The Quadratic Formula

Section 1.7 The Quadratic Formula

In fact, the method of completing the square can be used to develop a single formula which gives the solution to any quadratic equation, called the quadratic formula.

Subsection 1.7.1 The Quadratic Formula

Consider a general quadratic equation $ax^2 + bx + c = 0$ (where $a \neq 0$). We will complete the square. First, divide both sides by the leading coefficient,

\begin{align*} ax^2 + bx + c \amp = 0\\ x^2 + \frac{b}{a}x + \frac{c}{a} \amp = 0 \amp\amp \text{dividing both sides by $a$}\\ x^2 + \frac{b}{a}x \amp = -\frac{c}{a} \amp\amp \text{subtracting $c/a$ from both sides}\\ x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \amp = \frac{b^2}{4a^2} - \frac{c}{a} \amp\amp \text{completing the square by adding $\brac{\frac{1}{2} \cdot \frac{b}{a}}^2 = \frac{b^2}{4a^2}$ to both sides}\\ \brac{x + \frac{b}{2a}}^2 \amp = \frac{b^2 - 4ac}{4a^2} \amp\amp \text{factoring on the left, simplifying on the right}\\ x + \frac{b}{2a} \amp = \pm \sqrt{\frac{b^2 - 4ac}{4a^2}} \amp\amp \text{using the square root property}\\ x + \frac{b}{2a} \amp = \pm \frac{\sqrt{b^2 - 4ac}}{2\abs{a}} \amp\amp \text{simplifying the square root} \end{align*}

Here, $\sqrt{4a^2} = 2\abs{a}\text{,}$ which equals $2a$ if $a > 0$ and $-2a$ if $a \lt 0\text{.}$ However, in either case, the plus/minus symbol in front will not be affected, and so we can simply write $\sqrt{4a^2} = 2a\text{.}$ Then,

\begin{align*} x + \frac{b}{2a} \amp = \pm \frac{\sqrt{b^2 - 4ac}}{2a}\\ x \amp = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \amp\amp \text{subtracting $\frac{b}{2a}$ from both sides} \end{align*}

In summary,

Theorem 1.7.1. The quadratic formula.

Let $ax^2 + bx + c = 0$ be a quadratic equation, $a \neq 0\text{.}$ Then, the solutions are given by,

\begin{equation*} \boxed{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}} \end{equation*}

This is the quadratic formula. Substituting the values of $a, b\text{,}$ and $c$ from the equation into the formula will give you the solutions for $x\text{.}$ Note that the equation only has real number solutions if the expression under the square root is non-negative, i.e. $b^2 - 4ac \geq 0\text{.}$ If $b^2 - 4ac \lt 0\text{,}$ there are no real number solutions.

Subsection 1.7.2 Discriminant

In the quadratic formula,

\begin{equation*} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{equation*}

the expression under the square root sign, $b^2 - 4ac\text{,}$ provides information about the nature of the soltuions to the equation.

If $b^2 - 4ac > 0\text{,}$ then there are two real solutions, given by the quadratic formula, one with the $+$ sign and the other with the $-$ sign. If instead $b^2 - 4ac = 0\text{,}$ then $\sqrt{b^2 - 4ac} = 0\text{,}$ and the quadratic equation simplifies to $x = -\frac{b}{2a}\text{,}$ a single real solution. Finally, if $b^2 - 4ac \lt 0\text{,}$ then the square root $\sqrt{b^2 - 4ac}$ is not a real number, and so there are no real solutions.

This quantity $b^2 - 4ac$ is called the discriminant.

Definition 1.7.2.

The discriminant of a quadratic expression $ax^2 + bx + c$ is given by,

\begin{equation*} \Delta = b^2 - 4ac \end{equation*}

The discriminant is often denoted by the Greek letter delta $\Delta$, because delta and discriminant both start with “d”.

Theorem 1.7.3.

Let $ax^2 + bx + c = 0$ be a quadratic equation, $a \neq 0\text{.}$

If $b^2 - 4ac > 0\text{,}$ then the equation has 2 real solutions.
If $b^2 - 4ac = 0\text{,}$ then the equation has 1 real repeated solution.
If $b^2 - 4ac \lt 0\text{,}$ then the equation has 0 real roots and 2 non-real complex solutions.

In this way, the discriminant quite literally helps us “discriminate” (distinguish apart) between the possible types of roots of an equation.

Further, if the discriminant is positive and is a perfect square (that is, its square root is a rational number), then $\sqrt{b^2 - 4ac}$ is rational, and this means that both solutions provided by the quadratic formula are rational.

Theorem 1.7.4.

Let $ax^2 + bx + c = 0$ be a quadratic equation with integer coefficients (or rational coefficients). If $b^2 - 4ac > 0$ is a perfect square, then the equation has 2 rational solutions.

If the discriminant is not a perfect square, then the two real solutions are irrational. In particular, they are of the form,

\begin{equation*} x = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \ , \ \frac{-b - \sqrt{b^2 - 4ac}}{2a} \end{equation*}

Or, equivalently,

\begin{equation*} x = -\frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a} \ , \ -\frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a} \end{equation*}

These numbers are irrational conjugates, in that they are the same, except for differing signs on the second term.

Subsection 1.7.3 Remembering the Quadratic Formula

Also, there are various jingles and songs to help remember the quadratic formula,