Completing the Square – Grade 10 IGCSE Maths

Welcome to Completing the Square!

Completing the square is a powerful algebraic technique that rewrites any quadratic expression into the form (x + p)² + q. This reveals the vertex of a parabola, allows exact solutions to quadratics, and is the foundation for deriving the quadratic formula.

ax² + bx + c → a(x + p)² + q | Vertex at (−p, q) | Axis of symmetry: x = −p

Learning Objectives

Complete the square when a = 1
Complete the square when a ≠ 1
Find the vertex and axis of symmetry of a parabola
Solve quadratic equations by completing the square
Find the minimum or maximum value of a quadratic
Derive the quadratic formula from first principles

Method (a = 1)

x² + bx + c → (x + b/2)² − (b/2)² + c

When a ≠ 1

Factor out a from x-terms first

Applications

Solving, vertex, quadratic formula

Worked Examples

6 mark-scheme style examples

Common Mistakes

5 errors students make

Visualiser

Interactive parabola explorer

Learn 1 — Completing the Square (a = 1)

When the coefficient of x² is 1, completing the square follows a simple process. We rewrite x² + bx + c by creating a perfect square bracket.

x² + bx + c = (x + b/2)² − (b/2)² + c

Step-by-Step Process

Step 1: Identify b. Halve it to get b/2. This goes inside the bracket: (x + b/2).

Step 2: Square b/2 to get (b/2)². This must be subtracted outside the bracket to keep the expression equivalent.

Step 3: Bring along the original constant c. Combine the constant terms: −(b/2)² + c.

Result: (x + b/2)² + [c − (b/2)²] — this is the completed square form.

Worked Example A: x² + 6x + 2

b = 6, so b/2 = 3. Write (x + 3)².

(b/2)² = 3² = 9. Subtract 9: (x + 3)² − 9.

Add the constant +2: (x + 3)² − 9 + 2 = (x + 3)² − 7

Check: expand (x+3)² − 7 = x² + 6x + 9 − 7 = x² + 6x + 2 ✓

Worked Example B: x² − 4x + 9

b = −4, so b/2 = −2. Write (x − 2)².

(b/2)² = (−2)² = 4. Subtract 4: (x − 2)² − 4.

Add constant +9: (x − 2)² − 4 + 9 = (x − 2)² + 5

Reading off the Vertex and Axis of Symmetry

From the form (x + p)² + q:
• Vertex is at (−p, q) — note the sign change on p!
• Axis of symmetry is x = −p
• Example: (x + 3)² − 7 → vertex at (−3, −7), axis x = −3
• Example: (x − 2)² + 5 → vertex at (2, 5), axis x = 2

Learn 2 — Completing the Square (a ≠ 1)

When a ≠ 1, we must factor out a from the x-terms first, then complete the square inside the bracket. The key challenge is remembering that the factored coefficient multiplies the correction term.

General Method

ax² + bx + c = a(x² + (b/a)x) + c = a[(x + b/2a)² − (b/2a)²] + c

Step 1: Factor out a from the x² and x terms only — leave the standalone constant outside.

Step 2: Inside the bracket, complete the square as normal (now coefficient of x² is 1).

Step 3: Re-expand the outer factor a to affect the correction term — this is where students most often make errors.

Step 4: Combine all constant terms outside the bracket.

Worked Example A: 2x² + 12x + 5

Factor 2 from x-terms: 2(x² + 6x) + 5

Inside bracket: x² + 6x. b/2 = 3, (b/2)² = 9. So: (x + 3)² − 9

Substitute back: 2[(x + 3)² − 9] + 5

Expand the 2: 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13

Common Error! Writing 2(x + 3)² − 9 + 5 = 2(x + 3)² − 4. Wrong! The 2 multiplies the −9 to give −18, not leaving it as −9.

