The set of all values of k > - 1 , for which the equation (3 x ^ 2...

Let $t = \left(3{x}^{2} + 4x + 2\right)$

$⇒ {\left(t + 1\right)}^{2} - \left(k + 1\right)\left(t\right)\left(t + 1\right) + k{\left(t\right)}^{2} = 0$

$⇒ {t}^{2} + 2t + 1 - ( k + 1 ) {t}^{2} - ( k + 1 ) t + k{t}^{2} = 0$

$⇒ 2t + 1 - kt - t = 0$

$⇒ t\left(1 - k\right) = - 1$

Replacing the value of $t$ again

$3{x}^{2} + 4x + 2 = t = \dfrac{1}{k - 1}$

$⇒ 3{x}^{2} + 4x + \left(\dfrac{2k - 3}{k - 1}\right) = 0$

For real roots, $D \ge 0$

$⇒ 16 - 4 \times 3 \times \left(\dfrac{2k - 3}{k - 1}\right) \ge 0$

$⇒ \dfrac{6k - 9}{k - 1} - 4 \le 0$

$⇒ \dfrac{\left(2k - 5\right)}{\left(k - 1\right)} \le 0$

$⇒ k \in ( 1, \dfrac{5}{2} ]$