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:


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.

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. MAKE SURE YOU DON'T REPEAT A COMMENT ALREADY MADE BY SOMEONE ELSE. Nothing very wrong with the discussion, it just became a bit too long and was also losing focus. I'm simply responding to several flags. $\endgroup$ May 30, 2017 at 6:32
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    $\begingroup$ > I asked my teacher, and she said she agrees with me; the radical can be simplified. I'd suggest asking her again. Maybe she didn't understand you, or was pre-occupied with something else. If she still says your simplification is correct, I'd say you need a new teacher. $\endgroup$
    – bubba
    May 30, 2017 at 13:00
  • $\begingroup$ @bubba I agree. I'm hoping she was just really busy.... As for the question, the radical portion is as simple as it gets and cannot be simplified further. You can't just tear apart the radical across an addition or subtraction as a basic rule of simplifying radicals. If there was a middle term as produced from something like the FOIL process, then you could eliminate the radical and pull things apart, but here you cannot. $\endgroup$
    – MrJman006
    May 30, 2017 at 14:38
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    $\begingroup$ By the way, it's okay to keep the radical $\sqrt{b^2 - 4ac}$, because it actually involves an expression of the discriminant of the polynomial whose two values of $x$ are the roots.That expression could be written $\frac{-b \pm \sqrt{\Delta}}{2a}$, which is simpler and commonly used. $\endgroup$
    – Right Leg
    May 30, 2017 at 14:53
  • $\begingroup$ Why don't you try out your simplified formula on some problems from your textbook and see if it works? It is easy to test the solutions you get by plugging them into the original equation. The result will be zero if the solution is correct. Depending on the outcome of your trials, you go to the teacher and thank her for a great idea, or go to the Principal and denounce her. $\endgroup$ May 31, 2017 at 17:22

6 Answers 6


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 $\sqrt{(-1)^2} = \sqrt{1} = 1 = \left|-1\right|\neq -1$.


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*}

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    $\begingroup$ I don't think that this answer address the source of the problem: It simply says: You're wrong, and I'll give you a specific example where it does not hold. $\endgroup$
    – amWhy
    May 28, 2017 at 22:15
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    $\begingroup$ Oh, I see my mistake. It has to do with how multiplying polynomials (FOIL) works, right? +1. $\endgroup$
    – AAM111
    May 28, 2017 at 22:36
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    $\begingroup$ "While it is true in general that $\sqrt{a^2b} = a\sqrt{b}$...". Not quite. $\sqrt{a^2b} = |a|\sqrt{b}$. That absolute value sign becomes very important when $a$ is a negative real number. It's a small thing, but it's usually the small things that trip up those learning math (like the original asker). So it's good to teach them rigour at all stages. $\endgroup$
    – Deepak
    May 29, 2017 at 1:21
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    $\begingroup$ @amwhy And that is exactly how you show something is wrong. Guessing what the OP thought and trying to explain why it's wrong is an awful way to teach and learn math. In life everything that is not forbidden is allowed and thank god for it, but in math everything that's not allowed is forbidden and the earlier students learn that the better off they will be when they get to the non-trivial stuff. $\endgroup$
    – DRF
    May 29, 2017 at 18:32
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    $\begingroup$ We shall agree to disagree. Studies in cognition overwhelmingly show that learning occurs best when, in the case of a mistake on the part of the user, it is identified, without judgment, and explained, but importantly, when a mistake is found, it is met with a suggestion, example, and direction which the learner has access to. At any rate, I think my comment is a valid perspective, among many perspectives, some more informed than others. $\endgroup$
    – amWhy
    May 29, 2017 at 18:51

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.)

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."

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.

  • $\begingroup$ I credit you, because in addition to noting the simplification of the asker is wrong, and providing a good "counterexample, I think you nailed the source of confusion for the op in your last sentence. $\endgroup$
    – amWhy
    May 28, 2017 at 22:20
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    $\begingroup$ @amWhy To be fair, Stahl's answer also does that in the last paragraph (our answers are pretty much isomorphic as far as I can tell - including timing!). $\endgroup$ May 28, 2017 at 22:21
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    $\begingroup$ Noah I did NOT downvote @Stahl, but I did upvote your answer, because you are far more explicit about the source of the problem, at least more immediately so. $\endgroup$
    – amWhy
    May 28, 2017 at 22:23
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    $\begingroup$ @projectilemotion My penultimate paragraph still addresses the issue (and I've edited to clarify that). $\endgroup$ May 28, 2017 at 22:38
  • $\begingroup$ IMHO this remains the better of the two answers, before and after the asker's edit. I can only upvote once, which I did earlier, so I added this comment to note the correctness of, and the educational strength of this answer. $\endgroup$
    – amWhy
    May 28, 2017 at 22:46

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.

  • $\begingroup$ This was how I was taught it: first get it into the the form above... and then apply the formula. If this really is a simplification can be argued in that it makes the forumla simpler but it makes the preparatory work less simple. You cannot apply it to any quadratic equation without first modifying it. $\endgroup$
    – MichaelK
    May 30, 2017 at 13:53

\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.


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


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".

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    $\begingroup$ I didn't downvote, but this doesn't actually address the question, which was about simplifying the discriminant $\sqrt{b^2-4ac}$. $\endgroup$ Jun 6, 2017 at 18:55
  • $\begingroup$ @NoahSchweber when I wrote this answer, the title was different -- in fact it wasn't edited by OP, so it's still not clear OP's original intent. OP wrote "However, this strikes me as not being simplified," which suggests to me OP is more generally struggling with parsing the solution and looking to simplify it. Simplifying it means not just shrinking the radical, but making it easier to understand/remember. $\endgroup$ Jun 7, 2017 at 14:46

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