Worked Example B: 3x² − 6x + 4

Factor 3 from x-terms: 3(x² − 2x) + 4

Inside bracket: x² − 2x. b/2 = −1, (b/2)² = 1. So: (x − 1)² − 1

Substitute back: 3[(x − 1)² − 1] + 4

Expand: 3(x − 1)² − 3 + 4 = 3(x − 1)² + 1

Vertex of 3(x − 1)² + 1 is at (1, 1). Minimum value is 1. Since a = 3 > 0, it's a minimum (U-shape).

Worked Example C: −2x² + 8x − 3

Factor −2 from x-terms: −2(x² − 4x) − 3

Inside bracket: (x − 2)² − 4

Expand: −2[(x − 2)² − 4] − 3 = −2(x − 2)² + 8 − 3 = −2(x − 2)² + 5

Vertex at (2, 5). Since a = −2 < 0, it's a maximum (∩-shape). Maximum value is 5.

Learn 3 — Applications of Completing the Square

Application 1: Solving Quadratic Equations

Once in the form (x + p)² = k, take the square root of both sides — remember ±.

Solve x² + 6x − 2 = 0

Complete the square: (x + 3)² − 9 − 2 = 0 → (x + 3)² = 11

Take square root: x + 3 = ±√11

Answer: x = −3 + √11 or x = −3 − √11

Always give both values (±). Write in exact form unless asked for decimal. Larger root: −3 + √11 ≈ 0.317. Smaller root: −3 − √11 ≈ −6.317.

Application 2: Finding Minimum / Maximum Values

From a(x + p)² + q:
• If a > 0: minimum value is q, occurring when x = −p
• If a < 0: maximum value is q, occurring when x = −p
• Example: minimum of 2x² + 12x + 5 = 2(x + 3)² − 13 is −13 at x = −3

Application 3: Deriving the Quadratic Formula

Starting from ax² + bx + c = 0, we complete the square to find x.

Divide through by a: x² + (b/a)x + c/a = 0

Move constant: x² + (b/a)x = −c/a

Add (b/2a)² to both sides: (x + b/2a)² = b²/4a² − c/a = (b² − 4ac)/4a²

Take square root: x + b/2a = ±√(b² − 4ac) / 2a

Rearrange: x = [−b ± √(b² − 4ac)] / 2a ✓ — the quadratic formula!

Application 4: Sketching the Parabola

From a(x + p)² + q:
1. Vertex: (−p, q)
2. Direction: U-shape if a > 0; ∩-shape if a < 0
3. y-intercept: substitute x = 0 into original equation
4. x-intercepts (if any): set expression = 0, solve
5. Axis of symmetry: x = −p (vertical line through vertex)

Example 1 — Complete the square: x² + 8x + 3 M1A1

M1: b/2 = 4, write (x + 4)², subtract 4² = 16: (x + 4)² − 16 + 3

A1: (x + 4)² − 13

Vertex: (−4, −13). Axis of symmetry: x = −4. Minimum value = −13.

Example 2 — Complete the square: 2x² − 12x + 7 M1M1A1

M1: Factor 2 from x-terms: 2(x² − 6x) + 7

M1: Inside bracket: (x − 3)² − 9. So: 2[(x − 3)² − 9] + 7

A1: 2(x − 3)² − 18 + 7 = 2(x − 3)² − 11

Minimum value = −11, occurring at x = 3.

Example 3 — Solve x² + 6x − 2 = 0 (exact answers) M1A1A1

M1: Complete the square: (x + 3)² − 9 − 2 = 0 → (x + 3)² = 11

A1: x + 3 = ±√11

A1: x = −3 + √11 or x = −3 − √11

Example 4 — Solve 3x² − 12x + 5 = 0 M1M1A1

M1: 3(x² − 4x) + 5 = 0 → 3(x² − 4x) = −5

M1: 3[(x − 2)² − 4] = −5 → 3(x − 2)² − 12 + 5 = 0 → 3(x − 2)² = 7 → (x − 2)² = 7/3

A1: x − 2 = ±√(7/3) → x = 2 ± √(7/3)

Rationalising: x = 2 ± √21/3 ≈ 2 ± 1.528. So x ≈ 3.528 or x ≈ 0.472.

