# Why can ALL quadratic equations be solved by the quadratic formula?

In algebra, all quadratic problems can be solved by using the quadratic formula. I read a couple of books, and they told me only HOW and WHEN to use this formula, but they don't tell me WHY I can use it. I have tried to figure it out by proving these two equations are equal, but I can't.

Why can I use $x = \frac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$ to solve all quadratic equations?

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Look at Wikipedia. –  t.b. Jul 3 '11 at 16:16
This is false. You can't use the quadratic formula to solve quadratic equations in fields of characteristic 2. But maybe you don't know what that means yet. –  KCd Jul 3 '11 at 17:31
@KCd: Not to mention the existence of a square root. –  Phira Jul 3 '11 at 18:56
KCd and thei mention some important conditions which become crucial in general theory. The formula requires you to be able to divide by 2 and by $a$ and to take a square root. So $b^2-4ac$, also called the Discriminant, has to be a square. "characteristic 2" is one of the settings in which you can't divide by 2. If you can't take the square root of the Discriminant there are algebraic ways of adding the square root so you can solve the equation. Likewise, if you can't divide by $a$ there are sometimes ways of extending to a situation where you can. That is where more advanced algebra begins. –  Mark Bennet Jul 3 '11 at 19:16
@idonno I admire your curiosity! If only precalc students were more inquisitive! –  Pedro Tamaroff Feb 23 '12 at 3:38

I would like to prove the Quadratic Formula in a cleaner way. Perhaps if teachers see this approach they will be less reluctant to prove the Quadratic Formula.

Added: I have recently learned from the book Sources in the Development of Mathematics: Series and Products from the Fifteenth to the Twenty-first Century (Ranjan Roy) that the method described below was used by the ninth century mathematician Sridhara. (I highly recommend Roy's book, which is much broader in its coverage than the title would suggest.)

We want to solve the equation $$ax^2+bx+c=0,$$ where $a \ne 0$. The usual argument starts by dividing by $a$. That is a strategic error, division is ugly, and produces formulas that are unpleasant to typeset.

Instead, multiply both sides by $4a$. We obtain the equivalent equation $$4a^2x^2 +4abx+4ac=0.$$ Note that $4a^2x^2+4abx$ is almost the square of $2ax+b$. More precisely, $$4a^2x^2+4abx=(2ax+b)^2-b^2.$$ So our equation can be rewritten as $$(2ax+b)^2 -b^2+4ac=0$$ or equivalently $$(2ax+b)^2=b^2-4ac.$$ Now it's all over. We find that $$2ax+b=\pm\sqrt{b^2-4ac}$$ and therefore $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
No fractions until the very end!

Added: I have tried to show that initial division by $a$, when followed by a completing the square procedure, is not a simplest strategy. One might remark additionally that if we first divide by $a$, we end up needing a couple of additional "algebra" steps to partly undo the division in order to give the solutions their traditional form.

Division by $a$ is definitely a right beginning if it is followed by an argument that develops the connection between the coefficients and the sum and product of the roots. Ideally, each type of proof should be presented, since each connects to an important family of ideas. And a twice proved theorem is twice as true.

