 +q^2 +r^2 =β pq=r

Solveforp,q,rp+q+r=αp2+q2+r2=βpq=r

Answered by mr W last updated on 08/Jun/24

p^2 +q^2 +(pq)^2 =β (p+q)^2 +(pq)^2 −2(pq)=β (α−pq)^2 +(pq)^2 −2(pq)=β 2(pq)^2 −2(α+1)(pq)+α^2 −β=0 ⇒pq=((α+1±(√(1+2α−α^2 +2β)))/2)=r ⇒p+q=α−r p, q are roots of x^2 −(α−r)x+r=0 ⇒p, q=((α−r±(√((α−r)^2 −4r)))/2)

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

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