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Completing The Square

Q: What happens if I get a negative number inside the square root?

If you reach a step like $(x-2)^2 = -9$, you will need to take the square root of a negative number. In Algebra 1, this means there are no real solutions to the equation. In Algebra 2 and beyond, you will learn about imaginary numbers to find complex solutions.

Q: Does this method work if the 'b' term is an odd number?

Yes, it works perfectly. When $b$ is odd, your calculation for $(\frac{b}{2})^2$ will result in a fraction, as shown in Example 3. While the arithmetic is a bit more complex, the steps and the underlying principle remain exactly the same.

Q: What if the 'b' term is zero?

If $b=0$, the equation is something like $ax^2 + c = 0$. You don't need to complete the square here. You can solve it directly by isolating the $x^2$ term and taking the square root, for example: $2x^2 - 18 = 0 \implies 2x^2 = 18 \implies x^2 = 9 \implies x = \pm 3$.

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Ever faced a quadratic equation that just won't factor? Completing the square is a powerful algebraic technique that can solve any quadratic equation. This guide will show you how to transform expressions into a 'perfect square,' making them simple to solve and understand.

Completing The Square — an original Algebra911 reference diagram defining completing the square with its key formula and a worked example. — Completing the Square: A Step-by-Step Guide

What Is Completing the Square?

Completing the square is an algebraic method used to rewrite a quadratic expression of the form $a x^{2} + b x + c$ into the vertex form $a (x - h)^{2} + k$ . The primary goal is to create a perfect square trinomial on one side of the equation, which is an expression that can be factored into a binomial squared, like $(x + d)^{2}$ or $(x - d)^{2}$ .

Think of it geometrically. The expression $x^{2} + b x$ can be visualized as an incomplete square. You have a large square piece with area $x^{2}$ (side length $x$ ) and two rectangular pieces that represent $b x$ (which we can split into two rectangles of area $\frac{b}{2} x$ ). To make a complete, larger square, you are missing a small corner piece. The area of this missing piece is exactly $(\frac{b}{2})^{2}$ . By adding this missing piece, you 'complete the square.' This technique transforms a difficult-to-solve problem into a simple one involving a squared term, which can be easily solved by taking the square root.

Why Is This Method So Important?

You might wonder why you need another method to solve quadratics when you already know factoring or the quadratic formula. Completing the square is more than just a solving technique; it's a foundational concept in algebra with several key applications:

It Solves *Any* Quadratic Equation: Factoring only works for some equations. Completing the square, like the quadratic formula, works for all of them, whether the solutions are integers, fractions, or irrational numbers.
It Derives the Quadratic Formula: The famous quadratic formula isn't magic! It is derived by applying the method of completing the square to the general quadratic equation $a x^{2} + b x + c = 0$ . Understanding this process gives you a deeper insight into how the formula works.
It Finds the Vertex of a Parabola: When you graph a quadratic equation, you get a parabola. By converting an equation to vertex form $y = a (x - h)^{2} + k$ using completing the square, you can instantly identify the vertex at the point $(h, k)$ . This is essential for graphing and optimization problems.

The Key: Understanding Perfect Square Trinomials

The entire method hinges on one idea: creating a perfect square trinomial. This is a special type of trinomial that results from squaring a binomial. Let's look at the pattern:

$(x + d)^{2} = (x + d) (x + d) = x^{2} + d x + d x + d^{2} = x^{2} + 2 d x + d^{2}$

Notice the relationship between the middle term's coefficient ( $2 d$ ) and the last term ( $d^{2}$ ). The last term is the square of half the middle coefficient.

If you have $x^{2} + 10 x$ , what constant do you need to add to make it a perfect square?

The middle coefficient is $10$ .
Take half of it: $\frac{10}{2} = 5$ .
Square that result: $5^{2} = 25$ .

So, $x^{2} + 10 x + 25$ is a perfect square trinomial because it factors into $(x + 5)^{2}$ .

This gives us our magic formula for finding the term needed to complete the square. For any expression $x^{2} + b x$ , the constant term $c$ that completes the square is:

c = \left(\frac{b}{2}\right)^2

This is the most important formula in the entire process. Once you have this, you can turn any quadratic into a perfect square.

How to Solve Equations by Completing the Square

Here is a reliable, step-by-step process for solving a quadratic equation using this method. Let's use the general form $a x^{2} + b x + c = 0$ .

