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Q.
The set of all values of $k >-1$, for which the equation $\left(3 x^{2}+4 x+3\right)^{2}-(k+1)\left(3 x^{2}+4 x+3\right)$ $\left(3 x^{2}+4 x+2\right)+k\left(3 x^{2}+4 x+2\right)^{2}=0$ has real roots, is :
JEE MainJEE Main 2021Complex Numbers and Quadratic Equations
Solution:
$\left(3 x^{2}+4 x+3\right)^{2}-(k+1)\left(3 x^{2}+4 x+3\right)\left(3 x^{2}+4 x+2\right)$ $+k\left(3 x^{2}+4 x+2\right)^{2}=0$
Let $3 x^{2}+4 x+3=a$
and $3 x^{2}+4 x+2=b $
$\Rightarrow b=a-1$
Given equation becomes
$\Rightarrow a^{2}-(k+1) a b+k b^{2}=0$
$\Rightarrow a(a-k b)-b(a-k b)=0$
$\Rightarrow (a-k b)(a-b)=0$
$ \Rightarrow a=k b$ or $a=b($ reject $)$
$\because a=k b$
$\Rightarrow 3 x^{2}+4 x+3=k\left(3 x^{2}+4 x+2\right)$
$\Rightarrow 3(k-1) x^{2}+4(k-1) x+(2 k-3)=0$
for real roots $D \ge 0$
$\Rightarrow 16( k -1)^{2}-4(3( k -1))(2 k -3) \geq 0$
$\Rightarrow 4( k -1)\{4( k -1)-3(2 k -3)\} \geq 0$
$\Rightarrow 4( k -1)\{-2 k +5\} \geq 0$
$\Rightarrow -4( k -1)\{2 k -5\} \geq 0$
$\Rightarrow ( k -1)(2 k -5) \leq 0$
$\therefore k \in\left[1, \frac{5}{2}\right]$
$\because k \neq 1$
$\therefore k \in\left(1, \frac{5}{2}\right]$
