Question

\(\alpha, \beta\) are zeroes of the polynomial \(3x^2 - 8x + k\). Find the value of \(k\), if \(\alpha^2 + \beta^2 = \dfrac{40}{9}\)

Hint:

Use identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) and convert all to same denominator.

Answer 1

Given: \(a = 3,\ b = -8,\ c = k\) We use identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] \[ \alpha + \beta = \dfrac{-b}{a} = \dfrac{8}{3},\quad \alpha\beta = \dfrac{c}{a} = \dfrac{k}{3} \] \[ \Rightarrow \alpha^2 + \beta^2 = \left( \dfrac{8}{3} \right)^2 - 2 \cdot \dfrac{k}{3} = \dfrac{64}{9} - \dfrac{2k}{3} \] Set equal to \(\dfrac{40}{9}\): \[ \dfrac{64}{9} - \dfrac{2k}{3} = \dfrac{40}{9} \Rightarrow \dfrac{24}{9} = \dfrac{2k}{3} \Rightarrow \dfrac{8}{3} = \dfrac{2k}{3} \Rightarrow k = 4 \]