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Step-by-Step Solution
Step 1: Rewrite the expression in simpler variables
Let
$3x^2 + 4x + 3 = a \quad \text{and} \quad 3x^2 + 4x + 2 = b.$
Then notice that
$b = (3x^2 + 4x + 3) - 1 = a - 1.$
This helps transform the given equation into a simpler form in terms of $a$ and $b$.
Step 2: Substitute into the given equation
The original equation
$ (3x^2 + 4x + 3)^2 \;-\; (k+1)(3x^2 + 4x + 3)(3x^2 + 4x + 2)\; +\; k(3x^2 + 4x + 2)^2 \;=\; 0$
becomes
$ a^2 \;-\; (k+1)\,a\,b \;+\; k\,b^2 \;=\; 0.$
Step 3: Factorize the simplified expression
Observe that
$ a^2 \;-\; (k+1)\,a\,b \;+\; k\,b^2
\;=\; a(a - kb) \;-\; b(a - kb)
\;=\; (a - kb)\,(a - b).$
So the equation
$ (a - kb)\,(a - b) = 0 $
implies either
$ a = kb \quad \text{or} \quad a = b.$
Step 4: Reject the invalid case
If
$a = b,$
then
$3x^2 + 4x + 3 = 3x^2 + 4x + 2,$
which would give
$3 = 2,$
a contradiction.
Hence, the only valid case is
$ a = kb \,\Longrightarrow\, 3x^2 + 4x + 3 = k(3x^2 + 4x + 2).$
Step 5: Form the resulting quadratic in x
From
$3x^2 + 4x + 3 = k(3x^2 + 4x + 2),$
we get
$3(k - 1)\,x^2 \;+\; 4(k - 1)\,x \;+\; (2k - 3) \;=\; 0.
Step 6: Impose the condition for real roots
The above expression is a quadratic in $x.$ For it to have real roots, its discriminant $D$ must be nonnegative:
$D = 16\,(k - 1)^2 \;-\; 4 \cdot 3(k - 1)\,(2k - 3) \;\ge\; 0.
Factor out $4(k - 1)$ to simplify:
$4(k - 1)\Bigl[4(k - 1) \;-\; 3(2k - 3)\Bigr] \;\ge\; 0.
Inside the bracket:
$4(k - 1) - 3(2k - 3) = 4(k-1) - (6k - 9) = 4k - 4 - 6k + 9 = -2k + 5.$
So the condition becomes
$4(k - 1)\,[-2k + 5] \;\ge\; 0
\;\Longrightarrow\;(k - 1)\,(2k - 5) \;\le\; 0.
This inequality holds when
$k \in [1, \tfrac{5}{2}].$
However, from the factorization step, $k = 1$ makes the entire quadratic degenerate (it does not yield a valid expression for real roots when substituted back because it would force $a$ and $b$ to be equal, which was contradictory). Thus we exclude $k = 1.$
Step 7: Combine the conditions to find the final range
Also, the question stipulates $k > -1,$ which is automatically satisfied for the derived range. Hence the final valid range of $k$ is
$\displaystyle k \in \bigl(1, \tfrac{5}{2}\bigr].
Final Answer
The set of all values of $k > -1$ for which the original equation has real roots is
$\displaystyle \bigl(1, \tfrac{5}{2}\bigr].
