Completing the Square

Completing the square is a technique that transforms any quadratic expression into a perfect square trinomial — an expression you can write as a single squared binomial. This lets you solve quadratic equations using the square root method (which you already know) and converts quadratics into vertex form, which reveals the highest or lowest point of the parabola.

Completing the square is also the method used to derive the quadratic formula, so understanding it gives you insight into where that formula comes from.

What Is a Perfect Square Trinomial?

A perfect square trinomial is a trinomial that factors as a binomial squared:

$x^2 + 6x + 9 = (x + 3)^2$

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

Notice the pattern. In $x^2 + bx + c$, the trinomial is a perfect square when $c$ equals $\left(\dfrac{b}{2}\right)^2$:

• $x^2 + 6x + \mathord{?}$: half of 6 is 3, and $3^2 = 9$. The missing term is 9.
• $x^2 - 10x + \mathord{?}$: half of $-10$ is $-5$, and $(-5)^2 = 25$. The missing term is 25.

The completing the square process adds exactly the right constant to create this pattern.

The Method: Step by Step

To solve $ax^2 + bx + c = 0$ by completing the square:

1. Move the constant term to the right side of the equation.
2. If $a \neq 1$, divide every term by $a$ so the leading coefficient is 1.
3. Take half of the coefficient of $x$ and square it.
4. Add that value to both sides of the equation.
5. Factor the left side as a perfect square binomial.
6. Solve using the square root method ($\pm$).

Examples with Leading Coefficient 1

Example 1: $x^2 + 8x + 7 = 0$

Step 1 — Move the constant to the right:

$x^2 + 8x = -7$

Step 2 — Take half of 8 and square it:

$\left(\frac{8}{2}\right)^2 = 4^2 = 16$

Step 3 — Add 16 to both sides:

$x^2 + 8x + 16 = -7 + 16$

$x^2 + 8x + 16 = 9$

Step 4 — Factor the left side as a perfect square:

$(x + 4)^2 = 9$

Step 5 — Take the square root of both sides:

$x + 4 = \pm 3$

Step 6 — Solve both cases:

$x = -4 + 3 = -1 \quad \text{or} \quad x = -4 - 3 = -7$

Check $x = -1$: $(-1)^2 + 8(-1) + 7 = 1 - 8 + 7 = 0$ . Correct.

Check $x = -7$: $(-7)^2 + 8(-7) + 7 = 49 - 56 + 7 = 0$ . Correct.

Answer: $x = -1$ or $x = -7$

Example 2: $x^2 - 6x + 2 = 0$

Step 1 — Move the constant:

$x^2 - 6x = -2$

Step 2 — Half of $-6$ is $-3$; square it: $(-3)^2 = 9$.

Step 3 — Add 9 to both sides:

$x^2 - 6x + 9 = -2 + 9 = 7$

Step 4 — Factor:

$(x - 3)^2 = 7$

Step 5 — Square root:

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

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

As decimals: $x \approx 3 + 2.646 = 5.646$ or $x \approx 3 - 2.646 = 0.354$.

Answer: $x = 3 + \sqrt{7}$ or $x = 3 - \sqrt{7}$

Notice that this equation does not factor neatly — the solutions are irrational. Completing the square handles these cases perfectly.

Example 3: $x^2 + 5x - 6 = 0$

Step 1 — Move the constant:

$x^2 + 5x = 6$

Step 2 — Half of 5 is $\dfrac{5}{2}$; square it: $\left(\dfrac{5}{2}\right)^2 = \dfrac{25}{4}$.

Step 3 — Add $\dfrac{25}{4}$ to both sides:

$x^2 + 5x + \frac{25}{4} = 6 + \frac{25}{4} = \frac{24}{4} + \frac{25}{4} = \frac{49}{4}$

Step 4 — Factor:

$\left(x + \frac{5}{2}\right)^2 = \frac{49}{4}$

Step 5 — Square root:

$x + \frac{5}{2} = \pm\frac{7}{2}$

Step 6 — Solve:

$x = -\frac{5}{2} + \frac{7}{2} = \frac{2}{2} = 1$

$x = -\frac{5}{2} - \frac{7}{2} = \frac{-12}{2} = -6$

Answer: $x = 1$ or $x = -6$

Odd coefficients of $x$ produce fractions — that is normal and expected.

Examples with Leading Coefficient Other Than 1

Example 4: $2x^2 + 12x + 10 = 0$

Step 1 — Divide every term by 2:

$x^2 + 6x + 5 = 0$

Step 2 — Move the constant:

$x^2 + 6x = -5$

Step 3 — Half of 6 is 3; $3^2 = 9$. Add 9 to both sides:

$x^2 + 6x + 9 = -5 + 9 = 4$

Step 4 — Factor:

$(x + 3)^2 = 4$

Step 5 — Solve:

$x + 3 = \pm 2$

$x = -3 + 2 = -1 \quad \text{or} \quad x = -3 - 2 = -5$

Answer: $x = -1$ or $x = -5$

Example 5: $3x^2 - 18x + 6 = 0$

Step 1 — Divide by 3:

$x^2 - 6x + 2 = 0$

This is the same as Example 2 above: $x = 3 \pm \sqrt{7}$.

