Question

If the sum of the roots of the equation $3x^2-(3k-2)x-(k-6)=0$ is equal to the product of its roots, then k = ?

(a) 1
(b) −1
(c) 0
(d) 2

Answer 1

(d) 2

Let the roots of the equation be $\alpha \mathrm{and}\beta$.
Then,
$\alpha +\beta =\frac{-\left[-\left(3k-2\right)\right]}{3}=\frac{3k-2}{3}$ and $\alpha \beta =\frac{-\left(k-6\right)}{3}=\frac{-k+6}{3}$
Therefore,
$\frac{3k-2}{3}=\frac{-k+6}{3}$
$\Rightarrow 3k-2=-k+6$
$\Rightarrow 4k=8\Rightarrow k=2$