Box 5.3a: Derivation of the Quadratic Formula

Where the Quadratic Formula Comes From

It is believed that the ancient Babylonians were the first to solve problems on which we, today, would apply a quadratic equation. They didn't use the quadratic formula, but rather a method called completing the square. The quadratic formula, though, is simply a formalization of this ancient method.

Suppose you have a quadratic in the form of

   ax² + bx + c  =  0                                             eq. q-1

If a = 0 the solution is easy -- just subtract c from both sides, then divide by b to get

           c
   x  =  -                                                        eq. q-2
           b

So we focus on the case where a is nonzero. In this case, we can divide through by a to get an equivalent quadratic

   x² + px + q  =  0                                              eq. q-3a

where

         b
   p  =                                                           eq. q-3b
         a

and

         c
   q  =                                                           eq. q-3c
         a

This simplifies it a little -- we have only two coefficients to worry about instead of three.

But if only we could massage this into

   (x + u)² - v  =  0                                             eq. q-4

then it's simple algebra to add v to both sides, take the square root of both sides, then subtract u from both sides to get

            _
   x  =  ± √v - u                                                 eq. q-5

which are the solutions for x. Converting a quadratic into the form given in eq. q-4 is called completing the square.

What the ancient Babylonians realized is that you can always complete the square in any quadratic of the form given in eq. q-3a. And by doing so, you convert it into the form given in eq. q-4.

To complete the square, you must find expressions in terms of p and q for u and v. In the search for these expressions, you start by multiplying out eq. q-4. What you get is

   x² + 2ux + u² - v  =  0                                        eq. q-6

We can see immediately that the only coefficient of the x term is 2u. In eq. q-3a, the only coefficient of the x term was p. So if it is at all possible to massage eq. q-3a into eq. q-4, then the x term coefficients must be equal, hence it must be true that

   2u  =  p                                                       eq. q-7a

and therefore

         p
   u  =                                                           eq. q-7b
         2

You can make the same argument with the constant coefficient (that is the stuff that is not multiplied by x or x²). In order for eq. q-3a and eq. q-4 to be the same, their constant coefficients must be equal. Hence

   q  =  u² - v                                                   eq. q-8a

and substituting our newly found expression for u into this we have

         p²
   q  =     - v                                                   eq. q-8b
          4

From this it is easy to write v in terms of p and q.

         p²
   v  =     - q                                                   eq. q-9
          4

So if you substitute the expressions we now have for u and v into eq. q-5, which is what we had hoped to use in order to get a solution, we do indeed get a solution:



   x  =  ±


     p
  -                                          eq. q-10a
     2

or with a little algebraic munging

            _______
         ± √p² - 4q - p
   x  =                                                           eq. q-10b
                2

I'll leave it to you to make the substitutions suggested by equations q-3b and q-3c. When you do that together with just a little more algebraic munging, the classic form of the quadratic formula emerges.

Alternative Derivation of the Quadratic Formula

If you didn't like that derivation, here's another, starting again with

   x² + px + q  =  0                                              eq. q-3a

Make the following substitution for x:

            p
  x  =  u −                                                       eq. q-11
            2

So the new quadratic equation is


    p
u −  
    2


²
  +  p


    p
u −  
    2



 + q  =  0                               eq. q-12a

When you multiply it out you get

           p    p²          p
   u² − 2u   +     + pu − p   + q  =  0                           eq. q-12b
           2 4     2

When you take all the cancellations you end up with

        p²
   u² −    + q  =  0                                              eq. q-12c
     4

To solve equation q-12c for u, simply get u² by itself on the left. Then u will be plus and minus the square root of the other side. Once you have u, substitute back to x using equation q-11. It should start looking familiar at that point.

Of course this derivation begs the question, "How did you know to make the substitution in equation q-11?" All I can say is that if you walk along the beach long enough you will find a pretty shell or two. One of those shells is that you can eliminate the second-highest power term of any polynomial using a substitution similar to this.

Also see Historical Info on the Solutions to Quadratic, Cubic and Quartic Polynomials.

Return to Main Text

You can email me by clicking this button: