Find the value of k if the coefficient of x² in the binomial expansion of (1 + kx)²(2 – x)² is 6.

Steps to Solve

1. Expand $(1 + kx)^2$ Expanding $(1 + kx)^2$ gives $1 + 2kx + k^2x^2$.

2. Expand $(2 - x)^2$ Expanding $(2 - x)^2$ gives $4 - 4x + x^2$.

3. Multiply the two expansions We need to multiply $(1 + 2kx + k^2x^2)(4 - 4x + x^2)$ and identify the terms that result in $x^2$. $1(x^2) + 2kx(-4x) + k^2x^2(4)$ This gives us $x^2 - 8kx^2 + 4k^2x^2$.

4. Combine the $x^2$ terms Combining these terms, we get $(1 - 8k + 4k^2)x^2$.

5. Set the coefficient of $x^2$ equal to 6 We are given that the coefficient of $x^2$ is 6, so we set $1 - 8k + 4k^2 = 6$.

6. Rearrange the equation Subtracting 6 from both sides, we get $4k^2 - 8k - 5 = 0$.

7. Solve the quadratic equation for k We can use the quadratic formula to solve for $k$: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = -8$, and $c = -5$.

8. Apply the formula $k = \frac{8 \pm \sqrt{(-8)^2 - 4(4)(-5)}}{2(4)}$ $k = \frac{8 \pm \sqrt{64 + 80}}{8}$ $k = \frac{8 \pm \sqrt{144}}{8}$ $k = \frac{8 \pm 12}{8}$

9. Find the two possible values for k $k_1 = \frac{8 + 12}{8} = \frac{20}{8} = \frac{5}{2}$ $k_2 = \frac{8 - 12}{8} = \frac{-4}{8} = -\frac{1}{2}$