TITLE: Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$?
QUESTION [14 upvotes]: I am currently taking Algebra 1 (the school year's almost over ), and we just learned the quadratic formula, another method to solve quadratic equations:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
However, this strikes me as not being simplified. Isn't it more proper to write it like this?
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$
$$x=\frac{-b\pm((b-2)\sqrt{ac})}{2a}$$
$$x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$$
Why isn't $x=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$ more commonly used as the quadratic formula??
I'm sorry for my typo, I have edited it.
I have now edited in my steps, per request of commenters.

REPLY [5 votes]: (I think the original post has been edited to change the location of the error.)
You have this incorrect step (known as "The Freshman's Dream"):

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
  $$x=\frac{-b\pm(\sqrt{b^2}-\sqrt{4ac})}{2a}$$

We can show by example that $$\sqrt{a-b} \neq \sqrt{a}-\sqrt{b}$$
Using a=25 and b = 16
$$\sqrt{25-16} = \sqrt{9} = 3$$
$$\sqrt{25}-\sqrt{16} = 5-4 = 1$$
$$3 \neq 1$$