Completing the Square

Solve quadratics by completing the square

Completing the Square

What is Completing the Square?

Completing the square is a method to rewrite a quadratic expression as a perfect square trinomial.

Goal: Transform ax² + bx + c into a(x - h)² + k form

Why learn it?

  • Solve quadratic equations
  • Find vertex of parabola
  • Derive quadratic formula
  • Convert to vertex form

Key idea: Add and subtract a special value to make a perfect square

Perfect Square Trinomials Review

Pattern: (x + a)² = x² + 2ax + a²

Examples:

  • (x + 3)² = x² + 6x + 9
  • (x - 5)² = x² - 10x + 25
  • (x + 7)² = x² + 14x + 49

Key observation: The constant term is the square of half the coefficient of x

In x² + 6x + 9:

  • Coefficient of x is 6
  • Half of 6 is 3
  • 3² = 9 (the constant!)

General pattern for x² + bx + ?:

Complete the square by adding (b/2)²

The Completing the Square Process

For x² + bx:

Step 1: Find half of b → b/2

Step 2: Square it → (b/2)²

Step 3: Add and subtract (b/2)²

Step 4: Factor the perfect square

Step 5: Simplify

Example 1: Complete the square for x² + 8x

Step 1: Half of 8 = 4 Step 2: 4² = 16 Step 3: x² + 8x + 16 - 16 Step 4: (x + 4)² - 16

Check: (x + 4)² - 16 = x² + 8x + 16 - 16 = x² + 8x ✓

Example 2: Complete the square for x² - 6x

Half of -6 = -3 (-3)² = 9

x² - 6x + 9 - 9 = (x - 3)² - 9

Example 3: Complete the square for x² + 10x

Half of 10 = 5 5² = 25

x² + 10x + 25 - 25 = (x + 5)² - 25

Solving Equations by Completing the Square

Steps:

  1. Move constant to right side
  2. Complete the square on left side
  3. Factor perfect square trinomial
  4. Take square root of both sides
  5. Solve for x

Example 1: Solve x² + 6x + 5 = 0

Step 1: Move constant x² + 6x = -5

Step 2: Complete the square Half of 6 = 3, 3² = 9 Add 9 to both sides: x² + 6x + 9 = -5 + 9 x² + 6x + 9 = 4

Step 3: Factor (x + 3)² = 4

Step 4: Square root both sides x + 3 = ±2

Step 5: Solve x + 3 = 2 or x + 3 = -2 x = -1 or x = -5

Check: (-1)² + 6(-1) + 5 = 1 - 6 + 5 = 0 ✓ (-5)² + 6(-5) + 5 = 25 - 30 + 5 = 0 ✓

Example 2: Solve x² - 4x - 12 = 0

Move constant: x² - 4x = 12

Complete the square: Half of -4 = -2, (-2)² = 4 x² - 4x + 4 = 12 + 4 x² - 4x + 4 = 16

Factor: (x - 2)² = 16

Square root: x - 2 = ±4

Solve: x = 2 + 4 = 6 or x = 2 - 4 = -2

Example 3: Solve x² + 8x + 7 = 0

x² + 8x = -7

Half of 8 = 4, 4² = 16 x² + 8x + 16 = -7 + 16 (x + 4)² = 9

x + 4 = ±3 x = -4 + 3 = -1 or x = -4 - 3 = -7

When Leading Coefficient is Not 1

If ax² + bx + c where a ≠ 1:

Step 1: Factor out a from x² and x terms Step 2: Complete the square inside parentheses Step 3: Distribute a back through Step 4: Simplify

Example 1: Complete the square for 2x² + 12x + 10

Step 1: Factor out 2 from first two terms 2(x² + 6x) + 10

Step 2: Complete inside parentheses Half of 6 = 3, 3² = 9 2(x² + 6x + 9 - 9) + 10 2(x² + 6x + 9) - 18 + 10

