Mathematics - Quadratic Equation Question with Solution | TestHub

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

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

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

$\Rightarrow 2t+1-kt-t=0$

$\Rightarrow t\left(1-k\right)=-1$

Replacing the value of $t$ again

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

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

For real roots, $D\ge 0$

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

$\Rightarrow \frac{6k-9}{k-1}-4\le 0$

$\Rightarrow \frac{\left(2k-5\right)}{\left(k-1\right)}\le 0$

$\Rightarrow k\in (1,\frac{5}{2}]$