Solving second degree equations (deriving the quadratic formula)

The general form of a second degree equation is

$$x^2 + px + q = 0 \tag{0}$$

where $p$ and $q$ can be any real number and $x$ is the roots of the equation. In equations of this form $q$ is the product of the roots and $p$ is the sum of the roots.

Equations of the form $x^2 + q = 0$ are easy to solve because $p = 0$. The roots are simply $±\sqrt{x} - q$. The graphical representation of equations of this form is a curve that is symmetric about the $x$ axis. For instance the curve of $y = x^2 - 5$ looks like:

Now consider a more complicated equation like $y = x^2 - 6x - 5$. If we rewrite this equation as $y = x(x - 6) - 5$ we can see that $y = -5$ when $x(x - 6) = 0$. This could happen when $x = 0$ or when $x = 6$. The axis around which the curve is symmetric is half way between these values at the line $x = 3$. We can confirm this by plotting the curve:

If we shifted the curve to the left by 3 units the axis of symmetry would be the $x$ axis. It is easy to find the roots of the equation of that line. We can do the shift algebraically by replacing $x$ with $x + 3$ in the equation for $y$.

In the general case, we are introducing a new equation whose roots are equal to the roots of the general second degree equation plus half the coeeficient of the $x$ term:

$$y = x + \frac{p}{2}$$

Therefore

$$x = y - \frac{p}{2} \tag{1}$$

Substituting this value of $x$ into the general equation gives:

$$\begin{aligned} (y - \frac{p}{2})^2 + p(y - \frac{p}{2}) + q &= 0 \\ y^2 - \frac{py}{2} - \frac{py}{2} + \frac{p^2}{4} + py - \frac{p^2}{2} + q &= 0 \\ y^2 - py + \frac{p^2}{4} +py - \frac{p^2}{2} + q &= 0 \\ y^2 - \frac{p^2}{4} + q &= 0 \end{aligned}$$

Therefore

$$y^2 = \frac{p^2}{4} - q$$

So our values of y are:

$$y = ±\sqrt{\frac{p^2}{4} - q}$$

Substituting this into equation $(1)$ gives:

$$x = -\frac{p}{2} ± \sqrt{\frac{p^2}{4} - q}$$

If we have an equation of the form

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

We can divide by $a$ to get

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

This is the same form as equation $(0)$ above where $p = \frac{b}{a}$ and $q = \frac{c}{a}$

Substituting these values into the equation $(1)$ gives

$$\begin{aligned} x &= -\frac{\frac{b}{a}}{2} ± \sqrt{\frac{(\frac{b}{a})^2}{4} - \frac{c}{a}} \\ &= -\frac{b}{2a} ± \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}} \\ &= -\frac{b}{2a} ± \sqrt{\frac{b^2 - 4ac}{4a^2}} \\ &= -\frac{b ± \sqrt{b^2 - 4ac}}{2a} \tag{2} \end{aligned}$$

This is the quadratic formula.

Sources

• Mathematics for the Nonmathematician (pp 112-117) - Morris Kline
• The Symmetry That Makes Solving Math Equations Easy - Patrick Honner