If $\alpha$ and $\beta$ are the zeroes of polynomial $3 x^{2}+6 x+k$ such that $\alpha+\beta+\alpha

Question

If $\alpha$ and $\beta$ are the zeroes of polynomial $3 x^{2}+6 x+k$ such that $\alpha+\beta+\alpha \beta=-\frac{2}{3}$, then the value of $k$ is:\

Professor PC Thomas Classes · Answer

(d) : Compare $3 x^{2}+6 x+k$ with $a x^{2}+b x+c$, we get $a=3, b=6$ and $c=\mathrm{k}$\ $	herefore \alpha+\beta+\alpha \beta=\left(\frac{-b}{a}ight)+\frac{c}{a} \Rightarrow \frac{-2}{3}=\frac{-6}{3}+\frac{k}{3}$\ $\Rightarrow \frac{k}{3}=\frac{-2}{3}+\frac{6}{3} \Rightarrow \frac{k}{3}=\frac{4}{3} \Rightarrow k=4$