What does 'completing the square' actually mean?

It means rewriting x^2 + bx as (x + \frac{b}{2})^2 - \frac{b^2}{4}, turning the quadratic into a perfect square trinomial (x + \frac{b}{2})^2 plus a constant. The phrase comes from the geometric idea of filling in a missing square to make a larger square.

Why do I need to use \pm when taking the square root?

Because both 3^2 = 9 and (-3)^2 = 9. When you solve (x + 2)^2 = 9, both x + 2 = 3 and x + 2 = -3 are valid, giving you two different answers for x.

Can I use completing the square if the leading coefficient isn't 1?

Yes, but you must divide the entire equation by the leading coefficient first so that x^2 has coefficient 1. Otherwise the expression won't form a perfect square trinomial.

When should I use completing the square instead of the Quadratic Formula?

Use completing the square if you want to find the vertex of a parabola, or if factoring obviously won't work and you want to understand the structure. Use the Quadratic Formula if you just need numerical answers quickly, especially with messy coefficients.

Completing the Square

Completing the square rewrites a quadratic as a perfect square plus a constant, making it easier to solve and find the parabola's vertex.

x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \frac{b^2}{4}

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Type the problem. The solver will use Completing the Square where Completing the Square is the right tool, and tell you when it is not.

What each symbol means

What Completing the Square takes

Symbol	Meaning
$b$	The coefficient of the linear term $x$ in the quadratic; if $x$ is missing the linear term, $b = 0$ .
$x$	The variable you're solving for; if you use a different variable (like $t$ or $r$ ), substitute it for $x$ in the formula.

When to use it

Reach for completing the square when you need to solve a quadratic that doesn't factor easily or when you want to find a parabola's vertex form.

Level

Usually taught in: Algebra I · Appears on: SAT

Worked examples

1. Solve a quadratic with small integers

Problem

Solve

x^2 + 6x - 7 = 0

$x^2 + 6x = 7$
Rearrange by adding 7 to both sides of the original equation.
$x^2 + 6x + 9 = 7 + 9$
Complete the square: take half the coefficient of $x$ (which is $\frac{6}{2} = 3$ ), square it ( $3^2 = 9$ ), and add to both sides.
$(x + 3)^2 = 16$
Recognize that $x^2 + 6x + 9$ is the perfect square trinomial $(x + 3)^2$ .
$x + 3 = \pm 4$
Take the square root of both sides; use $\pm$ because both $4^2 = 16$ and $(-4)^2 = 16$ .
$x = 1 \quad \text{or} \quad x = -7$
Subtract 3 from each side: $x = -3 + 4 = 1$ and $x = -3 - 4 = -7$ .

Answer: $x = 1 \text{ or } x = -7$

This problem is set up perfectly for completing the square: the leading coefficient is 1 and the integers are small. We can see how each step transforms the expression into the standard form.

2. Solve a quadratic with a leading coefficient and irrational roots

Problem

Solve

2x^2 - 8x + 3 = 0

$2x^2 - 8x = -3$
Move the constant to the right by subtracting 3 from both sides.
$x^2 - 4x = -\frac{3}{2}$
Divide everything by the leading coefficient 2 to make the $x^2$ coefficient equal 1.
$x^2 - 4x + 4 = -\frac{3}{2} + 4$
Complete the square: half of $-4$ is $-2$ ; square it to get $(-2)^2 = 4$ . Add 4 to both sides.
$x^2 - 4x + 4 = \frac{5}{2}$
Simplify the right side: $-\frac{3}{2} + 4 = -\frac{3}{2} + \frac{8}{2} = \frac{5}{2}$ .
$(x - 2)^2 = \frac{5}{2}$
Recognize that $x^2 - 4x + 4 = (x - 2)^2$ .
$x - 2 = \pm\sqrt{\frac{5}{2}}$
Take the square root of both sides.
$x = 2 \pm \frac{\sqrt{10}}{2}$
Solve for $x$ : simplify $\sqrt{\frac{5}{2}} = \frac{\sqrt{5}}{\sqrt{2}} = \frac{\sqrt{10}}{2}$ by rationalizing.

Answer: $x = 2 + \frac{\sqrt{10}}{2} \text{ or } x = 2 - \frac{\sqrt{10}}{2}$

Here we have a leading coefficient that isn't 1 and irrational solutions, which makes completing the square essential—the quadratic formula might be quicker, but this method shows the structure.