Example 5 — Find the range of y = x² − 4x + 7 M1A1

M1: Complete the square: (x − 2)² − 4 + 7 = (x − 2)² + 3

A1: Since (x − 2)² ≥ 0 for all x, the minimum value is 3 at x = 2.

Range: y ≥ 3

Example 6 — Prove the quadratic formula from ax² + bx + c = 0 M1M1M1A1

M1: Divide by a: x² + (b/a)x + c/a = 0

M1: Complete the square: (x + b/2a)² − b²/4a² + c/a = 0

M1: (x + b/2a)² = b²/4a² − c/a = (b² − 4ac)/4a²

A1: x = [−b ± √(b² − 4ac)] / 2a ✓

Common Mistakes — Completing the Square

Mistake 1 — Halving b but forgetting to square it

❌ x² + 6x → (x + 3)² − 3 [student subtracted b/2 = 3, not (b/2)² = 9]

✅ x² + 6x = (x + 3)² − 9 Always subtract (b/2)², not b/2. Check: (x+3)² − 9 = x² + 6x + 9 − 9 = x² + 6x ✓

Mistake 2 — When a ≠ 1: forgetting the factor multiplies the correction term

❌ 2x² + 12x + 5 = 2(x + 3)² − 9 + 5 = 2(x + 3)² − 4 [forgot to multiply −9 by 2]

✅ 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13. The outer factor 2 multiplies everything inside the bracket, including the −9.

Mistake 3 — Sign errors with negative b

❌ x² − 6x → (x + 3)² − 9 [used +3 instead of −3 inside bracket]

✅ x² − 6x = (x − 3)² − 9. When b is negative (here b = −6), b/2 = −3, so the bracket is (x − 3). Check: (x−3)² − 9 = x² − 6x + 9 − 9 = x² − 6x ✓

Mistake 4 — Reading the vertex x-coordinate with wrong sign

❌ From (x + 4)² − 7, student writes vertex = (4, −7)

✅ (x + p)² + q → vertex at x = −p. From (x + 4)² − 7: p = 4, so x-coord = −4. Vertex = (−4, −7). The x-coord of the vertex is always the opposite sign of what's inside the bracket.

Mistake 5 — Confusing minimum and maximum

❌ Student says −x² + 4x − 1 has a minimum value of 3

✅ The leading coefficient determines the shape. a > 0 → U-shape → minimum. a < 0 → ∩-shape → maximum. Here a = −1 < 0, so completing square gives −(x−2)² + 3 → maximum value = 3 at x = 2.

Key Formulas

Given in Exam (Formula Sheet)

Quadratic formula: x = [−b ± √(b²−4ac)] / 2a
Discriminant: b² − 4ac

The quadratic formula is on the Cambridge IGCSE formula sheet — you do NOT need to memorise it, but you must know when and how to use it.

Must Memorise

x² + bx + c = (x + b/2)² − (b/2)² + c
ax² + bx + c = a(x + b/2a)² − b²/4a + c
Vertex from (x + p)² + q is at (−p, q)
a > 0 → minimum value = q at x = −p
a < 0 → maximum value = q at x = −p
Solve: (x + p)² = k → x = −p ± √k

Quick Reference — The Process

When a = 1: x² + bx + c → (x + b/2)² − (b/2)² + c

When a ≠ 1: ax² + bx + c → factor a → a[(x + b/2a)² − (b/2a)²] + c → expand

To solve: get (x + p)² = k → x = −p ± √k

Vertex: (−p, q) from a(x + p)² + q. Min/max = q.

Parabola Visualiser — Completing the Square

Adjust a, b, c to see y = ax² + bx + c. The completed square form and vertex are shown below.

a: 1 b: 4 c: 1

Adjust sliders to see the completed square form.

Exercise 1 — Complete the Square (a = 1) — Find the Vertex

Write each quadratic in completed square form. Enter the y-coordinate of the vertex (the constant term q).