Step 3: Factor and simplify 2(x + 3)² - 8

Example 2: Solve 3x² + 12x - 15 = 0

Move constant: 3x² + 12x = 15

Factor out 3: 3(x² + 4x) = 15

Divide by 3: x² + 4x = 5

Complete the square: Half of 4 = 2, 2² = 4 x² + 4x + 4 = 5 + 4 (x + 2)² = 9

x + 2 = ±3 x = -2 + 3 = 1 or x = -2 - 3 = -5

Example 3: 2x² - 8x + 6 = 0

2x² - 8x = -6 2(x² - 4x) = -6 x² - 4x = -3

Half of -4 = -2, (-2)² = 4 x² - 4x + 4 = -3 + 4 (x - 2)² = 1

x - 2 = ±1 x = 3 or x = 1

Converting to Vertex Form

Vertex form: y = a(x - h)² + k

  • Vertex at (h, k)
  • Opens up if a > 0, down if a < 0

Use completing the square to find vertex!

Example 1: Convert y = x² + 6x + 5 to vertex form

Complete the square: y = x² + 6x + 9 - 9 + 5 y = (x + 3)² - 4

Vertex form: y = (x + 3)² - 4 Vertex: (-3, -4)

Example 2: Convert y = x² - 8x + 10 to vertex form

y = x² - 8x + 16 - 16 + 10 y = (x - 4)² - 6

Vertex: (4, -6)

Example 3: Convert y = 2x² + 8x + 3 to vertex form

Factor out 2: y = 2(x² + 4x) + 3

Complete the square: y = 2(x² + 4x + 4 - 4) + 3 y = 2(x + 2)² - 8 + 3 y = 2(x + 2)² - 5

Vertex: (-2, -5)

Deriving the Quadratic Formula

Start with: ax² + bx + c = 0

Divide by a: x² + (b/a)x + c/a = 0

Move constant: x² + (b/a)x = -c/a

Complete the square: Half of b/a is b/(2a) Square: b²/(4a²)

x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²)

Factor: (x + b/(2a))² = -c/a + b²/(4a²)

Common denominator: (x + b/(2a))² = (-4ac + b²)/(4a²) (x + b/(2a))² = (b² - 4ac)/(4a²)

Square root: x + b/(2a) = ±√(b² - 4ac)/(2a)

Solve for x: x = -b/(2a) ± √(b² - 4ac)/(2a)

Quadratic Formula: x = (-b ± √(b² - 4ac))/(2a)

This is how we derive it!

When to Use Completing the Square

Best when:

  • Coefficient of x is even (makes fractions easier)
  • Converting to vertex form
  • Deriving formulas
  • Understanding parabola properties

Maybe not best when:

  • Easy to factor (use factoring instead)
  • Messy fractions (quadratic formula easier)
  • Leading coefficient is large

Always works: Even when factoring doesn't!

Completing the Square with Fractions

Example: x² + 5x + 3 = 0

x² + 5x = -3

Half of 5 = 5/2 (5/2)² = 25/4

x² + 5x + 25/4 = -3 + 25/4 x² + 5x + 25/4 = -12/4 + 25/4 (x + 5/2)² = 13/4

x + 5/2 = ±√(13/4) = ±√13/2

x = -5/2 ± √13/2 = (-5 ± √13)/2

Tip: Fractions are OK! Don't fear them.

Finding Maximum/Minimum Values

Once in vertex form y = a(x - h)² + k:

If a > 0 (opens up): Minimum value is k at x = h

If a < 0 (opens down): Maximum value is k at x = h

Example: y = x² - 6x + 11

Complete the square: y = x² - 6x + 9 - 9 + 11 y = (x - 3)² + 2

Opens up (a = 1 > 0) Minimum value: 2 (at x = 3)

Example 2: y = -2x² + 8x - 3

y = -2(x² - 4x) - 3 y = -2(x² - 4x + 4 - 4) - 3 y = -2(x - 2)² + 8 - 3 y = -2(x - 2)² + 5

Opens down (a = -2 < 0) Maximum value: 5 (at x = 2)

Real-World Applications

Example 1: Projectile Motion

Height: h(t) = -16t² + 64t + 5

Find maximum height:

h(t) = -16(t² - 4t) + 5 h(t) = -16(t² - 4t + 4 - 4) + 5 h(t) = -16(t - 2)² + 64 + 5 h(t) = -16(t - 2)² + 69