Standardize the Equation: Make sure the coefficient of the $x^{2}$ term (the value of $a$ ) is $1$ . If it's not $1$ , divide every single term in the equation by $a$ .
Isolate the $x$ Terms: Move the constant term (the number without a variable) to the right side of the equation. You should have something that looks like $x^{2} + b x = - c$ .
Find the Magic Number: Take the coefficient of the $x$ term (the value of $b$ ), divide it by $2$ , and then square the result. This is your $(\frac{b}{2})^{2}$ value.
Add to Both Sides: Add the magic number you just calculated to both sides of the equation. This is crucial for keeping the equation balanced.
Factor the Left Side: The left side of your equation is now a perfect square trinomial. Factor it into its binomial squared form, $(x + \frac{b}{2})^{2}$ .
Solve for $x$ : Take the square root of both sides of the equation. Don't forget to include the plus-or-minus symbol ( $\pm$ ) on the right side. Finally, isolate $x$ to find your two solutions.

Worked Examples: Completing the Square in Action

Let's walk through a few examples to see how these steps work in practice.

Example 1

Solve the equation $x^{2} + 6 x - 16 = 0$ .

Standardize: The coefficient of $x^{2}$ is already $1$ , so we can skip this step.
Isolate: Move the constant term, $- 16$ , to the right side by adding $16$ to both sides. $x^{2} + 6 x = 16$
Find the Magic Number: The $b$ coefficient is $6$ . We calculate $(\frac{b}{2})^{2}$ . ${(\frac{6}{2})}^{2} = 3^{2} = 9$
Add to Both Sides: Add $9$ to both sides of the equation. $x^{2} + 6 x + 9 = 16 + 9$ $x^{2} + 6 x + 9 = 25$
Factor: The left side is a perfect square trinomial. It factors to $(x + 3)^{2}$ . $(x + 3)^{2} = 25$
Solve for $x$ : Take the square root of both sides, remembering the $\pm$ . $\sqrt{(x + 3)^{2}} = \pm \sqrt{25}$ $x + 3 = \pm 5$ This gives us two separate equations to solve: $x + 3 = 5 or x + 3 = - 5$ $x = 2 or x = - 8$ The solutions are $x = 2$ and $x = - 8$ .

Example 2

Solve the equation $2 x^{2} - 4 x - 14 = 0$ .

Standardize: The coefficient of $x^{2}$ is $2$ , not $1$ . We must divide the entire equation by $2$ . $\frac{2 x^{2}}{2} - \frac{4 x}{2} - \frac{14}{2} = \frac{0}{2}$ $x^{2} - 2 x - 7 = 0$
Isolate: Move the constant term, $- 7$ , to the right side. $x^{2} - 2 x = 7$
Find the Magic Number: The $b$ coefficient is $- 2$ . ${(\frac{- 2}{2})}^{2} = (- 1)^{2} = 1$
Add to Both Sides: Add $1$ to both sides. $x^{2} - 2 x + 1 = 7 + 1$ $x^{2} - 2 x + 1 = 8$
Factor: Factor the left side. $(x - 1)^{2} = 8$
Solve for $x$ : Take the square root of both sides. $x - 1 = \pm \sqrt{8}$ The square root of $8$ can be simplified: $\sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$ . $x - 1 = \pm 2 \sqrt{2}$ Now, isolate $x$ by adding $1$ to both sides. $x = 1 \pm 2 \sqrt{2}$ The solutions are $x = 1 + 2 \sqrt{2}$ and $x = 1 - 2 \sqrt{2}$ . These are irrational solutions, which is why factoring wouldn't have worked.

Example 3

Solve the equation $x^{2} + 5 x + 3 = 0$ .

Standardize: The $a$ coefficient is $1$ , so we're good.
Isolate: Move the constant term, $3$ , to the right side. $x^{2} + 5 x = - 3$
Find the Magic Number: The $b$ coefficient is $5$ . This will involve fractions, but don't panic! The process is the same. ${(\frac{5}{2})}^{2} = \frac{25}{4}$
Add to Both Sides: Add $\frac{25}{4}$ to both sides. To add $- 3$ and $\frac{25}{4}$ , we need a common denominator. $- 3 = - \frac{12}{4}$ . $x^{2} + 5 x + \frac{25}{4} = - 3 + \frac{25}{4}$ $x^{2} + 5 x + \frac{25}{4} = - \frac{12}{4} + \frac{25}{4}$ $x^{2} + 5 x + \frac{25}{4} = \frac{13}{4}$
Factor: The left side factors to $(x + \frac{5}{2})^{2}$ . ${(x + \frac{5}{2})}^{2} = \frac{13}{4}$
Solve for $x$ : Take the square root of both sides. $x + \frac{5}{2} = \pm \sqrt{\frac{13}{4}}$ $x + \frac{5}{2} = \pm \frac{\sqrt{13}}{\sqrt{4}} = \pm \frac{\sqrt{13}}{2}$ Finally, isolate $x$ by subtracting $\frac{5}{2}$ from both sides. $x = - \frac{5}{2} \pm \frac{\sqrt{13}}{2}$ This can be written as a single fraction: $x = \frac{- 5 \pm \sqrt{13}}{2}$

