Completing the Square

Section 1.4 Completing the Square

Consider a quadratic equation, $ax^2 + bx + c = 0\text{.}$ Recall that we have solved quadratic equations of the form,

\begin{equation*} (x - h)^2 = k \end{equation*}

using the square root property (i.e. taking the square root of both sides). For example, to solve $(x - 2)^2 = 5\text{,}$

\begin{align*} (x - 2)^2 \amp = 5\\ x - 2 \amp = \pm \sqrt{5} \amp\amp \text{by the square root property}\\ x \amp = 2 \pm \sqrt{5} \amp\amp \text{solving for $x$} \end{align*}

Alternatively, we could have expanded out the equation first, and write it in standard form,

\begin{align*} (x - 2)^2 \amp = 5\\ x^2 - 4x + 4 \amp = 5\\ x^2 - 4x - 1 \amp = 0 \end{align*}

The equations $x^2 - 4x - 1 = 0$ and $(x - 2)^2 = 5\text{.}$ The equation $x^2 - 4x - 1 = 0$ can't be solved by factoring (you can confirm this). However, notice that all of the steps are reversible. That is,

\begin{align*} x^2 - 4x - 1 \amp = 0\\ x^2 - 4x + 4 \amp = 5 \amp\amp \text{adding 5 to both sides}\\ (x - 2)^2 \amp = 5 \amp\amp \text{rewriting the left-hand side}\\ x - 2 \amp = \pm \sqrt{5} \amp\amp \text{by the square root property}\\ x \amp = 2 \pm \sqrt{5} \amp\amp \text{solving for $x$} \end{align*}

This illustrates a general technique for solving quadratic equations, in particular those which can't be factored. The idea is to write the quadratic as a square of the form $(x - h)^2\text{,}$ and then use the square root property. This technique is called completing the square.

Subsection 1.4.1 Solving by Completing the Square

Recall that a perfect square trinomial is a trinomial that can be factored as the square of a sum/difference.

\begin{equation*} x^2 + 8x + 16 = (x + 4)^2 \qquad \text{and} \qquad x^2 - 10x + 25 = (x - 5)^2 \end{equation*}

In general, for a perfect square trinomial, the constant term is equal to the square of $\frac{1}{2}$ of the coefficient of the $x$ term. In other words,

\begin{equation*} x^2 + ax + \brac{\frac{a}{2}}^2 = \brac{x + \frac{a}{2}}^2 \end{equation*}

If the coefficient of $x$ is $a\text{,}$ then the constant term must be $\brac{\frac{a}{2}}^2\text{.}$

First, consider an equation of the form,

\begin{equation*} x^2 + bx + c = 0 \end{equation*}

First, move the constant term to the other side of the equation. Then, add $\brac{\frac{b}{2}}^2$ to both sides of the equation,

\begin{equation*} x^2 + bx + \brac{\frac{b}{2}}^2 = \brac{\frac{b}{2}}^2 - c \end{equation*}

Then, rewrite the left-hand side as a perfect square,

\begin{equation*} \brac{x + \frac{b}{2}}^2 = \brac{\frac{b}{2}}^2 - c \end{equation*}

Finally, take the square root of both sides of the equaton and use the square root property, to solve for $x\text{.}$

If the leading coefficient of the quadratic is not 1, then first divide by that coefficient, and continue as above.

This process is called completing the square, because it involves quite literally taking a polynomial $x^2 + bx\text{,}$ and adding an appropriate constant such that it becomes a perfect square trinomial.

Subsection 1.4.2 Summary of Solving an Equation by Completing the Square

Write the equation in the form,
\begin{equation*} ax^2 + bx = -c \end{equation*}
Divide both sides by $a\text{,}$ to make the leading coefficient equal to 1,
\begin{equation*} x^2 + \frac{b}{a}x = -\frac{c}{a} \end{equation*}
Add the square of half of the coefficient of $x\text{,}$ to both sides of the equation.
Factor the perfect square trinomial (you can check you added the constant correctly by expanding the perfect square, and confirming it is equal to the original trinomial).
Use the square root property to take the square root of both sides of the equation.
Solve for $x\text{.}$
Check the solutions, by substituting them for $x$ in the original equation.

Subsection 1.4.3 Completing the Square Geometrically

Subsection 1.4.4 Historical Aside

The method of completing the square was known as early as 1600BC to the ancient Babylonians, well before algebraic notation was even invented.