I'm currently going through Introduction to Classical and Quantum Computing, by Thomas Wong, and I'm struggling with exercise 2.29 (page 107):
Exercise 2.29. Say $\left| \psi \right> = \alpha\left| 0 \right> + \beta\left| 1 \right>$ is a normalized quantum state, i.e., $|\alpha|^2 + |\beta|^2 = 1$.
(a) Calculate $H\left| \psi \right>$.
Given the two following transformations the Hadamard gate does:
$$\begin{align} H\left| 0 \right> &= \frac{1}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>)\\ H\left| 1 \right> &= \frac{1}{\sqrt{2}}(\left| 0 \right> - \left| 1 \right>) \end{align}$$
I did the following:
$$\begin{align} H\left| \psi \right> &= \alpha H\left| 0 \right> + \beta H\left| 1 \right>\\ &= \frac{\alpha}{\sqrt{2}}(\left| 0 \right> + \left| 1 \right>) + \frac{\beta}{\sqrt{2}}(\left| 0 \right> - \left| 1 \right>)\\ &= \frac{\alpha}{\sqrt{2}}\left| 0 \right> + \frac{\alpha}{\sqrt{2}}\left| 1 \right> + \frac{\beta}{\sqrt{2}}\left| 0 \right> - \frac{\beta}{\sqrt{2}}\left| 1 \right>\\ &= \frac{\alpha + \beta}{\sqrt{2}}\left| 0 \right> + \frac{\alpha - \beta}{\sqrt{2}}\left| 1 \right> \end{align}$$
Though, the answer given at the back of the book (page 360) is (notice the $+$ sign): $$\begin{align} H\left| \psi \right> &= \frac{\alpha + \beta}{\sqrt{2}}\left| 0 \right> + \frac{\alpha + \beta}{\sqrt{2}}\left| 1 \right> \end{align}$$
Why is that? This is a simple exercise, yet I can't seem to find what I'm doing wrong.