20 Apr

Suppose that a, b and c are constants with ‘a’ nonzero. Consider the equation

$\large ax^2 + bx + c = 0$

It is well known that the values of ‘x’ satisfying this equation are given by the formula

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

There is a subtlety in the case when the quantity under the square root is negative, because in this case the solutions are not real numbers. However, if the reader wishes to, he or she may assume the quantity under the square root to be real.

There are many ways to see why the quadratic formula given above is true, but the most conceptual one, which generalizes, is as follows.

For simplicity, assume that ‘a’ is equal to 1. (If a is not equal to 1, we can divide the equation through by ‘a’ to get a new equation with ‘a’ equal to 1 without changing the values of x that satisfy the equation, though the values of ‘b’ and ‘c’ change). So consider the equation

$\large x^2 + bx + c = 0$

Let’s call the numbers that satisfy the equation ‘r’ and ‘s.’ Then the expression on the left can be written as

$\large (x - r)(x - s)$

Multiplying this expression out, we get

$\large x^2 -(r + s)x + rs = 0$

Comparing this with the form of the expression involving ‘b’ and ‘c’, we can conclude that

$\large -b = r + s \hspace{0.5 cm} c = rs$

This gives expressions for ‘b’ and ‘c’ in terms of ‘r’ and ‘s.’ We want to express ‘r’ and ‘s’ in terms of ‘b’ and ‘c.’ The difficulty is that r + s and rs are symmetric, in the sense that they’re invariant when ‘r’ and ‘s’ are flipped, whereas ‘r’ and ‘s’ alone are not symmetric in this sense. So we somehow want to break the symmetry present in ‘b’ and ‘c.’ Well, if we square ‘b,’ we get

$\large b^2 = (r + s)^2 = r^2 + 2rs + s^2$

and subtracting ‘c’ from this four times gives the expression

$\large b^2 - 4c = r^2 - 2rs + s^2 = (r - s)^2$

which is symmetric, but which is the square of something that’s not symmetric. So if we take its square root, then we get something that’s not symmetric. Taking the square root, we get

$\large \sqrt{b^2 - 4c} = r - s$

and

$\large - \sqrt{b^2 - 4c} = s - r$

each of which is permuted when r and s are permuted. (Note that we’ve just made an arbitrary choice as to which root is called ‘r’ and which root is called ‘s’.)

Having broken the symmetry, we can solve for r and s by adding

$\large -b = r + s$

to each of the two preceding equations and dividing through by 2. This gives

$\large r = \frac{- b + \sqrt{b^2 - 4c}}{2}$

and

$\large s = \frac{- b - \sqrt{b^2 - 4c}}{2}$

which is the well known quadratic formula.

### 4 Responses to “The Quadratic Formula”

1. Andy D April 22, 2013 at 3:37 am #

Nice! Never thought about things this way…

“the expression on the right can be written” —> you want “on the left”, no?

• Jonah Sinick April 22, 2013 at 3:46 am #

Thanks Andy. I wish this were taught in school …