Deriving the Quadratic Formula (the easy way)

In Bowen’s post from April 25, he showed a great method for factoring non-monic quadratics. Using that same method, you can derive the quadratic formula pretty cleanly, without lots of fractions, rationalizations, and the like.

The goal is to find a solution to the generic quadratic equation:

ax^2 + bx + c = 0

We can’t factor using sums and products, so we resort to completing the square. The equation would be much easier to work with if it were monic. We could divide through by a, but then there’s lots of fractions to keep track of. Instead, let’s multiply both sides by a and make it a quadratic in ax:

a^2x^2 + abx + ac = 0

Hmmm. Again, things would be so much easier to complete the square if that middle term were even. We can multiply both sides by 2 to do that, but then the first term isn’t a perfect square. Fine, let’s multiply both sides by 4 then:

4a^2x^2 + 4abx + 4ac = 0

Now, just rewrite that a little bit:

(2ax)^2 + 2b(2ax) + 4ac = 0

and now we can put our fingers over 2ax and see this as a simpler monic:

F^2 + 2bF + 4ac = 0

And now, complete the square. First, let’s put that constant on the other side:

F^2 + 2bF = -4ac

To get a complete square on the left side, we need a b^2, so add it to each side:

F^2 + 2bF + b^2 = b^2 - 4ac

The left side is now a perfect square… and the right side looks familiar… Let’s factor the left side:

(F + b)^2 = b^2 - 4ac

To solve for F, take the square root of each side:

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

and subtract b from each side:

F = - b \pm \sqrt{b^2 - 4ac}

What was F again? Lift that finger…oh, yeah, 2ax:

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

So finish this off by dividing each side by 2a:

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

And there you have it… the quadratic formula, with no fractions until the final step.  Enjoy!

6 Responses to Deriving the Quadratic Formula (the easy way)

  1. I was taught how to do factoring of quadratic polynomials which were not monic in high school back in
    1949-50. I do not know when it disappeared from school mathematics, but it is a useful device, and I
    have used it for other problems in my research. With that background, maybe my comment on using
    this to derive the quadratic formula will not be taken as from someone who wants to see his schooling
    replicated now. I think it is a serious mistake to use this to derive the quadratic formula. One reason
    given was the lack of fractions until the final step. Students have to learn to work with fractions, and
    using them in this setting shows that they can be used to help derive something general and important.
    A more important reason is that students need to learn how to complete a square, which is used in
    many other settings than in deriving the quadratic formula. It is a general method which needs to natural
    to use in appropriate settings. The trick used in the derivation of the quadratic formula is not what I want
    students to think of when they need to complete a square.

  2. Bowen Kerins says:

    The method presented here isn’t actually the one that appears in CME Algebra 1 — that method first solves quadratics for monic cases, then transforms ax^2 + bx + c = 0 into x^2 + \frac b a x + \frac c a = 0. I am curious about opinions on that development, and whether people think it is appropriate for the text to show more than one derivation. One thought for the next version of CME (if such a thing happens) is to use the same development for the quadratic formula, then (perhaps in a later lesson) show how the “chunking” method can also be used to derive it.

    My opinion is that the derivation here has the most muscle with students who already know the quadratic formula, because you get to see it being built here: the 4ac is constructed, then the b^2, then the square root, and so on. It’s impressive, but if you didn’t recognize those already as pieces of the quadratic formula, it wouldn’t be.

    You definitely bring up a good point about completing the square, and I agree that completing the square needs to be present in other, more general, settings, and should not be limited in scope to the step seen here. Thanks for the comment.

  3. Al Cuoco says:

    I like Kevin’s approach, because it amounts to a change of variable that makes the polynomial monic. This has many other applications.

    Here’s yet another derivation of the formula. Normalize the equation so that it looks like x^2+bx+c=0 (via the scaling method implicit in Kevin’s post or by dividing both sides by a). If r and s are the roots, you know that

    r+s = -b and rs = c

    Since you know r+s, you could fund r and s if you knew r-s. But

    (r-s)^2= (r+s)^2 – 4rs = b^2-4c, so r-s is the square root of that, and you’re done.

    This works for cubics, too, if you use the cube roots of unity instead of plus or minus 1 (the square roots of unity).


