If α, β are the roots of ax2 + bx + c = 0, and α + k, β + k are th

1.	If α, β are the roots of ax2 + bx + c = 0, and α + k, β + k are the roots of px2 + qx + r = 0, then k =(a) \(-\frac{1}{2}\) (a/b - p/q)(b) (a/b – p/q) (c) \(\frac{1}{2}\) (b/a - q/p)(d) (ab – pq)
Answer» (c) \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\) As α, β are the roots of the equation ax² + bx + c = 0, so α + β = \(-\frac{b}{a}\), αβ = \(\frac{c}{a}\) Also, (α + x), (β + x) are the roots of the equation px² + qx + r = 0, then α + \(x\) + β + \(x\) = – \(\frac{q}{p}\) and (α + x) (β + x) = \(\frac{r}{p}\) ⇒ α + β + 2x = – \(\frac{q}{p}\) ⇒ \(\frac{-b}{a}\) + 2x = – \(\frac{q}{p}\) ⇒ K = \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\)

If α, β are the roots of ax2 + bx + c = 0, and α + k, β + k are the roots of px2 + qx + r = 0, then k =(a) \(-\frac{1}{2}\) (a/b - p/q)(b) (a/b – p/q) (c) \(\frac{1}{2}\) (b/a - q/p)(d) (ab – pq)

Answer»

(c) \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\)

As α, β are the roots of the equation ax² + bx + c = 0, so

α + β = \(-\frac{b}{a}\), αβ = \(\frac{c}{a}\)

Also, (α + x), (β + x) are the roots of the equation

px² + qx + r = 0, then α + \(x\) + β + \(x\) = – \(\frac{q}{p}\)

and (α + x) (β + x) = \(\frac{r}{p}\) ⇒ α + β + 2x = – \(\frac{q}{p}\)

⇒ \(\frac{-b}{a}\) + 2x = – \(\frac{q}{p}\) ⇒ K = \(\frac{1}{2}\bigg(\frac{b}{a}-\frac{q}{p}\bigg).\)