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# Let α1, α2 and β1, β2 be the roots of ax2 + bx + c = 0 and px2 + qx + r = 0 respectively. If the system of equations α1 y + α2 z = 0 and β1 y +  β2 z = 0 has nontrivial solution, then prove that b2/q2 = ac/pr.

Navjyot Kalra
7 years ago
Sol. We are given that α1, α2 are the roots of
ax2 + bx + c = 0
∴ α1 + α2 = -b/a; α1 α2 = c/a . . . . . . . . . . . . . (1)
And β1, β2 are the roots of px2 + qx + r = 0
∴ β1 + β2 = -q/p ; β1 β2 = r/p . . . . . . . . . . . . . . . . . (2)
The system of equations, α1 y + α2 z = 0
And β1 y + β2 z = 0 has a non trivial solution.
∴ we must have |α1 β1 α2 β2| = 0
NOTE THIS STEP :
⇒ α1 β2 – α2 β1 = 0 ⇒ α1/ α2 = β12
By componendo and dividend, we get
α1 + α2/ α1 - α2 = β1 + β2/ β1 - β2
⇒ (α1 + α2) (β1 - β2) = (α1 - α2) (β1 + β2)
⇒ (α1 + α2)2 [(β1 - β2)2 - 4β1 β2]
= [(α1 + α2)2 – 4 α1α2] (β1 + β2)2
Using equations (1) and (2) we get
b2/a2 [ q2/p2 – 4r/p] = q2/p2 [b2/a2 – 4c/a]
⇒ b2 q2/a2 p2 – 4b2r/a2r = q2 b2/q2 a2 – 4cq2/ap2 ⇒ -4b2 r/a2 p = -4cq2/ap2
⇒ b2 r/a = sq2/p ⇒ b2/q2 = ac/pr Hence Proved