Maximum height: 69 feet (at t = 2 seconds)

Example 2: Profit Function

P(x) = -2x² + 80x - 600

Find maximum profit:

P(x) = -2(x² - 40x) - 600 P(x) = -2(x² - 40x + 400) + 800 - 600 P(x) = -2(x - 20)² + 200

Maximum profit: 200 dollars (when x = 20 items)

Example 3: Fencing Problem

Area: A = x(100 - 2x) = 100x - 2x² = -2x² + 100x

A = -2(x² - 50x) A = -2(x² - 50x + 625) + 1250 A = -2(x - 25)² + 1250

Maximum area: 1,250 square feet (when x = 25 feet)

Common Mistakes to Avoid

  1. Forgetting to add to both sides Must maintain equation balance!

  2. Wrong value to add Must be (b/2)², not b/2!

  3. Sign errors (x - 3)² ≠ (x + 3)²

  4. Not factoring out leading coefficient first If a ≠ 1, factor it out before completing square

  5. Arithmetic errors with fractions Be careful with (b/2)² when b is odd

  6. Forgetting ± when taking square root √4 gives ±2, not just 2!

  7. Distributing a incorrectly 2(x² + 4x + 4) - 8 ≠ 2(x + 2)² - 8 Should be 2(x + 2)² - 8 (the 2×4 = 8 came out!)

Completing the Square: Step-by-Step Summary

For solving ax² + bx + c = 0:

  1. If a ≠ 1, divide everything by a
  2. Move constant to right side
  3. Take half of coefficient of x
  4. Square that value
  5. Add to both sides
  6. Factor left side as perfect square
  7. Take square root of both sides (don't forget ±)
  8. Solve for x

For converting to vertex form:

  1. If a ≠ 1, factor it out from x² and x terms
  2. Complete the square inside parentheses
  3. Remember: a(x² + bx + (b/2)²) = a(x + b/2)² but you added a·(b/2)²
  4. Adjust constant outside to compensate
  5. Write as a(x - h)² + k

Practice Problems Strategy

Level 1: Complete the square for x² + bx (even b)

  • x² + 4x, x² + 10x, x² - 8x

Level 2: Solve equations with a = 1

  • x² + 6x + 5 = 0

Level 3: Complete the square with odd b

  • x² + 3x, x² + 7x (practice fractions!)

Level 4: Solve with a ≠ 1

  • 2x² + 8x + 6 = 0

Level 5: Convert to vertex form

  • y = 3x² - 12x + 5

Level 6: Applications

  • Maximum/minimum problems

Comparing Methods

Completing the Square vs Factoring:

  • Factoring: Faster when factors are obvious
  • Completing square: Works even when can't factor

Completing the Square vs Quadratic Formula:

  • Quadratic formula: Memorize once, use always
  • Completing square: Understand process, find vertex

All three methods work! Choose based on situation.

Verification

Always check by expanding back:

If you get (x + 3)² - 5 from x² + 6x + 4:

Expand: (x + 3)² - 5 = x² + 6x + 9 - 5 = x² + 6x + 4 ✓

Or substitute solutions back into original equation

Quick Reference

To complete x² + bx: Add (b/2)²

Perfect square result: (x + b/2)²

If a ≠ 1: Factor out a first

Vertex form: y = a(x - h)² + k Vertex: (h, k)

Don't forget ± When taking square roots!

Value added = value subtracted To maintain balance

Tips for Success

  • Practice finding (b/2)² quickly
  • Master working with fractions
  • Check work by expanding
  • Remember to maintain balance (add to both sides)
  • Factor out leading coefficient before completing square
  • Don't forget the ± when square rooting
  • Use completing the square to understand vertex form
  • Connect to graphing parabolas
  • Verify vertex by substitution
  • Practice with even and odd coefficients
  • Understand why it works, not just how
  • Apply to real-world optimization problems
  • Compare with other solution methods

Completing the square is a powerful technique that unlocks understanding of quadratics, parabolas, and optimization. Master it and you'll have deep insight into quadratic functions!

