Completing the Square, Step by Step

Quick answer

Completing the square rewrites a quadratic as a perfect square plus a constant. Take half the coefficient of x, square it, and add that number to both sides. For x squared plus 6x = 7, half of 6 is 3 and 3 squared is 9, so add 9 to both sides to get (x + 3) squared = 16, giving x = 1 or x = -7.

What you'll learn

Complete the square to solve a quadratic equation
Explain geometrically why the method is called completing the square
Rewrite a quadratic in vertex form to find its maximum or minimum

What completing the square does

Completing the square rewrites a quadratic so the variable appears only once, inside a squared bracket:

x^2 + 6x - 7 = 0 \quad\longrightarrow\quad (x + 3)^2 = 16

That matters because once the variable appears once, you can undo the square directly and solve. It is the same “isolate the variable” goal you have had since one-step equations, and this is the detour that gets you there.

Why it is called completing the square

The name is literal. Drawing the picture shows you where the algebra comes from.

Think of $x^2 + 6x$ as area. Draw a square with side $x$ — area $x^2$ . Attach the $6x$ as two rectangles of width $3$ along two adjacent sides, each of area $3x$ .

You now have an L-shape. It is almost a bigger square with side $x + 3$ , but the corner is missing. That corner is a $3 \times 3$ square, area $9$ .

\underbrace{x^2 + 6x}_{\text{the L-shape}} + \underbrace{9}_{\text{missing corner}} = \underbrace{(x+3)^2}_{\text{complete square}}

Adding $9$ completes the square. The half-of- $b$ rule comes straight from the width of those rectangles: you split $6x$ into two strips of $3x$ , so the corner is $3^2 = 9$ .

The algebra behind the rule

(x + k)^2 = x^2 + 2kx + k^2

Compare the middle term with $bx$ : you need $2k = b$ , so $k = \tfrac{b}{2}$ . The constant you must add is $k^2 = \left(\tfrac{b}{2}\right)^2$ .

Take half of the coefficient of $x$ , square it, and add it to both sides.

How to complete the square

Move the constant to the right side.
If the leading coefficient is not $1$ , divide every term by it.
Take half the coefficient of $x$ , square it, add to both sides.
Factor the left side as a perfect square.
Take the square root of both sides — remember $\pm$ .
Solve for $x$ .

Worked examples

Common mistakes

Where the quadratic formula comes from

Completing the square on the general equation $ax^2 + bx + c = 0$ produces the quadratic formula — the same five steps, done once with letters instead of numbers:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

That is the payoff. Do this once symbolically and you never have to do it again numerically.

Practice problems

What number completes the square for $x^2 + 10x$ ?

Hint

Half the coefficient of $x$ , then square it.

Answer

$25$

Full solution

Half of $10$ is $5$ , and $5^2 = 25$ . So $x^2 + 10x + 25 = (x+5)^2$ .
Solve $x^2 + 4x - 12 = 0$ by completing the square.

Answer

$x = 2$ or $x = -6$

Full solution

$x^2 + 4x = 12$ . Half of $4$ is $2$ , $2^2 = 4$ .

$x^2 + 4x + 4 = 16$ , so $(x + 2)^2 = 16$ and $x + 2 = \pm 4$ .

$x = 2$ or $x = -6$ . Check: $4 + 8 - 12 = 0$ ✓
Solve $x^2 - 6x + 2 = 0$ .

Answer

$x = 3 \pm \sqrt{7}$

Full solution

$x^2 - 6x = -2$ . Half of $-6$ is $-3$ , $(-3)^2 = 9$ .

$x^2 - 6x + 9 = 7$ , so $(x - 3)^2 = 7$ and $x - 3 = \pm\sqrt{7}$ .

$x = 3 \pm \sqrt{7}$ .
Solve $x^2 + 3x - 1 = 0$ .

Hint

Half of $3$ is $\tfrac{3}{2}$ . Expect fractions.

Answer

$x = \tfrac{-3 \pm \sqrt{13}}{2}$

Full solution

$x^2 + 3x = 1$ . Add $\tfrac{9}{4}$ to both sides: $\left(x + \tfrac{3}{2}\right)^2 = \tfrac{13}{4}$ .

$x + \tfrac{3}{2} = \pm\tfrac{\sqrt{13}}{2}$ , so $x = \tfrac{-3 \pm \sqrt{13}}{2}$ .
Solve $3x^2 - 12x + 9 = 0$ .

Hint

Divide through by $3$ before completing the square.

Answer

$x = 3$ or $x = 1$

Full solution

Divide by $3$ : $x^2 - 4x + 3 = 0$ , so $x^2 - 4x = -3$ .

Half of $-4$ is $-2$ , $(-2)^2 = 4$ : $(x-2)^2 = 1$ , so $x - 2 = \pm 1$ .

$x = 3$ or $x = 1$ . Check: $3(9) - 12(3) + 9 = 0$ ✓
Write $y = x^2 + 2x - 5$ in vertex form.

Answer

$y = (x + 1)^2 - 6$

Full solution

Half of $2$ is $1$ , $1^2 = 1$ . Add and subtract $1$ :

$y = (x^2 + 2x + 1) - 1 - 5 = (x + 1)^2 - 6$ . Vertex $(-1, -6)$ .
What is the minimum value of $y = x^2 - 10x + 30$ , and where does it occur?

Answer

Minimum $5$ at $x = 5$

Full solution

Half of $-10$ is $-5$ , $(-5)^2 = 25$ .

$y = (x^2 - 10x + 25) - 25 + 30 = (x - 5)^2 + 5$ .

Since $(x-5)^2 \ge 0$ , the minimum is $5$ , at $x = 5$ .
Solve $x^2 + 2x + 5 = 0$ by completing the square. What happens?

Hint

Complete the square as usual and look at the right-hand side.

Answer

No real solutions

Full solution

$x^2 + 2x = -5$ . Add $1$ : $(x + 1)^2 = -4$ .

A real number squared is never negative, so there are no real solutions. The parabola sits entirely above the $x$ -axis.

Frequently asked questions

Why do you take half of b and then square it?

Expanding (x + k) squared gives x squared plus 2kx plus k squared. Matching the middle term 2kx to bx means k is half of b, and the constant you need is k squared. So halving b and squaring is not a trick — it is reading off the value of k that makes the trinomial a perfect square.

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

Use the quadratic formula when you only need the roots. Complete the square when you need vertex form, when you are finding a maximum or minimum, or when deriving a result. The quadratic formula is itself completing the square done once in general.

What to learn next

The Quadratic Formula: How and Why It WorksLearn the quadratic formula, where it comes from, and how to use it to solve any quadratic equation, with worked examples and practice problems.

Key terms in this lesson

Constant: A constant is a term with a fixed value and no variable attached. In 3x + 7, the constant is 7, because it stays the same whatever x turns out to be.
Quadratic equation: A quadratic equation is one that can be written as ax squared plus bx plus c = 0, with a not equal to zero. The squared term is what makes it quadratic rather than linear.

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSA.REI.B.4aReasoning with Equations and InequalitiesUse the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)² = q that has the same solutions. Derive the quadratic formula from this form.

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