3. Find optimal dimensions of a rectangular garden

Problem

A rectangular garden plot has a perimeter of 30 meters. If you want to maximize the area, what dimensions should the plot have?

$A = w(15 - w)$
Set up the area function: if width is $w$ and perimeter is 30, then length is $15 - w$ .
$A = 15w - w^2$
Expand the product using the distributive property.
$A = -(w^2 - 15w)$
Factor out $-1$ to prepare for completing the square.
$A = -\left(w^2 - 15w + \frac{225}{4}\right) + \frac{225}{4}$
Complete the square inside the parentheses: half of $-15$ is $-\frac{15}{2}$ , and $\left(-\frac{15}{2}\right)^2 = \frac{225}{4}$ .
$A = -\left(w - \frac{15}{2}\right)^2 + \frac{225}{4}$
Write the perfect square trinomial in factored form.
$w = \frac{15}{2} = 7.5$
The maximum area occurs when the squared term equals zero, which happens at $w = 7.5$ metres.
$\text{length} = 15 - 7.5 = 7.5$
The length is also 7.5 metres, so the garden is a square.

Answer: $\text{The garden should be } 7.5 \text{ m} \times 7.5 \text{ m}$

Completing the square transforms the area function into vertex form, revealing immediately that the maximum occurs at $w = 7.5$ . This shows us both the optimal dimensions and that the area function is a downward-opening parabola. Without completing the square, we'd need calculus to find this maximum.

Common mistakes

Where Completing the Square usually goes wrong

The mistake	Why it is wrong	The fix
I added $(\frac{b}{2})^2$ to only one side: $x^2 + 6x + 9 = 5$ instead of $x^2 + 6x + 9 = 5 + 9$ .	Equations must stay balanced; whatever you add to one side must be added to the other.	Add the constant to BOTH sides: $x^2 + 6x = 5$ becomes $x^2 + 6x + 9 = 5 + 9 = 14$ .
I found $(x + 3)^2 = 16$ , so I wrote $x + 3 = 4$ and got $x = 1$ only.	Squaring gives two solutions; you must use $\pm$ when taking the square root.	Write $x + 3 = \pm 4$ , giving $x = 1$ or $x = -7$ .
I had $2x^2 + 8x = 5$ and added $4^2$ to get $(2x + 4)^2 = 5 + 16$ .	You must divide by the leading coefficient FIRST so the coefficient of $x^2$ is 1, or you won't have a perfect square trinomial.	Divide by 2 first: $x^2 + 4x = \frac{5}{2}$ , then add $\left(\frac{4}{2}\right)^2 = 4$ to both sides.

Tips and when to use something else

Always move the constant to the right side before completing the square.
If the leading coefficient isn't 1, divide the entire equation by it before adding $(\frac{b}{2})^2$ .
Remember $\pm$ when you take the square root; quadratics usually have two solutions.
If the solutions are messy fractions or surds, try the Quadratic Formula instead—it's faster for complicated cases.

Related formulas

Quadratic FormulaThe Quadratic Formula solves any equation of the form ax^2 + bx + c = 0 in one step, giving both solutions at once without factoring.Vertex FormVertex Form reveals a parabola's vertex and axis of symmetry at a glance, making it ideal for transformations and optimization problems.FOIL MethodThe FOIL Method expands products of two binomials by multiplying each term in the first binomial by each term in the second.Combining Like TermsCombining like terms merges coefficients of identical variables to simplify expressions and solve equations more efficiently.

Frequently asked questions

What does 'completing the square' actually mean?: It means rewriting $x^2 + bx$ as $(x + \frac{b}{2})^2 - \frac{b^2}{4}$ , turning the quadratic into a perfect square trinomial $(x + \frac{b}{2})^2$ plus a constant. The phrase comes from the geometric idea of filling in a missing square to make a larger square.
Why do I need to use $\pm$ when taking the square root?: Because both $3^2 = 9$ and $(-3)^2 = 9$ . When you solve $(x + 2)^2 = 9$ , both $x + 2 = 3$ and $x + 2 = -3$ are valid, giving you two different answers for $x$ .
Can I use completing the square if the leading coefficient isn't 1?: Yes, but you must divide the entire equation by the leading coefficient first so that $x^2$ has coefficient 1. Otherwise the expression won't form a perfect square trinomial.
When should I use completing the square instead of the Quadratic Formula?: Use completing the square if you want to find the vertex of a parabola, or if factoring obviously won't work and you want to understand the structure. Use the Quadratic Formula if you just need numerical answers quickly, especially with messy coefficients.

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