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I'm not really appreciating the difference here. In order for the quadratic formula to make any sense, you need $2a$ to be invertible in your ring, so why is it a "strategic error" to divide by $a$ in the beginning? (For instance, we could write the equation as $a(x^2 + \frac{b}{a}x+ \frac{c}{a}) = 0 \iff x^2 + Bx + C = 0$, solve that equation, and then substitute back $B = \frac{b}{a}$, $C = \frac{c}{A}$. This seems to give equally clean algebra.) And what more general setting do you have in mind? –  Pete L. Clark Jul 3 '11 at 20:31
@Pete L. Clark: The setting of the original question was presumably the reals. I wrote that it was a strategic error, not that it was an error. It is non-optimal as a calculation strategy. Dividing by $a$ produces a welter of fractions that is a hurdle for weak students. So from a didactic point of view, dividing is not good. Certainly your suggestion about $B$ and $C$ would help, though $2B=b/a$ might be better. New letters may, however, increase the "mystery math" quotient. As to other settings, maybe modulo $m$ where $(m,2a)=1$. –  André Nicolas Jul 3 '11 at 20:53
I'm still trying to understand the mathematical advantages, if any, of your approach. (By the way, you say it is well-known to anyone in number theory, and I am a number theorist.) Both your argument, the standard argument, and the form of the result require that $2a$ be invertible. Both arguments work verbatim over any domain $R$ for which $2a$ is invertible. And both arguments require more attention in rings which are not domains, since then $a^2 = b^2$ need not imply $a = \pm b$ (or even that $a$ and $b$ generate the same principal ideal). So I'm still not understanding your claims. –  Pete L. Clark Jul 3 '11 at 21:28
@Pete L. Clark: My claim is that division is ugly and produces formulas that are unpleasant to typeset. There is no mathematical difference, though the "standard" way can produce a small hiccup when we try to take $4a^2$ outside the square root sign. –  André Nicolas Jul 3 '11 at 21:37
I wish you'd been my Maths Teacher :-( –  5arx Jul 4 '11 at 10:21

Here is a slightly less ad-hoc approach to deriving the formula.

You look at the polynomial $ax^2+bx+c$ and you think of it as being composed of two kinds of indeterminates: coefficients $a$,$b$,$c$, and variable $x$. What you wish to do is if $ax^2+bx+c=a(x-r_1)(x-r_2)$ you want find an expression for $r_1$ and $r_2$ in terms of $a,b,c$ involving only the operations $+,-,\times,\div$ and $\sqrt[n]{}$.

But how are $r_1$ and $r_2$ related to $a,b$ and $c$? If you look at the expression $ax^2+bx+c=a(x-r_1)(x-r_2)$, it is easy to compute that $b=-a(r_1+r_2)$ and $c=ar_1r_2$.

Intuitively because you know that $(r_1+r_2)=-\frac ba$, determining $r_1$ and $r_2$ is the same as determining $(r_1-r_2)$. Let $E=(r_1-r_2)$ and note that $2r_1=(r_1+r_2)+(r_1-r_2)=-\frac ba+E=$ and $2r_2=(r_1+r_2)-(r_1-r_2)=-\frac ba-E$, so we already have most of our quadratic formula: $$r_1,r_2=\frac{-b}{2a}\pm\frac{E}2$$

All we need to do then, is express $E=(r_1-r_2)$ using $+,-,\times,\div,\sqrt[n]{}$ in terms of $a,b,c$. In order to do this, we need to take a small detour to see what expressions in $+,-,\times,\div$ and $a,b,c$ could possible be.

Note that the coefficients $b=-a(r_1+r_2)$ and $c=r_1r_2$ are symmetric functions in $r_1$ and $r_2$ in the sense that if you exchange $r_1$ with $r_2$ for each other, the values of $b$ and $c$ do not change. Furthermore, $b$ and $c$ are in fact scalar multiples of the so-called elementary symmetric functions, which have the property that any symmetric function (in $2$ variables) can be expressed uniquely as a polynomial (quotient of polynomials for our purposes) in them.

In particular, we can "symmetrize" the quantity $E=(r_1-r_2)$ to obtain the discriminant $D=(r_1-r_2)^2$ which is in some sense "the smallest" symmetric function of $r_1$ and $r_2$ that becomes 0 if $r_1=r_2$. Technically, though, the above is the discriminant only when $a=1$ because our coefficients $b$ and $c$ are elementary symmetric functions scaled by $a$, so we define the general discriminant to be $D=a^2(r_1-r_2)^2$. Because $D$ is symmetric and $b$ and $c$ are (up to a multiplicative factor) elementary symmetric, we should be able to express $D$ as a polynomial in $b$ and $c$.

