How to find the range of $\sqrt {x^2-5x+4}$ where x is real. How to find the range of $$\sqrt {x^2-5x+4}$$ where $x$ is real.
What I've tried:
Let $\sqrt {x^2-5x+4}=y$, solving for real $x$, as $x$ is real discriminant must be $D\geq0$. Solving I get $y^2\geq\frac{-9}{4}$. Which I suppose implies all real . But on wolfram alpha it says $y\geq0$. What am I doing wrong.
Any help appreciated.
Thank you.

Note that the domain of the function is $(-\infty,1]\cup [4, \infty)$, since 
$$ y=\sqrt{(x-1)(x-4)}, $$
which tells you that $y\geq 0$
Note: Compare with the function

$$y=\sqrt{x}.$$