Converting to Vertex Form

Completing the square also converts the standard form $y = ax^2 + bx + c$ into vertex form $y = a(x - h)^2 + k$, where $(h, k)$ is the vertex (the turning point) of the parabola. This is useful for graphing.

Example 6: Convert $y = x^2 - 4x + 7$ to vertex form

Step 1 — Group the $x$ terms:

$y = (x^2 - 4x) + 7$

Step 2 — Complete the square inside the parentheses. Half of $-4$ is $-2$; $(-2)^2 = 4$. Add and subtract 4:

$y = (x^2 - 4x + 4 - 4) + 7$

$y = (x^2 - 4x + 4) - 4 + 7$

Step 3 — Factor the perfect square trinomial:

$y = (x - 2)^2 + 3$

Answer: Vertex form is $y = (x - 2)^2 + 3$. The vertex is at $(2, 3)$, and since the coefficient of the squared term is positive, the parabola opens upward. The minimum value of $y$ is 3.

Real-World Application: HVAC — Optimizing Duct Airflow

An HVAC technician is sizing a rectangular duct where the perimeter of the cross-section must be 40 inches (a constraint based on the available chase space). The technician wants to maximize the cross-sectional area to achieve the best airflow.

Let $w$ = width. Since the perimeter is $2w + 2h = 40$, the height is $h = 20 - w$.

The area is:

$A = w(20 - w) = 20w - w^2 = -w^2 + 20w$

To find the maximum, convert to vertex form by completing the square:

$A = -(w^2 - 20w)$

Half of $-20$ is $-10$; $(-10)^2 = 100$. Add and subtract 100 inside:

$A = -(w^2 - 20w + 100 - 100) = -(w^2 - 20w + 100) + 100$

$A = -(w - 10)^2 + 100$

The vertex is at $(10, 100)$. Maximum area is 100 square inches, achieved when $w = 10$ inches (making $h = 10$ inches too — a square duct).

Answer: The technician should specify a 10-inch by 10-inch duct opening. This gives 100 square inches of cross-sectional area, the maximum possible for a 40-inch perimeter. A square cross-section always maximizes area for a given perimeter — a useful principle HVAC techs apply regularly.

A Carpentry Application: Maximizing a Garden Border

A carpenter is building a rectangular raised garden bed against a wall. The homeowner has 24 feet of border material for the three open sides (the wall serves as the fourth side). What dimensions maximize the planting area?

Let $w$ = the side perpendicular to the wall. The border uses $2w$ for the two sides and $24 - 2w$ for the front.

$A = w(24 - 2w) = 24w - 2w^2 = -2(w^2 - 12w)$

Complete the square: half of $-12$ is $-6$; $(-6)^2 = 36$.

$A = -2(w^2 - 12w + 36 - 36) = -2(w - 6)^2 + 72$

Maximum area is 72 square feet when $w = 6$ feet. The front length is $24 - 2(6) = 12$ feet.

Answer: The bed should be 6 feet deep by 12 feet wide, yielding 72 square feet of planting area.

Common Mistakes to Avoid

1. Forgetting to add to both sides. When you add $\left(\dfrac{b}{2}\right)^2$ to complete the square, you must add it to both sides of the equation (or add and subtract on the same side if converting to vertex form). Missing this changes the equation entirely.

2. Not dividing by the leading coefficient first. Completing the square requires the $x^2$ coefficient to be 1. If it is not, divide everything by that coefficient before proceeding.

3. Sign errors with half the coefficient. If the coefficient of $x$ is $-6$, then half is $-3$ and the square is $+9$ (positive). The squared result is always positive regardless of the sign of $b$.

4. Forgetting the $\pm$ when taking the square root. Just as with the square root method, you need both the positive and negative root.

5. Arithmetic errors with fractions. When $b$ is odd (like 5), you get fractions ($\frac{25}{4}$). Work carefully and convert constants to the same denominator before adding.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• Completing the square transforms $x^2 + bx$ into a perfect square by adding $\left(\dfrac{b}{2}\right)^2$
• The leading coefficient must be 1 — divide everything by $a$ first if it is not
• Whatever you add to one side of the equation, add to the other side too
• This method works on every quadratic equation, including ones that do not factor over the integers
• Completing the square also converts standard form to vertex form $y = a(x - h)^2 + k$, revealing the vertex of the parabola
• Vertex form is essential for optimization problems — the vertex gives the maximum or minimum value

Return to Algebra 1 for more topics in this section.