Common Mistakes to Avoid

Completing the square is a process with several steps, which means there are a few places where students commonly make errors. Be on the lookout for these pitfalls:

Forgetting to Add to Both Sides: The most common mistake is calculating the magic number $(\frac{b}{2})^{2}$ and only adding it to the left side of the equation. This unbalances the equation and leads to a wrong answer. Always add it to both sides.
Missing the $\pm$ Sign: When you take the square root of both sides to undo the square (e.g., $(x + 3)^{2} = 25$ ), you must include the plus-or-minus symbol. The square root of $25$ is not just $5$ , it's $\pm 5$ . Forgetting this will cause you to lose one of the two solutions.
Errors with the $a$ Coefficient: If $a$ is not $1$ , you must divide every term by $a$ at the very beginning. A common error is to only divide the first two terms and forget the constant or the zero on the other side.
Sign Errors with $b$ : Be careful when calculating $(\frac{b}{2})^{2}$ if $b$ is negative. For example, if $b = - 8$ , then $\frac{b}{2} = - 4$ . But $(- 4)^{2} = 16$ , not $- 16$ . The number you add is always positive. However, the factored form will use the negative: $(x - 4)^{2}$ .

Quick Summary and Reference

Here is a quick reference table summarizing the process of solving $a x^{2} + b x + c = 0$ by completing the square.

Step	Action	Example: $x^{2} + 8 x - 9 = 0$
1	Ensure $a = 1$ and move the constant $c$ to the right side.	$x^{2} + 8 x = 9$
2	Find the magic number: $(\frac{b}{2})^{2}$ .	$(\frac{8}{2})^{2} = 4^{2} = 16$
3	Add this number to both sides of the equation.	$x^{2} + 8 x + 16 = 9 + 16$
4	Factor the left side as a perfect square and simplify the right.	$(x + 4)^{2} = 25$
5	Take the square root of both sides (remember $\pm$ ).	$x + 4 = \pm 5$
6	Isolate $x$ to find the two solutions.	$x = - 4 \pm 5 ⟹ x = 1, x = - 9$

Frequently Asked Questions

Why is it called 'completing the square'?

The name comes from a geometric interpretation. The expression $x^{2} + b x$ represents the area of an incomplete square. By adding the term $(\frac{b}{2})^{2}$ , you are adding the area of a small corner piece that physically completes a larger square, making it easy to find the side length.

Can I use this method if the equation factors easily?

Yes, you absolutely can. Completing the square will always give you the same answers as factoring. However, if an equation is simple to factor, that method is usually much faster and requires fewer steps.

Is completing the square the same as the quadratic formula?

They are not the same, but they are deeply related. The quadratic formula is actually derived by performing the completing the square process on the general quadratic equation $a x^{2} + b x + c = 0$ . Think of the quadratic formula as a shortcut that gives you the final result of completing the square without showing the steps.

What happens if I get a negative number inside the square root?

If you reach a step like $(x - 2)^{2} = - 9$ , you will need to take the square root of a negative number. In Algebra 1, this means there are no real solutions to the equation. In Algebra 2 and beyond, you will learn about imaginary numbers to find complex solutions.

When should I complete the square instead of using the quadratic formula?

The quadratic formula is often faster for just finding solutions. However, completing the square is essential when you need to convert an equation into vertex form, $y = a (x - h)^{2} + k$ , to find the vertex of a parabola for graphing or other applications.

Does this method work if the 'b' term is an odd number?

Yes, it works perfectly. When $b$ is odd, your calculation for $(\frac{b}{2})^{2}$ will result in a fraction, as shown in Example 3. While the arithmetic is a bit more complex, the steps and the underlying principle remain exactly the same.

What if the 'b' term is zero?

If $b = 0$ , the equation is something like $a x^{2} + c = 0$ . You don't need to complete the square here. You can solve it directly by isolating the $x^{2}$ term and taking the square root, for example: $2 x^{2} - 18 = 0 ⟹ 2 x^{2} = 18 ⟹ x^{2} = 9 ⟹ x = \pm 3$ .