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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.`
5 years ago Navjyot Kalra
654 Points
```
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 = β1/β2

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

```
5 years ago
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