1. x² + 4x + 1 = (x + 2)² + q. Find q.

2. x² + 6x + 5 = (x + 3)² + q. Find q.

3. x² − 8x + 10. Vertex y-coordinate (minimum value)?

4. x² + 10x + 20. Vertex y-coordinate?

5. x² − 2x − 3. Vertex x-coordinate?

6. x² + 8x + 3. Minimum value?

7. x² − 6x + 11. Vertex y-coordinate?

8. x² + 12x + 40. Minimum value?

Exercise 2 — Complete the Square (a ≠ 1)

Write in form a(x + p)² + q. Find the minimum or maximum value q.

1. 2x² + 8x + 1 = 2(x + 2)² + q. Find q.

2. 3x² + 6x + 5. Minimum value?

3. 2x² − 12x + 7. Minimum value?

4. 4x² + 16x + 9. Minimum value?

5. −x² + 4x − 1. Maximum value?

6. −2x² + 8x − 3. Maximum value?

7. 5x² − 20x + 13. Minimum value?

8. 3x² + 12x − 1. Minimum value?

Exercise 3 — Solving by Completing the Square

Solve by completing the square. Give the larger root to 2 decimal places.

1. x² + 4x − 1 = 0. Larger root (2 d.p.)?

2. x² − 6x + 4 = 0. Larger root (2 d.p.)?

3. x² + 2x − 5 = 0. Larger root (2 d.p.)?

4. x² − 8x + 3 = 0. Larger root (2 d.p.)?

5. x² + 10x + 7 = 0. Larger root (2 d.p.)?

6. x² − 4x − 2 = 0. Larger root (2 d.p.)?

7. 2x² + 4x − 3 = 0. Larger root (2 d.p.)?

8. 3x² − 6x − 1 = 0. Larger root (2 d.p.)?

Exercise 4 — Minimum / Maximum Values (Exam Focus)

Find the minimum or maximum value using completing the square.

1. Minimum value of x² + 6x + 2?

2. Minimum value of x² − 10x + 30?

3. Maximum value of −x² + 6x − 5?

4. Minimum value of 2x² − 4x + 7?

5. Maximum value of −3x² + 12x − 8?

6. x value at which x² + 8x + 3 is minimum?

7. x value at which −x² + 4x + 5 is maximum?

8. Minimum value of 4x² − 24x + 41?

Exercise 5 — Mixed and Word Problems

Apply completing the square in context.

1. A rectangle has perimeter 20 cm. Width = x, length = 10 − x. Area = x(10−x) = −x²+10x. Maximum area (cm²)?

2. A ball's height: h = −5t² + 20t + 2. Maximum height?

3. Minimum of x² − 3x + 4? (Give as fraction if needed — enter decimal to 2 d.p.)

4. Solve x² + 6x + 1 = 0. Larger root (2 d.p.)?

5. If (x + p)² + q = x² − 10x + 26, find p + q.

6. 3x² − 6x + 5 = 3(x − 1)² + q. Find q.

7. Range of y = 2x² − 8x + 11 is y ≥ k. Find k.

8. Maximum of −2x² + 12x − 7?

Practice — 25 Questions

🔵 = Non-calculator 🟢 = Calculator allowed

🔵 1. x² + 2x + 5 → (x+1)² + q. Find q.

🔵 2. Minimum of x² + 4x − 1?

🔵 3. x² − 8x + 20. Vertex x-coordinate?

🔵 4. x² − 8x + 20. Minimum value?

🔵 5. 2x² + 4x + 9. Minimum value?

🔵 6. −x² + 6x − 5. Maximum value?

🔵 7. x² + 14x + 50. Vertex y-coord?

🔵 8. (x + 5)² + q = x² + 10x + 3. Find q.

🔵 9. 3x² − 12x + 15. Minimum value?

🔵 10. x at which x² − 10x + 31 is minimum?

🟢 11. Solve x² + 6x − 1 = 0. Larger root (2 d.p.)?