We do so in a somewhat ad-hoc matter (though there are algorithms that will do this procedurally): $$D=a^2(r_1-r_2)^2$$ so $$D=a^2(r_1^2-2r_1r_2+r_2^2)$$ hence $$D=a^2(r_1^2+2r_1r_2+r_2^2-4r_1r_2)$$ and finally $$D=a^2(r_1+r_2)^2-a^24r_1r_2$$ giving us $$D=b^2-4ac$$

Evidently, now we have that $\sqrt{D}=a(r_1-r_2)=aE$ and so $E=\frac{\sqrt{D}}a$. This allows us to rewrite our formula so far to get from $$r_1,r_2=\frac{-b}{2a}\pm\frac{E}{2}$$ to $$r_1,r_2=\frac{-b}{2a}\pm\frac{\sqrt{D}}{2a}$$ and finally $$r_1,r_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The only strange question is: why did we only have to take one square root in order to get the formula, i.e. why did the quantity $E=(r_1-r_2)$ turn out to be a square root of a nice polynomial in $a,b,c$? That is where modern Galois theory comes in.

What's really happening is this: the first four suggest that you think of the coefficients as living in the field $F$ (a set of expressions such that adding, subtracting, multiplying, or dividing any two of them gives another expression in the set) consisting of $\{\dfrac {p(a,b,c)}{q(a,b,c)}\}$ where $p$ and $q$ are polynomials in three variables (and rational coefficients). Then $r_1$ and $r_2$ will generate an extension field $E$ of $F$, that is, the smallest field $E$ that contains $F$ and also $r_1$ and $r_2$. Galois theory says that this extension field $E$ will be a $2!=2$-dimensional vector space over $F$ and hence a single square root will be sufficient to generate $E$. Thus we need an expression in the coerfficicients (symmetric expression in the roots) whose square root is an expression in the roots, but not symmetric, and a natural choice then is the most elementary anti-symmetric function known as the Vandermonde determinant which is precisely $(r_1-r_2)$ in this case (anti-symmetric=swapping two variables flips the sign, obviously the square of an anti-symmetric function is a symmetric function).

For general polynomials, the extension field will be of higher dimension, and so you will need to take possibly several roots of different orders. Galois theory allows us to compute what these roots ought to be and in what order (giving us the cubic and quartic formulas in a way that is not ad-hoc at all), and also shows that the general degree $5$ and above polynomial does not have a formula involving only $+,-,\times,\div,\sqrt[n]{}$. (Some people feel frightened by this, because taking roots should invert the raising of powers, but this is not the case because the order of operations matters...) Now, if the coefficients of the higher degree polynomial satisfy some additional relations (i.e. are not completely independent from each other), then Galois theory also gives procedures for computing formulas for those cases and also for determining what such relations ought to be.

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Nice introduction to the Lagrange viewpoint. –  André Nicolas Jul 3 '11 at 18:41
This is why I hate it when teachers simply throw out Bhaskara's formula out there, from the blue. Although the last explanation is not really in the high school scope, the former is really useful. (+1) –  Pedro Tamaroff May 2 '12 at 23:03
I love this answer. I had always found Galois theory too advanced; but this answer makes it seem extremely logical and beautiful. –  000 May 21 '12 at 21:44
Fantastic answer! –  Sammy Black Jan 30 '14 at 19:42

Probably the easiest way to understand where the quadratic formula comes from is by 'completing the square': solving equations of the form '$x^2$=whatever' is easy, so let's see if we can put our quadratic equation ($ax^2+bx+c=0$) in that form.

The first thing to do is divide by $a$; of course this doesn't work if $a=0$, but then if that's the case our formula wasn't quadratic in the first case! This gives us $x^2+{b\over a}x+{c\over a}=0$. Now, that $b\over a$ term keeps us from having a clean square - but if we remember how to square a sum of two numbers - $(m+n)^2=m^2+2mn+n^2$ - then by substituting $x$ for $m$, we can see that our $n$ should be half of the linear term: $(x+{b\over 2a})^2 = x^2+{b\over a}x + {b^2\over 4a^2}$. But now the constant term isn't right; we have to adjust it to make it $c\over a$. A correction of $({c\over a}-{b^2\over 4a^2})$ will do this; we get $(x+{b\over 2a})^2+({c\over a}-{b^2\over 4a^2}) = 0$.

