How do I find all the coefficients of $x^0$ in this: $(x-\frac{2}{\sqrt x})^8$

Question

How do I find all the coefficients of $x^0$ in this: $(x-\frac{2}{\sqrt x})^8$

I got to $8 - k + (-0.5)k = 0$ and then $ 16 - 3k = 0 $ and thus, I can't find $k$ that will solve this equation. What do u guys think?

Answer 1

The Binomial Theorem states that $$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k} a^{n - k}b^k$$ Here, $a = x$, $b = -\dfrac{2}{\sqrt{x}}$, and $n = 8$. Hence, $$\left(x - \frac{2}{\sqrt{x}}\right)^8 = \sum_{k = 0}^{8} \binom{8}{k} x^{8 - k}\left(-\frac{2}{\sqrt{x}}\right)^k$$ For the $x^0$ term to exist, we require that $$x^{8 - k}\left(\frac{1}{\sqrt{x}}\right)^k = x^{8 - k}(x^{-\frac{1}{2}})^k = x^{8 - k}x^{-\frac{k}{2}} = x^{8 - \frac{3k}{2}} = x^0$$ Since the bases are the same, the exponents must be equal. Hence, \begin{align*} 8 - \frac{3k}{2} & = 0\\ 16 - 3k & = 0\\ -3k & = 16\\ k & = \frac{16}{3} \end{align*} Since $k$ is not an integer, there is no $x^0$ term in the binomial expansion of $$\left(x - \frac{2}{\sqrt{x}}\right)^8$$