Let α1, α2, β1, β2 be the roots of ax2+bx+c=0 and px2+qx+r=0, respectively. If the system of equations α1y+α2z=0 and β1y+β2z=0 has a non - trivial solution, then
b2q2=acpr
Since, α1, α2 are the roots pf ax2+bx+c=0.
⇒ α1+α2=−ba and α1α2=ca . . . (i)
Also, β1,β2 are the roots of px2+qx+r=0
⇒ β1+β2=−qp and β1β2=rp . . . (ii)
Given system of equations
α1y+α2z=0
and β1y+β2z=0, has non - trivial solution.
∴ ∣∣∣α1α2β1β2∣∣∣=0 ⇒ α1α2=β1β2
Applying componendo - dividendo
α1+α2α1−α2=β1+β2β1−β2⇒ (α1+α2)(β1−β2)=(α1−α2)(β1+β2)⇒ (α1+α2)2 {(β1+β2)2−4β2β2}=(β1+β2)2{(α1+α2)2−4α1α2}
From Eqs. (i) and (ii), we get
b2a2(q2p2−4rp)=q2p2(b2a2−4ca)⇒ b2ra2p=q2cap2⇒ b2ra=q2cp⇒ b2q2=acpr