Question

solve 13 question please ​

Answer 1

We know that in a Quadratic equation

$\alpha + \beta = \frac{ - b}{a} \\ \\ \\ \alpha \beta = \frac{c}{a}$
here
$3 {x}^{2} + 2kx + x - k - 5 \\ \\ 3 {x}^{2} + x(2k + 1) - k - 5 \\ \\ \alpha + \beta = \frac{ - 2k - 1}{3} \\ \\ \alpha \beta = \frac{ - k - 5}{3}$
ATQ

$2( \alpha + \beta ) = \alpha \beta \\ \\ 2( \frac{ - 2k - 1}{3} ) = \frac{ - k - 5}{3} \\ \\ \frac{ - 4k - 2}{3} = \frac{ - k - 5}{3} \\ \\ - 4k + k = - 5 + 2 \\ \\ - 3k = - 3 \\ \\ k = 1$
Hope it helps you