Roots of factorization with expressions for constants

Question

When faced with the factorization $$(\alpha - 1 -\lambda)(\mu - 1 -\lambda)=0$$ the roots of $\lambda$ can simply be said to be $\alpha - 1$ and $\mu - 1$. Why is that answer different to the following working? $$\begin{align*} \alpha\mu - \alpha - \alpha\lambda - \mu +1 + 2\lambda - \lambda\mu + \lambda^2 &=0 \\ \lambda^2 + (2-\alpha - \mu) \lambda + \alpha\mu - \alpha - \mu + 1 &=0 \end{align*}$$ $$\begin{align*} \lambda &= \frac{\alpha + \mu - 2 \pm\sqrt{(2-\alpha-\mu)^2 - 4(\alpha\mu - \alpha -\mu + 1)}}2 \\ &= \frac{\alpha + \mu - 2 \pm\sqrt{4 - 4\alpha -4\mu + \alpha^2 + 2\alpha\mu + \mu^2 - 4(\alpha\mu - \alpha -\mu + 1)}}2 \\ &= \frac{\alpha + \mu - 2 \pm\sqrt{\alpha^2 - 2\alpha\mu + \mu^2}}2 \\ &= \frac{\alpha + \mu - 2 \pm\sqrt{(\alpha - \mu)^2}}2\\ &= \frac{-\alpha + 3\mu }2\qquad\text{or}\qquad \frac {3\alpha - \mu}2 \end{align*}$$

Answer 1

$$\lambda_{1,2}=\frac{\alpha+\mu-2\pm\sqrt{(\alpha-\mu)^2}}{2}=\frac{\alpha+\mu-2\pm(\alpha-\mu)}{2}\\ \implies\lambda_1=\frac{2\alpha-2}2=\alpha-1\\ \implies\lambda_2=\frac{2\mu-2}2=\mu-1$$ What seems to be the problem? I cannot be sure how you ended up with the solutions that you did.