Menu Close

Solve-for-p-q-r-p-q-r-p-2-q-2-r-2-pq-r-




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)22(pq)=β(αpq)2+(pq)22(pq)=β2(pq)22(α+1)(pq)+α2β=0pq=α+1±1+2αα2+2β2=rp+q=αrp,qarerootsofx2(αr)x+r=0p,q=αr±(αr)24r2
Commented by Frix last updated on 08/Jun/24
��

Leave a Reply

Your email address will not be published. Required fields are marked *