Binomial theorem question. Find the value of the constant $k$

Question

$$\left[(k+x)\left(2-\frac{x}{2}\right)\right]^6$$

where the coefficient of $x^{2}$ is $84$.Find the value of the constant $k$. I tried to expand the equation but got a equation of degree 6 for some reason.

Answer 1

By multiplying the two binomials inside the parentheses, we get: $$\left(-\frac{x^2}{2}+\left(2-\frac k 2\right)x+2k\right)^6$$ Here, we can't use the binomial theorem since this is a trinomial. Instead, we need to get creative.

When you think about it, there are two ways to get $x^2$:

• Five of the terms we multiply are $2k$ and one of them is $-\frac{x^2}{2}$. There are $\frac{6!}{5!1!}=6$ ways to do this.
• Four of the terms we multiply are $2k$ and two of them are $\left(2-\frac k 2\right)x$. There are $\frac{6!}{4!2!}=15$ ways to do this.

Thus, the $x^2$ term of this expansion must be: $$6\cdot(2k)^5\frac{-x^2}{2}+15(2k)^4\left(\left(2-\frac k 2\right)x\right)^2=-96k^5x^2+(60k^6-480k^5+960k^4)x^2=(60k^6-576k^5+960k^4)x^2$$

Now, you said the coefficient of $x^2$ is $84$, so we get: $$60k^6-576k^5+960k^4=84$$ There's no way we're solving this by hand. According to Wolfram Alpha, we get the real solutions of: $$x \approx -0.50361$$ $$x \approx 0.603321$$ $$x \approx 2.13401$$ $$x \approx 7.45339$$ Since this is a polynomial of sixth degree, it's likely that solving this polynomial using just radicals is actually impossible, so there's no way I can give you an exact answer. However, I have checked all of these and they are correct because they all give us a coefficient of $x^2$ that is very close to $84$ when typed into Wolfram Alpha.

• Here is the check for the first solution.
• Here is the check for the second solution.
• Here is the check for the third solution.
• Here is the check for the fourth solution.

Answer 2

We have $$\frac{1}{2^6}*(x+k)^6(x-4)^6\\(x+k)^6(x-4)^6=(P(x)+15x^2k^4+6xk^5+k^6)(Q(x)+15x^2*4^4-6x*4^5+4^6)\\15*4^4*k^6-36*4^5*k^5+4^6*15k^4$$ We have isolated the coefficient of $x^2$ so we have $$15*4k^6-36*4^2k^5+15*4^3k^4=84$$ hence $$5k^6-48k^5+80k^4=7$$ This equation is irreducible and Wolfram gives $$\color{red}{k\approx 0.603231}$$