If α, β are the roots of the quadratic equation ax2 + bx + c = 0,

1.	If α, β are the roots of the quadratic equation ax2 + bx + c = 0, then αβ2 + α2β + αβ equals :(a) \(\frac{bc}{-a^2}\) (b) 0 (c) abc (d) \(\frac{c(a-b)}{a^2}\)
Answer» (d) \(\frac{c(a-b)}{a^2}\) Given, α, β are the roots of the equation ax² + bx + c = 0. Then, α + β = – \(\frac{b}{a}\) , αβ = \(\frac{c}{a}\) Then, αβ² + α²β + αβ = αβ (α + β) + αβ = \(\big(\frac{c}{a}\big)\) \(-\frac{b}{a}\) + \(\frac{c}{a}\) = - \(\frac{bc}{a^2}+\frac{c}{a}=\frac{-bc+ac}{a^2}=\)\(\frac{c(a-b)}{a^2}\)

If α, β are the roots of the quadratic equation ax2 + bx + c = 0, then αβ2 + α2β + αβ equals :(a) \(\frac{bc}{-a^2}\) (b) 0 (c) abc (d) \(\frac{c(a-b)}{a^2}\)

Answer»

(d) \(\frac{c(a-b)}{a^2}\)

Given, α, β are the roots of the equation ax² + bx + c = 0. Then,

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

Then, αβ² + α²β + αβ = αβ (α + β) + αβ = \(\big(\frac{c}{a}\big)\) \(-\frac{b}{a}\) + \(\frac{c}{a}\)

= - \(\frac{bc}{a^2}+\frac{c}{a}=\frac{-bc+ac}{a^2}=\)\(\frac{c(a-b)}{a^2}\)