But this is exactly what we wanted; we can move that second term over to the right and get $(x+{b\over 2a})^2 = {b^2\over 4a^2}-{c\over a}$. Getting the right-hand-side cleaned up a little bit makes it ${b^2-4ac\over 4a^2}$ - just multiply the numerator and denominator of $c\over a$ by $4a$ and combine terms. Now, we can go ahead and take the square root of both sides: $x+{b\over 2a} = \pm \sqrt{b^2-4ac\over 4a^2} = \pm {\sqrt{b^2-4ac}\over\sqrt{4a^2}} = {\pm\sqrt{b^2-4ac}\over 2a}$. The last step is to subtract $b\over 2a$ from both sides, finally giving the familiar: $x = {-b\pm\sqrt{b^2-4ac}\over 2a}$

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Almost at the end $\sqrt{4a^2}$ becomes $2a$, which is technically slightly off if $a$ is negative. Of course we get saved by the $\pm$ on top. –  André Nicolas Jul 3 '11 at 17:01
If a where negative to begin with, why wouldn't we multiply both sides by -1 as the first step? –  dbasnett Jul 3 '11 at 17:51
@dbasnett: Certainly that can be done. I was just pointing out that the "standard" divide immediately by $a$ approach runs into a small technical problem. That problem can be resolved by using $\sqrt{4a^2}=2|a|$. –  André Nicolas Jul 3 '11 at 18:03
And I thought, being a 56 year old that just completed College Algebra, I would never have a reason for all of that math.;) –  dbasnett Jul 3 '11 at 18:07
It is not necessary to think in terms of absolute values (or even in terms of real numbers). The point is that if $x^2 = y^2$, then $x = \pm y$. Therefore in algebra whenever you extract a square root you have precisely a $\pm$ ambiguity and should put a $\pm$ in front of it. But we already have a $\pm$ in front of the square root, so it is okay to take any one square root of $4a^2$, namely $2a$. Or, thinking in terms of real numbers and that $\sqrt{}$ always denotes the positive square root, $\pm |a| = \{a,-a\} = \pm a$. –  Pete L. Clark Jul 3 '11 at 22:04

Complete the square,

If you dare,

Will be there!

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What value does this add? –  Peter Taylor Jul 4 '11 at 12:29
@Peter Taylor: It adds perspective. It suggests, with a light touch, that the OP has not struggled with this question nearly enough before presenting it to the community. (This will probably come back to haunt me later, but anyway...) So, I'm upvoting this answer. –  Mike Jones Jul 4 '11 at 19:43

The other answers tell you where the formula "comes from" (namely, from completing the square). If you are just happy checking that the formula gives the correct solutions whatever $a$, $b$ and $c$, you may verify that the identity $$aX^2+bX+c=a\left(X-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(X-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$$ holds for every $a$, $b$ and $c$.

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In fact, the verification is very simple using the formula $(x-y)(x+y)=x^2-y^2$. –  Martin Brandenburg Jul 26 '14 at 9:05

Pre-note: Since the asker needs an insight, I'd present a non-rigorous proof/intuition.

Try working backwards!

Assuming that the quadratic formula holds true,$$x = {-b \pm\sqrt{b^2 - 4ac} \over 2a}$$Isolate $x$.\begin{align}2ax &=& -b \pm \sqrt{b^2 - 4ac} \\ 2ax + b & = &\pm\sqrt{b^2 - 4ac} \\ (2ax + b)^2 & = & b^2 - 4ac \\ 4a^2x^2 + 4abx + b^2& = & b^2 - 4ac \\ 4a^2x^2 + 4abx + 4ac & = & 0 \\ ax^2 + bx + c & = & 0 \end{align} Note that we divided both sides by $4a$ in the last step assuming a $\ne$ $0$, which is what we have learnt all along—in the polynomial $a_nx^n + a_{n - 1}x^{n - 1}\cdots a_0x^0$, $a_n \ne 0$.

Also, if you see everything from bottom to top, you'd almost get André Nicolas' proof!

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## Proof without words.

This one shows that $$ax^2+bx+c=a\left(x+\dfrac b{2a}\right)^2+c-\dfrac{b^2}{4a}$$ from which the quadratic formula can be easily derived.