🟢 12. Solve x² − 4x − 7 = 0. Larger root (2 d.p.)?

🟢 13. Solve x² + 8x + 10 = 0. Larger root (2 d.p.)?

🟢 14. Solve 2x² − 4x − 1 = 0. Larger root (2 d.p.)?

🔵 15. −3x² + 6x + 1. Maximum value?

🔵 16. 5x² + 10x + 8. Minimum value?

🔵 17. x² + 20x + 100. Vertex x-coord?

🔵 18. x² + 20x + 100. Vertex y-coord?

🟢 19. Solve x² − 2x − 6 = 0. Larger root (2 d.p.)?

🔵 20. 4x² + 8x + 5. Minimum value?

🔵 21. What is (b/2)² when b = 7? (Enter as decimal)

🔵 22. x² − 5x + 6. Minimum value (enter as decimal)?

🟢 23. Ball height h = −t² + 6t + 2. Maximum height?

🔵 24. 2x² + 8x + k. If minimum is 0, find k.

🟢 25. Solve 3x² + 6x − 2 = 0. Larger root (2 d.p.)?

Challenge — 12 Multi-Step Questions

These require combining multiple skills. Give answers to 2 d.p. where needed.

1. The parabola y = x² + bx + 5 has vertex at x = 3. Find b.

2. y = ax² − 8x + 13 has minimum at (2, 5). Find a.

3. Solve x² + 6x + 7 = 0 by CTS. Larger root (2 d.p.)?

4. Write x² − 3x + 4 in form (x − p)² + q. Find p × q (enter decimal)?

5. The minimum of kx² + 4x + 1 is 0. Find k.

6. Two solutions of x² + bx + c = 0 are x = 1 ± √5. Find c.

7. Rectangle area = x(12 − x). Maximum area?

8. Solve 2x² + 5x − 3 = 0 by CTS. Larger root?

9. x² + px + q = (x − 3)² − 4. Find p + q.

10. Minimum of 2x² − kx + 8 is 0. Find k (positive value).

11. If x = 2 + √3 is a root of x² + bx + c = 0, and the other root is 2 − √3, find b.

12. Height h = −4.9t² + 14t + 1. Maximum height (2 d.p.)?

Exam Style — 5 Structured Questions

Multi-part questions in IGCSE style. Enter numeric answers.

Q1 [4 marks] — f(x) = x² − 6x + 14

(a) Write f(x) in the form (x + p)² + q. Enter p (the number inside the bracket).

(b) Enter q (the constant term outside the bracket).

(c) State the minimum value of f(x).

(d) State the x-value at which this minimum occurs.

Q2 [3 marks] — Solve x² + 8x + 11 = 0 by completing the square. Give answers in the form a ± b√c.

(a) After completing the square: (x + 4)² = k. Find k.

(b) Larger root to 2 d.p.?

(c) Smaller root to 2 d.p.?

Q3 [4 marks] — g(x) = 3x² − 12x + 7

(a) Write in form a(x + p)² + q. Enter a × p (product).

(b) State the minimum value of g(x).

(c) State the x-value at which minimum occurs.

(d) Does g(x) have 0, 1, or 2 real roots? (Enter 0, 1 or 2)

Q4 [5 marks] — A ball is thrown and its height h = −t² + 4t + 5 (metres, t seconds).

(a) Maximum height (m)?

(b) Time when maximum height is reached (s)?

(c) Height when t = 0 (initial height)?

(d) Solve h = 0 by CTS. Positive t value (2 d.p.)?

(e) When t = 3, find h.

Q5 [4 marks] — Completing the square: 2x² + 8x + k.

(a) Write in form 2(x + p)² + q. Enter p.

(b) Express q in terms of k.

(Hint: 2(x+2)² + (k−8). Enter the value of q when k = 3.)

(c) If the minimum is −6, find k.

(d) For k = 2, how many real roots does 2x² + 8x + k = 0 have? (Enter 0, 1 or 2)