If one root of the equation ax2+px + q = 0 be n times the other root, then
Let the roots are α and nα
Sum of roots, α and nα=−ba⇒α=−ba(n+1) ........(i)
and product, α.nα=ca⇒α2=cna .......(ii)
From (i) and (ii), we get
⇒[−ba(n+1)]2=cna⇒b2a2(n+1)2=cna
⇒nb2=ac(n+1)2