Credits to LucasVB.

I hope this helps.
Best wishes, $\mathcal H$akim.

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First, let's examine an analogous simpler special case. Why does the difference of squares formula $\rm\: x^2 - a^2\ =\ (x-a)\ (x+a)\:$ always work? Well, let's consider the obvious proof. Expanding the RHS we obtain $\rm\ (x-a)\ (x+a)\ =\ x^2 - a\ x + x\ a - a^2\$ which indeed equals $\rm x^2 - a^2\$ as long as $\rm\ a\ x = x\ a\$ for all $\rm\:x\:,\:$ i.e. as long as $\rm\:a\:$ commutes with all elements of the ring. Because the proof employed only the commutative law in addition to the standard ring axioms (most notably the distributive law) this difference of squares formula works in all commutative rings. However, generally it fails in noncommutative rings, e.g. rings involving difference or differential operators.

In fact, if we consider both $\rm\:x\:$ and $\rm\:a\:$ as indeterminates, then we can specialize the "generic" factorization $\rm\: x^2 - a^2 = (x-a)\ (x + a)\:$ in the ring $\rm\:\mathbb Z[x,a]\:$ to any commutative ring by using an evaluation homomorphism mapping $\rm\:x,a\:$ to specific values in the target ring (such an evaluation map always exists by the universal property of polynomial rings). So this formula is an identity of commutative rings, a formula universally true, i.e. true in every commutative ring. Many other well-known formulas and proofs are of this sort, e.g. the binomial theorem, resultant formulas which determine if polynomials have a common root, the Cayley-Hamilton theorem, etc.

A somewhat similar remark holds true for the well-known derivation of the quadratic formula (see e.g. André's answer here). However, it is not truly a universal formula for commutative rings because, in addition to the commutative ring axioms, we have invoked some special properties in its derivation. Namely, we have assumed the $\rm\:2\:a\:$ is invertible, and we have assumed that the discriminant has a square root in the ring. So the proof of the quadratic formula goes through in any commutative ring satisfying these two additional hypotheses. More technically we could carry out the proof generically in a ring where such elements exist, say $\rm\:e = 1/(2a),\ d^2 = b^2 - 4ac\:,\:$ so the proofs works naturally in the ring $\rm\:\mathbb Z[a,b,c,d,e]/(2ae-1, d^2-b^2-4ac)\:.\:$ Therefore any invocation of the quadratic formula can be obtained simply by specializing the proof in this generic ring, just as we did above for the difference of squares formula.

Such "generic" or "universal" proofs can yield quite nontrivial results, e.g. one can "generically" algebraically cancel "apparent singularities" in one fell swoop, before evaluation - thus avoiding alternative dense topological arguments. For example, see this slick proof of Sylvester's determinant identity $\rm\ det\ (I+AB)=det\ (I+BA)\$ that proceeds by universally cancelling $\rm\ det\ A\$ from the $\rm\ det\$ of $\rm\ \ \ (1+A\ B)\ A\ =\ A\ (1+B\ A)\:.$ Such proofs exploit to the hilt the universal properties of formal polynomials (vs. less general polynomial functions - see here for much more on this distinction).

REMARK $\$ It's worth emphasizing that in general rings it is possible for quadratic equations to have more than $2$ roots, e.g. $\rm\:x^2 = 1\:$ has roots $\rm\:\pm1,\:\pm3\ \in\ \mathbb Z/8\:,\:$ the ring of integers modulo $8\:.\:$ Thus plugging one root of the discriminant into the quadratic formula doesn't necessarily yield all the roots of a quadratic. Such anomalies cannot occur in domains, i.e. rings without zero divisors, where $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0\:.\;$ Indeed, a polynomial $\rm\ f(x)\in D[x]\$ has at most $\rm\ deg\ f\$ roots in the ring $\rm\:D\$ iff $\rm\ D\:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $\rm\ \mathbb Z/m\$ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.

