How to Solve: Completing the Square

How to Solve: Completing the Square

A Complete Guide for Students & Teachers

Introduction

"Completing the square turns a messy quadratic into a perfect vertex form—so you can find the vertex, solve equations, and ace graphing questions in under 60 seconds!

What You Need To Know First

Before starting, you must understand:
1. Expanding brackets – Especially squaring binomials like (x + 3)².
2. Quadratic equations – Standard form: ax² + bx + c = 0.
3. Square roots – Solving x² = 9 gives x = ±3.

If any of these are shaky, review them first.

Key Vocabulary

Formulas To Know

1. Standard Quadratic Form

Formula: ax² + bx + c = 0 - a = coefficient of x² (must be 1 for basic completing the square). - b = coefficient of x. - c = constant term.

MEMORISE THIS – You’ll rewrite every quadratic in this form first.

2. Vertex Form (Completed Square)

Formula: a(x + h)² + k = 0 - (h, k) = vertex of the parabola. - a = same as in standard form (determines width/direction).

MEMORISE THIS – This is your goal when completing the square.

3. Completing the Square "Magic Number"

Formula: (b/2)² - b = coefficient of x in ax² + bx + c. - This number turns x² + bx into a perfect square.

MEMORISE THIS – You’ll use it in Step 3 below.

Step-by-Step Method

Goal: Rewrite ax² + bx + c in vertex form: a(x + h)² + k.

Step 1: Ensure a = 1

• If a ≠ 1, factor a out of the first two terms. Example: 2x² + 8x + 5 → 2(x² + 4x) + 5.
• If a = 1, skip to Step 2.

Step 2: Move the constant term

• Shift c to the other side (if solving an equation). Example: x² + 6x + 5 = 0 → x² + 6x = –5.
• If just rewriting (not solving), leave c where it is.

Step 3: Find the "magic number"

• Take half of b, then square it: (b/2)². Example: For x² + 6x, b = 6 → (6/2)² = 9.

Step 4: Add and subtract the magic number

• Add (b/2)² inside the brackets (or to both sides if solving). Example: x² + 6x + 9 – 9 = –5 → (x² + 6x + 9) – 9 = –5.
• If you factored a in Step 1, add/subtract a × (b/2)² instead. Example: 2(x² + 4x + 4 – 4) + 5 → 2(x² + 4x + 4) – 8 + 5.

Step 5: Rewrite as a perfect square

• The first three terms become (x + b/2)². Example: x² + 6x + 9 = (x + 3)².
• If you factored a, keep it outside: 2(x + 2)².

Step 6: Simplify

• Combine constants on the other side (if solving). Example: (x + 3)² – 9 = –5 → (x + 3)² = 4.
• If rewriting, leave in vertex form: 2(x + 2)² – 3.

Step 7: Solve (if needed)

• Take the square root of both sides, then solve for x. Example: (x + 3)² = 4 → x + 3 = ±2 → x = –3 ± 2 → x = –1 or x = –5.

Clean Worked Example (Using Steps Above)

Problem: Rewrite x² + 8x + 3 in vertex form.

1. a = 1 → No factoring needed.
2. Move +3 to the other side: x² + 8x = –3.
3. Magic number: (8/2)² = 16.
4. Add/subtract 16: x² + 8x + 16 – 16 = –3 → (x² + 8x + 16) – 16 = –3.
5. Rewrite as square: (x + 4)² – 16 = –3.
6. Simplify: (x + 4)² = 13.
7. Vertex form: (x + 4)² – 13.

Final Answer: (x + 4)² – 13.

Worked Examples

Example 1 – Basic (a = 1, no factoring)

Problem: Solve x² + 6x + 2 = 0 by completing the square.

Solution:
1. a = 1 → No factoring.
2. Move +2: x² + 6x = –2.
3. Magic number: (6/2)² = 9.
4. Add/subtract 9: x² + 6x + 9 – 9 = –2 → (x + 3)² – 9 = –2.
5. Rewrite: (x + 3)² = 7.
6. Solve: x + 3 = ±√7 → x = –3 ± √7.

What we did and why: - We turned x² + 6x into a perfect square by adding 9. - This let us solve for x using square roots.

Example 2 – Medium (a ≠ 1, factoring needed)

Problem: Rewrite 2x² + 12x + 5 in vertex form.

Solution:
1. Factor 2 from first two terms: 2(x² + 6x) + 5.
2. Magic number: (6/2)² = 9.
3. Add/subtract 9 inside the brackets: 2(x² + 6x + 9 – 9) + 5.
4. Rewrite: 2((x + 3)² – 9) + 5.
5. Distribute 2: 2(x + 3)² – 18 + 5.
6. Simplify: 2(x + 3)² – 13.

Final Answer: 2(x + 3)² – 13.

What we did and why: - We factored 2 first because a ≠ 1. - Added 9 inside the brackets, but had to subtract 2×9 = 18 to keep the equation balanced.

Example 3 – Exam Style (Disguised, time pressure)

Problem: The equation y = –x² + 4x + 1 represents a parabola. Find its vertex.

Solution:
1. Factor –1 from first two terms: y = –(x² – 4x) + 1.
2. Magic number: (–4/2)² = 4.
3. Add/subtract 4 inside brackets: y = –(x² – 4x + 4 – 4) + 1.
4. Rewrite: y = –((x – 2)² – 4) + 1.
5. Distribute –1: y = –(x – 2)² + 4 + 1.
6. Simplify: y = –(x – 2)² + 5.
7. Vertex is (h, k) = (2, 5).

Final Answer: Vertex at (2, 5).

What we did and why: - The negative a flips the parabola downward. - We kept the –1 factored until the end to avoid sign errors.

Common Mistakes

Exam Traps

1-Minute Recap

"Night before the exam? Here’s the 60-second version:

1. Start with ax² + bx + c. If a isn’t 1, factor it out of the first two terms.
2. Move c to the other side (if solving). If not, leave it.
3. Find the magic number: (b/2)². Add and subtract it inside the brackets.
4. Rewrite as a square: (x + b/2)². If you factored a, keep it outside.
5. Simplify constants. If solving, take square roots. If rewriting, stop at vertex form: a(x + h)² + k.
6. Double-check signs! Negative b or a flips things—don’t rush.

Common traps? Forgetting to factor a, messing up signs, or stopping too early. Practice one problem tonight—you’ve got this!