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

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.

Answer 1

Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have \begin{align*} \sqrt{b^2 - 4ac} &= \sqrt{b^2}\\ &= \left|b\right|\\ &\neq 0, \end{align*} so you can see that this simplification cannot be correct.

It seems that in your proposed simplification, you have completely disregarded the subtraction occurring in the radical. Moreover, $\sqrt{b^2} = \left|b\right|$, not just $b$. To see this with an example, take $b = -1$. Then $\sqrt{(-1)^2} = \sqrt{1} = 1 = \left|-1\right|\neq -1$.

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EDIT:

Again, the simplification is incorrect. While it is true in general that $\sqrt{a^2b} = \left|a\right|\sqrt{b}$, this is not the situation you are in here: $$ b^2 - 4ac\neq (b - 2)^2 ac = (b^2 - 4b + 4)ac = b^2 ac - 4abc + 4ac. $$ You seem to have made a few mistakes here (if I'm to take a stab at the reasoning behind the simplification): first you've incorrectly simplified $b^2 - 4ac$ as $(b^2 - 4)ac$ (which is not true, because the first term in the former has no $ac$), and then you've simplified $b^2 - 4$ as $(b - 2)^2$, which is also not true (take $b = 0$ to see why). In general, $(x + y)^n\neq x^n + y^n$: this is a common mistake algebra learners make! Remember that when expanding $(x + y)^2$, we need to use the distributive property, and not simply regard squaring as linear: \begin{align*} (x + y)^2 &= (x + y)(x + y)\\ &= x^2 + yx + xy + y^2\\ &= x^2 + 2xy + y^2. \end{align*}

Answer 2

The "simplification" is incorrect: $\sqrt{b^2-4ac}\not=2b\sqrt{ac}$.

For example, take $b=1, a=c=0$. Then the former expression is $1$ but the latter is $0$.

What is true is that $2\vert b\vert\sqrt{ac}=\sqrt{b^2\cdot4ac}$, but note the replacement of "$-$" with "$\cdot$", there: that's a major change! (Also note the absolute value, which is important but less fundamental in this case.)

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EDIT: the question has now been changed to reflect a new simplification - namely, $$\sqrt{b^2-4ac}=(2-b)\sqrt{ac}.$$ However, this one is also false: again, set $a=c=0$, $b=1$ to see the difference.

Note that this example really shows that the discriminant can't (in general) be written in the form $[stuff]\sqrt{ac}$. Note that this applies to the new edit, which replaced "$b-2$" with "$2-b$" (as well as the very first version) - the "simplification" is still wrong, for the same reason. In a certain sense, any expression of the form you are looking at gives $a$ and $c$ "too much power" over the value; no choice of $a$ and $c$ can guarantee that the discriminant is zero regardless of what $b$ is.

This time it's not clear to me what the algebra error is; can you explain why you thought this simplification worked? EDIT: Stahl's answer takes a stab at guessing what happened; if that's not an accurate interpretation, please explain how you came by this "simplification."

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FURTHER EDIT: You've added your reasoning; you make two fundamental mistakes. The gist of your argument is $$\sqrt{b^2-4ac}=\sqrt{b^2}-\sqrt{4ac}=(b-2)\sqrt{ac}.$$ Both of these equalities are false.

In the first case, we do not have $\sqrt{X+Y}=\sqrt{X}+\sqrt{Y}$, any more than we have $(X+Y)^2=X^2+Y^2$. For an explicit counterexample, take $X=Y=2$, where $\sqrt{X+Y}=\sqrt{4}=2$ but $\sqrt{X}+\sqrt{Y}=2\sqrt{2}>2$.

For the second one, it is true that $\sqrt{b^2}-\sqrt{4ac}=b-2\sqrt{ac}$ (assuming $b$ is positive, that is); however, this is not the same as $(b-2)\sqrt{ac}$! The parentheses definitely matter.

These are both the same "species" of error - they both involve misunderstanding how the various algebraic operations interact with each other. You can't rearrange operations willy-nilly: e.g. "adding, then squaring" is very different from "squaring, then adding", and so on.

Answer 3

There is a small simplification that can be made. Let's rewrite the quadratic equation as

$$ x^2 + 2B x+C=0. $$

Then the quadratic formula reduces to

$$ x= -B \pm \sqrt{B^2 - C}, $$

which is somewhat more palatable. It's occasionally more convenient in physics.

Answer 4

\begin{align} (b-2)\sqrt{ac} & = \sqrt{(b-2)^2} \sqrt{ac} & & \text{ if } b-2\ge 0 \\[10pt] & = \sqrt{(b-2)^2 ac}. \end{align} Is $(b-2)^2ac$ the same as $b^2-4ac\,$?

If one were to think that $(b-2)^2$ is the same as $b^2-4$ (and it is not) then one would have $(b-2)^2ac = b^2ac-4ac,$ so that is still not the same as $b^2-4ac.$

Notice that $\sqrt{5^2 - 3^2} = 4$ and $\sqrt{5^2}-\sqrt{3^2} = 5 - 3 = 2,$ so $\sqrt{5^2-3^2}$ is different from $\sqrt{5^2}-\sqrt{3^2}.$

One should not generally ask why one cannot do things like this; but rather whether one can. Don't start from the presumption that it can be done. That puts the burden of proof in the wrong place.

Answer 5

(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 $$\sqrt{25-16} = \sqrt{9} = 3$$ $$\sqrt{25}-\sqrt{16} = 5-4 = 1$$ $$3 \neq 1$$

Answer 6

I think a better way to parse the quadratic equation is as:

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

The left term ($-\frac{b}{2a}$) is the x-coordinate of the axis of symmetry (often taught in Algebra as a formula, but easily recognized to students of calculus as the solution to the first-order condition of a quadratic).

The right-hand term is the "margin" around the axis of symmetry giving how far to the left/right of the axis we find the zeroes.

The numerator of the right-hand term is the discriminant, the sign of which determines the number of zeroes there are.

The denominator of the right-hand term shows that as $a$ increases, ceteris paribus, the zeroes will move closer to the axis of symmetry -- intuitively, $a$ is the "stretch" factor, and higher $a$ means the parabola will be more "squeezed" (relatively speaking). The more "squeezed" the parabola, the closer its zeros will be to the "middle".