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Also, I would have given this post a +1 if you did not compliment your own previous arguments while denigrating those of others. The technique you are espousing is a very powerful and interesting one and can speak for itself: it doesn't need to be promoted in this way. –  Pete L. Clark Jul 3 '11 at 23:30
@Pete In fact I intended to link to some of my prior posts to elaborate on such points when I have some spare time. But notice that I don't explicitly say anything above about uniqueness or number of roots. In particular I never mention $\sqrt{x}$ or $\pm$. So your critique is rather puzzling. It does not seem to apply to anything that I wrote above. –  Bill Dubuque Jul 3 '11 at 23:37
@Pete Please do not attempt to put pejorative words into my mouth. Nothing I wrote above "denigrates [arguments] of others". The slick universal arguments I mention are folklore. They were known long before you or I were born. To celebrate their beauty is to celebrate the beauty of mathematics. This does not "denigrate" anything. –  Bill Dubuque Jul 3 '11 at 23:42
$@$Bill: My remark about square roots was not a critique. We are talking about the quadratic formula over general rings, so my remark was directed towards that, and it was directed towards those who were thinking about your answer. As for putting words in your mouth: you wrote (not for the first time) "dense topological arguments". A careful reading shows that "dense" modifies "arguments", which I read as a denigration. If you do not intend a double entendre then I suggest a rewording: e.g. "topological density arguments". –  Pete L. Clark Jul 3 '11 at 23:56
@Pete Try as I may, I cannot even begin to imagine how you can seriously attempt to misconstrue said little mathematical pun as "denigrating the work of others". In any case, it is certainly your prerogative to withhold your upvote due to your misinterpretation of the pun. In the future could you please refrain from discussing such matters in comments to my answers. Comments are supposed to be about mathematics, not bizarre misinterpretations of puns. –  Bill Dubuque Jul 4 '11 at 14:05

The formula comes from the technique known as completing the square.

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I translate from my 1968 Algebra book by Sebastião e Silva and Silva Paulo, because I really loved it, and still love.

$$ax^{2}+bx+c=0,\qquad (a\neq 0).\qquad (1)$$

Multiplying both sides by $1/a$ we get the equivalent equation

$$x^{2}+\frac{b}{a}x+\frac{c}{a}=0.$$

We are going to show that it is possible to find $h$ and $\alpha$ such that

$$x^{2}+\frac{b}{a}x+\frac{c}{a}=(x+h)^{2}-\alpha .$$

Expanding the RHS gives

$$x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}+2hx+h^{2}-\alpha .$$

This means, applying the method of undetermined coefficients, that

$$\left\{ \begin{array}{l} \frac{b}{a}=2h \\ \frac{c}{a}=h^{2}-\alpha. \end{array} \right.$$

Hence

$$\left\{ \begin{array}{l}h=\frac{b}{2a} \\ \alpha=h^{2}-\frac{c}{a}=\frac{b^{2}}{4a^{2}}-\frac{c}{a}=\frac{b^{2}-4ac}{4a^{2}}.\end{array}\right. \qquad (2)$$

In this way the given equation is reduced to the binomial equation in $x+h$

$$\left( x+h\right) ^{2}-\alpha =0\qquad \text{equivalent to}\qquad \left( x+h\right) ^{2}=\alpha$$

and thus it is satisfied when

$$x+h=\sqrt{\alpha }\qquad \text{or}\qquad x+h=-\sqrt{\alpha },$$

i.e. when

$$x=-h+\sqrt{\alpha }\qquad \text{or}\qquad x=-h-\sqrt{\alpha }.$$

Denoting the first value of $x$ by $x_{1}$ and the second by $x_{2}$ and replacing $h$ and $\alpha$ by their expressions given by $(2)$, yields

$$x_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\qquad \text{or}\qquad x_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}.\qquad (3)$$

As it can be seen nowhere the method of completing the square was mentioned. Rather the method of undetermined coefficients was fully explained previously.