  4. Ladnor Geissinger says:

    I suggest that factoring quadratics could profitably be developed from the simple idea of difference of squares — which we might discover by multiplication and then consider to be one of the first “factoring” results: (R – 1)*(R + 1) = R^2 – 1 or, if R = A/D and we multiply by D^2, then we get the usual form:
    (A – D)*(A+D) = A^2 – D^2
    Let’s change our perspective and let T = A – D and U = A + D , then A =[T+U] /2 and D = [U – T] /2 so that A is the “average” of T and U while D is the “deviation” of T and U from A. Our new version of difference of squares now takes the form sometimes called the “polarization identity”:
    T * U = ([T+U] /2)^2 – ([U-T] /2)^2
    Any product can be expressed as a difference of squares (average)^2 – (deviation)^2 !

    Notice that this immediately yields the “arithmetic/geometric mean inequality”. The square of the geometric mean is T*U which is smaller than the square of the average ([T+U] /2)^2 by the amount
    ([U – T] /2)^2 which is the square of the deviation.

    Now suppose that we have a quadratic Q = a*X^2 +b*X +c and we use the “chunking” or scaling idea and write a*Q = (a*X + b)*(a*X) + a*c . Let T = a*X and U = a*X+b then the “average” is a*X +[b/2] and the “deviation” is b/2 so by our difference of squares result,
    (a*X + b)*(a*X) = (a*X + [b/2])^2 – [b/2]^2 and we have “completed the square”.
    Thus a* Q = (a*X + [b/2])^2 – [b/2]^2 + a*c = (a*X + [b/2])^2 – (b^2 – 4*a*c)/4

    The quantity (b^2 – 4*a*c) is called the discriminant of Q; it tells us important information about Q.
    If you want to check on the roots of Q = 0, you can look instead at a*Q = 0.
    So the problem is to determine values of X for which (a*X + [b/2])^2 = (b^2 – 4*a*c)/4 and for that we need to know about a square root of (b^2 – 4*a*c). Check out the case when b^2 – 4*a*c = 0.
    If b^2 – 4*a*c is negative then its square root is imaginary and any solution is a complex number.
    If b^2 – 4*a*c is positive let k be its positive real square root. Then the equation is
    (a*X + [b/2])^2 = [k/2]^2 which reduces to the “quadratic formula”:
    a*X + [b/2] = [k/2] or a*X + [b/2] = – [k/2] That is, X is a root of Q if and only if
    either a*X = – b/2 – k/2 or a*X = -b/2 +k/2

    We now have a*Q = (a*X + [b/2])^2 – (b^2 – 4*a*c)/4 = (a*X + [b/2])^2 – [k/2]^2 a difference of squares, so
    can factor to get
    a*Q = (a*X + [b+k] /2) * (a*X + [b – k] /2)

    We are especially interested in the case when each of a,b,c is an integer and b^2 – 4*a*c is positive. Then by a number theory result, the square root k is either an integer also, or it is an irrational number. When k is an integer, you can easily check from b^2 – 4*a*c = k^2 that b and k have the same parity, so that
    p = [b+k]/2 and s = [b-k]/2 are both integers.

    Notice that means a*Q = (a*X + p) * (a*X + s) where p and s are integers and p+s = b and p*s = a*c. This is what you get if you use the “key number method” as described by Mr. Kerins. There remains just one more problem. We need to factor a = m*h with integers m and h such that m is a divisor or p and h is a divisor of s. If we can do that, let p=m*p’ and s=h*s’ with p’ and s’ integers. Then finally we have the desired factorization:
    Q = (h*X + p’) * (m*X + s’)

    Note that since a*c = p*s, a is a divisor of p*s. So we can first let m =gcd (a,p) and then
    let a = m*h and p = m*p’ and h and p’ have no factors in common. Now m* h* c = m* p’ *s so factoring out m we have h *c =p’ *s. Since h is a divisor of p’ *s and h is relatively prime to p’, then h must divide s so we can write s = h* s’. Finished!

  5. Christopher says:

    Boy oh boy. I gotta say that I don’t get this at all. The original post was kind of fun; a clever inroad for a challenging symbolic derivation. But then things go haywire in the comments right away. Is the derivation of the quadratic formula so fundamentally important argument that the particular argument we use is worth all this sturm und drang?

    Speaking both as a middle school guy and a college math instructor, I just don’t get it. These are all lovely arguments, it seems to me.

  6. arnold omondi says:

    from all those am kind of the way am a kenyan nd live in kenya..its tht my countries xyxtem math disturbes mst students bt me am a kind of a genious coz i factord out the exact way bt the prob waz tht he did it am thnk full tht great mynds thnk a lyk..nd am only 16yrz…am glad tht math exists….i jst wanted 2 cheak if someone had found it out bt i finnaly cured my curiosity., .bt u cn get mi on facebook as arnold addi fracuss..thnx…alot…

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