Precalculus — Chapter 3

Completing the Square

The complete guide — vertex form, solving equations, circles, and how to derive the quadratic formula from scratch.

Three Reasons to Complete the Square

1

Vertex Form

Convert ax² + bx + c to a(x−h)² + k to find the vertex (h, k) — axis of symmetry and max/min value.

2

Solve Equations

Solve any quadratic that won't factor. A backup method when factoring fails — always works.

3

Circle Equations

Rewrite x² + y² + Dx + Ey + F = 0 into standard form (x−h)² + (y−k)² = r² to find center and radius.

The Key Idea: Perfect Square Trinomials

A perfect square trinomial factors as a binomial squared:

x² + bx + (b/2)² = (x + b/2)²

x² − bx + (b/2)² = (x − b/2)²

Pattern Examples

x² + 6x + 9 = (x + 3)²   [b=6, (b/2)²=9]

x² − 10x + 25 = (x − 5)²   [b=10, (b/2)²=25]

x² + 3x + 9/4 = (x + 3/2)²   [b=3, (b/2)²=9/4]

Rule: Take half the coefficient of x, square it — that's the term to add.

Method 1: a = 1 (Coefficient of x² is 1)

1

Make sure coefficient of x² is 1 (if not, divide everything by a)

2

Move constant (c) to the right side

3

Add (b/2)² to BOTH sides

4

Factor left side as (x + b/2)²

5

Take square root of both sides (include ± on right)

6

Solve for x

Worked Example: Solve x² + 8x − 9 = 0

Step 2: x² + 8x = 9

Step 3: (b/2)² = (8/2)² = 16 → x² + 8x + 16 = 9 + 16

Step 4: (x + 4)² = 25

Step 5: x + 4 = ±5

Step 6: x = −4 + 5 = 1 or x = −4 − 5 = −9

Method 2: a ≠ 1 (Coefficient of x² is not 1)

Same steps, but first divide the entire equation by a (or factor a out of the x terms).

Worked Example: Solve 2x² − 12x + 4 = 0

Divide by 2: x² − 6x + 2 = 0

Move constant: x² − 6x = −2

(b/2)² = (−6/2)² = 9 → x² − 6x + 9 = −2 + 9

(x − 3)² = 7

x − 3 = ±√7

x = 3 ± √7

Vertex Form Example: Write y = 3x² − 18x + 7 in vertex form

Factor 3 from x terms: y = 3(x² − 6x) + 7

Complete inside: (b/2)² = 9 → add and subtract inside:

y = 3(x² − 6x + 9 − 9) + 7

y = 3(x − 3)² − 3(9) + 7 = 3(x − 3)² − 27 + 7

y = 3(x − 3)² − 20 → Vertex: (3, −20)

Application: Circle Equations

Complete the square in both x and y to convert a general circle equation to standard form.

Example: x² + y² − 6x + 4y − 3 = 0

Group x and y: (x² − 6x) + (y² + 4y) = 3

Complete x: add (−6/2)² = 9 → (x² − 6x + 9)

Complete y: add (4/2)² = 4 → (y² + 4y + 4)

(x² − 6x + 9) + (y² + 4y + 4) = 3 + 9 + 4

(x − 3)² + (y + 2)² = 16

Center: (3, −2), Radius: √16 = 4

Deriving the Quadratic Formula

The quadratic formula comes from completing the square on the general form ax² + bx + c = 0:

ax² + bx + c = 0

x² + (b/a)x + c/a = 0 [÷ by a]

x² + (b/a)x = −c/a

x² + (b/a)x + (b/2a)² = −c/a + (b/2a)²

(x + b/2a)² = (b² − 4ac) / (4a²)

x + b/2a = ±√(b² − 4ac) / (2a)

x = (−b ± √(b² − 4ac)) / (2a) ✓

Frequently Asked Questions

What are the steps to complete the square?

For ax² + bx + c: (1) Divide by a if a ≠ 1; (2) Move constant to the right side; (3) Add (b/2)² to both sides; (4) Factor the left side as a perfect square trinomial (x + b/2)²; (5) Solve for x by taking the square root of both sides. Example: x² + 6x + 5 = 0 → x² + 6x = −5 → x² + 6x + 9 = −5 + 9 → (x+3)² = 4 → x = −3 ± 2 → x = −1 or x = −5.

How do you complete the square when a ≠ 1?

Divide the entire equation by a first, then proceed with the standard steps. Example: 2x² + 8x − 3 = 0 → x² + 4x − 3/2 = 0 → x² + 4x = 3/2 → x² + 4x + 4 = 3/2 + 4 → (x+2)² = 11/2 → x = −2 ± √(11/2). Alternatively, factor a out of only the x² and x terms before completing the square.

Why do we complete the square?

Completing the square has three main uses: (1) Converting a quadratic to vertex form y = a(x−h)² + k to find the vertex; (2) Solving quadratic equations that don't factor; (3) Writing circle equations in standard form (x−h)² + (y−k)² = r² to find center and radius. It's also how the quadratic formula is derived.

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