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Why? Because we let $a$, $b$, and $c$ be anything (usually real numbers).
So our result doesn't depend on the coefficients, only on the fact that we had a polynomial of degree two (i.e. Quadratic)

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$a$ can't be anything; rather it can be anything except $0$. –  Michael Hardy Jul 4 '11 at 3:08
I thought that was implied by "degree two"! –  The Chaz 2.0 Jul 4 '11 at 5:40

This question was raised over two years ago and already received a number of nice answers. Let me just mention a really elementary approach to the general quadratic equation (in $\mathbb{R}$ or $\mathbb{Q}$).

Step 1. The standard identity $$(x + y)^2 = (x-y)^2 + 4xy \qquad \qquad (1)$$

Step 2. Given the area of a rectangle and the difference between its two sides, determine its length and width. The difference between the length $x$ and the width $y$ is $x - y$, and the area is $xy$. Now by (1), $x +y$ is the positive square root of $(x-y)^2 + 4xy$. Finally, $x = \frac{1}{2}(x+y) + (x-y)$ and $y = \frac{1}{2}(x+y) + (x-y)$.

Step 3. Consider the equation $ax^2 + bx + c = 0$ and suppose that $a \not= 0$. Dividing by $a$ yields $x^2 + \frac{b}{a}x +\frac{c}{a} = 0$ or $x(x + \frac{b}{a}) = -\frac{c}{a}$. Setting $y = x + \frac{b}{a}$, we get $xy= -\frac{c}{a}$ and $x - y = -\frac{b}{a}$. The analogy with Step 2 is clear and the solution is the same. Putting everything together leads to the quadratic formula.

I found this approach as a child, although each step took me several years... I wonder whether it was ever used to teach the quadratic formula.

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The simplest quadratic equation is $x^2-a=0$, since it has the solutions $x=\pm \sqrt{a}$.

The good news is that every quadratic equation can be reduced to this simple quadratic equation: The equation $ax^2+bx+c=0$ means $x^2+\frac{b}{a} x + \frac{c}{a}=0$. Writing $p=\frac{b}{a}$, $q=\frac{c}{a}$, we only have to solve $x^2+px+q=0$. If we write $x=y-\frac{p}{2}$, then a little calculation shows $y^2 - \frac{p^2}{4}+q=0$.

$$\Rightarrow ~ y=\pm \sqrt{\frac{p^2}{4}-q} ~ \Rightarrow ~ x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4}-q} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

Geometrically, $x=y-\frac{p}{2}$ moves the parabola horizontally so that it becomes symmetric to the $y$-axis. After this shift the roots are easy to find.

This method also works for polynomials of higher degrees: Tschirnhaus transformation.

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I'll just let the algebra speak for itself.

For $a \ \ne \ 0$

If

$ax^2 = bx + c$

Then

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

PROOF:

$4aax^2 = 4abx + 4ac$

$(2ax)^2 + b^2 = 4abx + 4ac + b^2$

$(2ax)^2 - 4abx + b^2 = b^2 + 4ac$

$(2ax - b)^2 = b^2 + 4ac$

$2ax - b = \pm \sqrt{b^2 + 4ac}$

$2ax = b \pm \sqrt{b^2 + 4ac}$

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

Q.E.D.

SHORTCUT

Using the standard full definition , replace b with -b and c with -c to get the same thing.

For $a \ne 0$

If

$ax^2 + bx + c = 0$

Then

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

Make the replacements in the 'if'

$ax^2 + (-b)x + (-c) = 0$

$ax^2 = bx + c$

Make the replacements in the 'then'

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

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

Q.E.D

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The proof of the quadratic formula takes advantage of completing the square. $$ax^2 + bx + c = 0, \ a \neq 0, \ a, \ b, \ c \ \in \mathbb R$$ $$ax^2 + bx = -c$$ $$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$$ $$x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$$ $$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}$$ $$\left(x+\dfrac{b}{2a}\right)^2 = -\dfrac{4ac}{4a^2} + \dfrac{b^2}{4a^2}$$ $$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{-4ac+b^2}{4a^2}$$ $$\left(x+\dfrac{b}{2a}\right)^2 = \dfrac{b^2-4ac}{4a^2}$$ $$x+\dfrac{b}{2a} = \pm \sqrt{\dfrac{b^2-4ac}{4a^2}}$$ $$x+\dfrac{b}{2a} = \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$ $$x=-\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$ $$\boxed{x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}}$$ This concludes the proof.