📚 Practice Problems

1Problem 1easy

Question:

Complete the square: x² + 6x

💡 Show Solution

Step 1: Identify b (the coefficient of x): b = 6

Step 2: Find (b/2)²: (6/2)² = 3² = 9

Step 3: Add and subtract this value: x² + 6x + 9 - 9

Step 4: Factor the perfect square trinomial: (x + 3)² - 9

Step 5: Verify by expanding: (x + 3)² - 9 = x² + 6x + 9 - 9 = x² + 6x ✓

Answer: (x + 3)² - 9

2Problem 2easy

Question:

Solve by completing the square: x² + 8x - 9 = 0

💡 Show Solution

Step 1: Move the constant to the right side: x² + 8x = 9

Step 2: Find (b/2)²: b = 8, so (8/2)² = 4² = 16

Step 3: Add 16 to both sides: x² + 8x + 16 = 9 + 16 x² + 8x + 16 = 25

Step 4: Factor the left side as a perfect square: (x + 4)² = 25

Step 5: Take the square root of both sides: x + 4 = ±5

Step 6: Solve for x: x + 4 = 5 → x = 1 x + 4 = -5 → x = -9

Step 7: Check both solutions: x = 1: 1² + 8(1) - 9 = 1 + 8 - 9 = 0 ✓ x = -9: (-9)² + 8(-9) - 9 = 81 - 72 - 9 = 0 ✓

Answer: x = 1 or x = -9

3Problem 3medium

Question:

Convert to vertex form by completing the square: y = x² - 10x + 18

💡 Show Solution

Step 1: Group the x terms (leave space for completing): y = (x² - 10x ) + 18

Step 2: Find (b/2)²: b = -10, so (-10/2)² = (-5)² = 25

Step 3: Add and subtract 25 inside the parentheses: y = (x² - 10x + 25 - 25) + 18 y = (x² - 10x + 25) - 25 + 18

Step 4: Factor the perfect square: y = (x - 5)² - 7

Step 5: Identify the vertex: Vertex form is y = a(x - h)² + k where vertex is (h, k) Vertex: (5, -7)

Step 6: Verify by expanding: (x - 5)² - 7 = x² - 10x + 25 - 7 = x² - 10x + 18 ✓

Answer: y = (x - 5)² - 7; Vertex: (5, -7)

4Problem 4medium

Question:

Solve: 2x² + 12x - 10 = 0

💡 Show Solution

Step 1: Factor out the leading coefficient from the x² and x terms: 2(x² + 6x) - 10 = 0

Step 2: Move the constant and isolate the parentheses: 2(x² + 6x) = 10 Divide by 2: x² + 6x = 5

Step 3: Complete the square: (b/2)² = (6/2)² = 9 x² + 6x + 9 = 5 + 9 x² + 6x + 9 = 14

Step 4: Factor: (x + 3)² = 14

Step 5: Take square root: x + 3 = ±√14

Step 6: Solve for x: x = -3 + √14 ≈ 0.742 x = -3 - √14 ≈ -6.742

Answer: x = -3 ± √14

5Problem 5hard

Question:

A ball is thrown with height h(t) = -16t² + 32t + 5. Find the maximum height and when it occurs.

💡 Show Solution

Step 1: Convert to vertex form by completing the square: h(t) = -16t² + 32t + 5

Step 2: Factor out -16 from the first two terms: h(t) = -16(t² - 2t) + 5

Step 3: Complete the square inside parentheses: For t² - 2t, (b/2)² = (-2/2)² = 1 h(t) = -16(t² - 2t + 1 - 1) + 5 h(t) = -16(t² - 2t + 1) + 16 + 5

Step 4: Factor and simplify: h(t) = -16(t - 1)² + 21

Step 5: Identify the vertex: Vertex form: h(t) = a(t - h)² + k Vertex: (1, 21)

Step 6: Interpret: Since a = -16 < 0, the parabola opens downward, so the vertex is a maximum. Maximum height: 21 feet Time at maximum: 1 second

Answer: Maximum height is 21 feet at t = 1 second