Table of Contents
Understanding the Method of Completing the Square
Completing the square is a specific method for rewriting a quadratic expression or solving a quadratic equation by turning part of it into a perfect square. In this chapter, we assume you already know what a quadratic equation is and how to work with basic algebraic operations.
We focus here on:
- How and why we complete the square
- Step-by-step procedures for different forms
- Using the method to solve equations
- Using it to find the vertex form of a quadratic
Perfect Square Trinomials
The method of completing the square is based on a common pattern called a perfect square trinomial.
A basic example is:
$$(x + 3)^2 = x^2 + 6x + 9.$$
More generally, for any real number $b$,
$$(x + b)^2 = x^2 + 2bx + b^2.$$
So a quadratic expression of the form
$$x^2 + 2bx + b^2$$
is a perfect square trinomial, because it factors as $(x + b)^2$.
This pattern is what we create when we “complete the square.”
Completing the Square for Simple Expressions
Consider an expression where the coefficient of $x^2$ is 1, for example
$$x^2 + 6x.$$
We want to add a number so that the expression becomes a perfect square trinomial.
- Take half of the coefficient of $x$.
The coefficient of $x$ is 6, half of 6 is 3.
- Square that half.
$3^2 = 9$.
- Add that square to the expression.
$$x^2 + 6x + 9.$$
Now we recognize this as
$$(x + 3)^2.$$
So we say we have “completed the square”:
$$x^2 + 6x = (x + 3)^2 - 9,$$
because we added 9 and subtracted 9 to keep the expression equal.
More generally, for $x^2 + px$:
- Half of $p$ is $\dfrac{p}{2}$.
- Square it to get $\left(\dfrac{p}{2}\right)^2$.
- Then
$$x^2 + px + \left(\dfrac{p}{2}\right)^2 = \left(x + \dfrac{p}{2}\right)^2,$$
and so
$$x^2 + px = \left(x + \dfrac{p}{2}\right)^2 - \left(\dfrac{p}{2}\right)^2.$$
Completing the Square to Solve Equations (Leading Coefficient 1)
We now apply this idea to solve quadratic equations. Consider:
$$x^2 + 6x + 5 = 0.$$
We want to turn the left side into a perfect square.
- Move the constant term to the other side.
$$x^2 + 6x = -5.$$ - Complete the square on the left.
- Half of 6 is 3, $3^2 = 9$.
- Add 9 to both sides to keep the equation balanced:
$$x^2 + 6x + 9 = -5 + 9.$$ - Rewrite the left side as a square; simplify the right side.
- Left: $x^2 + 6x + 9 = (x + 3)^2$.
- Right: $-5 + 9 = 4$.
So:
$$(x + 3)^2 = 4.$$ - Take the square root of both sides (remember the $\pm$).
$$x + 3 = \pm \sqrt{4} = \pm 2.$$ - Solve for $x$.
- $x + 3 = 2 \Rightarrow x = -1$,
- $x + 3 = -2 \Rightarrow x = -5$.
So the solutions are $x = -1$ and $x = -5$.
The key is the step where we added the square of half the coefficient of $x$ to both sides.
Completing the Square When the Coefficient of $x^2$ Is Not 1
Often, a quadratic equation has a coefficient $a \neq 1$ in front of $x^2$:
$$ax^2 + bx + c = 0.$$
To use completing the square comfortably, we first factor out $a$ from the $x^2$ and $x$ terms so the inside has leading coefficient 1.
Example:
$$2x^2 + 8x - 10 = 0.$$
- Move the constant to the other side (optional but often helpful).
$x^2 + 8x = 10.$$ - Factor out the coefficient of $x^2$ from the left side.
$(x^2 + 4x) = 10.$$ - Complete the square inside the parentheses.
- Inside we have $x^2 + 4x$.
- Half of 4 is 2, $2^2 = 4$.
- We want $x^2 + 4x + 4$ inside.
Add and subtract $4$ inside the parentheses, but be careful: adding 4 inside the parentheses means adding $2 \cdot 4 = 8$ to the whole left side (because of the factor 2).
A clear way is:
$$2(x^2 + 4x + 4) - 2 \cdot 4 = 10.$$
That is:
$$2(x^2 + 4x + 4) - 8 = 10.$$
- Rewrite the completed square and simplify.
- Inside: $x^2 + 4x + 4 = (x + 2)^2$.
- So:
$(x + 2)^2 - 8 = 10.$$ - Solve for the squared expression.
Add 8 to both sides:
$(x + 2)^2 = 18.$$
Divide by 2:
$$(x + 2)^2 = 9.$$ - Take square roots.
$$x + 2 = \pm 3.$$ - Solve for $x$.
- $x = 1$,
- $x = -5$.
The general pattern is:
- Factor $a$ from $x^2 + \dfrac{b}{a}x$.
- Complete the square inside the parentheses.
- Adjust the outside of the parentheses to keep the equation balanced.
- Then take square roots and solve.
Quick General Recipe for Completing the Square (Solving)
For a quadratic equation of the form
$$ax^2 + bx + c = 0,$$
to solve by completing the square:
- If $a \neq 1$, divide both sides by $a$ (or factor $a$ out of the $x^2$ and $x$ terms).
- Move the constant term to the other side.
- Take half of the coefficient of $x$, square it.
- Add this square to both sides (or, if $a$ was factored out, add it inside the parentheses and compensate outside).
- Rewrite the left side as a squared binomial.
- Take the square root of both sides (put $\pm$ on the right).
- Solve the resulting linear equations.
This method works even when the solutions are fractions or involve square roots that do not simplify nicely.
Using Completing the Square to Get Vertex Form
Another important use of completing the square is to rewrite a quadratic function into vertex form. This shows the vertex (turning point) of the parabola directly.
Suppose you have a function
$$y = x^2 + 4x + 1.$$
We want to rewrite it as
$$y = (x - h)^2 + k,$$
where $(h, k)$ is the vertex.
- Group the $x$-terms and leave space to complete the square.
$$y = x^2 + 4x \;+\; 1.$$ - Complete the square on the $x$-part only.
- For $x^2 + 4x$:
- Half of 4 is 2, $2^2 = 4$.
- Add and subtract 4 inside the expression:
$$y = (x^2 + 4x + 4) - 4 + 1.$$ - Rewrite the completed square and combine constants.
- $x^2 + 4x + 4 = (x + 2)^2$.
- $-4 + 1 = -3$.
So:
$$y = (x + 2)^2 - 3.$$
This is vertex form. The vertex is $(-2, -3)$.
When the coefficient of $x^2$ is not 1, you factor it out from the $x^2$ and $x$ terms first, then complete the square inside the parentheses, similar to solving equations.
Connection to the Quadratic Formula (Conceptual Overview)
If you apply completing the square to the general equation
$$ax^2 + bx + c = 0,$$
and carefully carry the algebra through, you end up with
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$
which is the quadratic formula.
So, completing the square is not only a method for solving; it is also the technique that produces the quadratic formula.
In practice:
- Completing the square is especially convenient when $a = 1$ and $b$ is even.
- For more complicated coefficients, the quadratic formula may be quicker, but both are based on the same underlying idea of turning part of the quadratic into a perfect square.