To solve this problem, we need to compare the coefficients of $x^4$ in the two given binomial expansions.

Step 1: Coefficient of $x^4$ in $(3 + 2x)^6$

The general term in the expansion of $(a + bx)^n$ is given by:

$T_r = \binom{n}{r} a^{n-r} (bx)^r$

Here, $a = 3$, $b = 2$, and $n = 6$. For $x^4$, $r = 4$. Substituting these values:

$T_4 = \binom{6}{4} \cdot 3^{6-4} \cdot (2x)^4$

Simplify:

$T_4 = \binom{6}{4} \cdot 3^2 \cdot 2^4 \cdot x^4$

$T_4 = \binom{6}{4} \cdot 9 \cdot 16 \cdot x^4$

$T_4 = \binom{6}{4} \cdot 144 \cdot x^4$

The coefficient of $x^4$ is:

$\text{Coefficient} = \binom{6}{4} \cdot 144$

Now calculate $\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \cdot 5}{2 \cdot 1} = 15$. Therefore:

$\text{Coefficient} = 15 \cdot 144 = 2160$

Step 2: Coefficient of $x^4$ in $(k + 3x)^6$

Similarly, the general term is:

$T_4 = \binom{6}{4} \cdot k^{6-4} \cdot (3x)^4$

$T_4 = \binom{6}{4} \cdot k^2 \cdot 3^4 \cdot x^4$

$T_4 = \binom{6}{4} \cdot k^2 \cdot 81 \cdot x^4$

$\text{Coefficient} = \binom{6}{4} \cdot k^2 \cdot 81$

Using $\binom{6}{4} = 15$:

$\text{Coefficient} = 15 \cdot k^2 \cdot 81$

$\text{Coefficient} = 1215k^2$

Step 3: Set the coefficients equal

The problem states that the coefficients of $x^4$ in the two expansions are equal. Thus:

$2160 = 1215k^2$

Solve for $k^2$:

$k^2 = \frac{2160}{1215}$

Simplify:

$k^2 = \frac{72}{45} = \frac{8}{5}$

Take the square root:

$k = \pm \sqrt{\frac{8}{5}}$

Simplify further:

$k = \pm \frac{2\sqrt{10}}{5}$

Final Answer:

$k = \pm \frac{2\sqrt{10}}{5}$

Would you like me to explain any step further or go into additional detail?

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Related Questions:

1. How does the binomial theorem work for general cases?
2. Can the coefficient comparison method be applied to any power series?
3. What is the significance of the general term in a binomial expansion?
4. What are some common mistakes when applying the binomial theorem?
5. How does one handle expansions with fractional exponents?

Tip:

Always simplify binomial coefficients first, as it reduces errors and calculations later.