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Most answers are explaining the method of completing the square. Although its the preferred method, I'll take another approach.

Consider an equation $$~~~~~~~~~~~~~~~~~~~~~ax^{2}+bx+c=0~~~~~~~~~~~~~~~~~~~~(1)$$We let the roots be $\alpha$ and $\beta$. Now, $$~~~~~~~~~~~~~~~~~~~~~x-\alpha = x-\beta = 0~~~~~~~~~~~~~~~~~~~~~~~~$$ $$~~~~~~~~~~~~~~~~~~~~~k(x-\alpha)(x-\beta)=0~~~~~~~~~~~~~~~~~~~~(2)$$ Equating equation 1 and 2 (k is a constant), $$ax^{2}+b{x}+c=k(x-\alpha)(x-\beta)$$ $$ax^{2}+b{x}+c=k(x^{2}-\alpha x-\beta x+\alpha \beta)$$ $$ax^{2}+b{x}+c=kx^{2}-k(\alpha+\beta )x+k\alpha \beta)$$ Comparing both sides, we get $$a=k~;~b=-k(\alpha +\beta)~;~c=k\alpha \beta$$ From this, we get $$\alpha + \beta = \frac{-b}{a}~~;~~~\alpha \beta = \frac{c}{a}$$

Now, to get the value of $\alpha$, we follow the following procedure : First we take out the value of $\alpha - \beta$, so that we can eliminate one term and find out the value of another. $$(\alpha-\beta)^{2} = \alpha ^{2}+ \beta ^{2} - 2 \alpha \beta$$ Now we'll add $4 \alpha \beta$ on both the sides $$(\alpha-\beta)^{2} +4 \alpha \beta = \alpha ^{2}+ \beta ^{2} + 2 \alpha \beta$$ $$(\alpha-\beta)^{2} +4 \alpha \beta = (\alpha + \beta )^{2}$$ $$(\alpha-\beta)^{2} = (\alpha + \beta )^{2} -4 \alpha \beta$$ $$\alpha-\beta = \pm \sqrt{(\alpha + \beta )^{2} -4 \alpha \beta }$$ Substituting the values of $\alpha + \beta$ and $\alpha \beta$, we get, $$\alpha-\beta = \pm \sqrt{(\frac{-b}{a} )^{2} -\frac{4c}{a} }$$ $$\alpha-\beta = \pm \sqrt{\frac{b^{2}-4ac}{a^{2}} }$$ or $$~~~~~~~~~~~~~~~~~~~~~\alpha-\beta = \frac{\pm \sqrt{b^{2}-4ac}}{a} ~~~~~~~~~~~~~~~~~~~~~(3)$$ Adding $Eq^{n} (2)~and~(3)$, we get, $$2 \alpha = \frac{-b \pm \sqrt{b^{2}-4ac}}{a}$$ $$\alpha = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

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$x-\alpha = x-\beta = 0$ would imply $\alpha=\beta=0$. Your proof needs the theorem from vieta. I can't see how you get to the result by adding equations 2 and 3 as you stated in the text. Sorry, but the whole proof is rather complicated and rather boring. –  miracle173 Nov 25 '14 at 5:16
you could rather write it as $x_1-\alpha = x_2-\beta = 0$, just to differentiate between the roots. This proof might be boring but it's not at all complicated. –  The Pragmatick Nov 25 '14 at 11:35

All the answers are good and provide the method or algorithm by which the quadratic formula is derived. But i think the question has one more part in it. Why for quadratics and not for sth else?

So by methods of Galois Theory (mention it just for reference), the thing that makes the quadratic solvable by such formulas (called radical formulas) is that due ot its degree ($2$) and the properties of arithmetic fields (e.g the reals) one can step-by-step construct solutions by manipulating the equation without introducing ambiguity in the process (in the language of modern Galois theory this is stated as "the Galois extension of the equation is solvable", that is the group representing the extension has this property which allows to construct step-by-step without introducing ambiguity during the process)

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