EDEXCEL GCSE MATHS · HIGHER
Completing the square
Equations and inequalities · Lesson 4 of 7
Teacher copy - includes the notes for whoever is teaching from it.
Warm-up
Answer each one, then check.
1. Expand (x + 3)².
x² + 6x + 9
2. Expand (x − 4)².
x² − 8x + 16
3. Work out √11 to 2 decimal places.
3.32
4. Half of 6 is what?
3
5. What is the turning point of y = x²?
(0, 0)
Learning Objectives
1. Write x² + bx + c in the form (x + p)² + q.
2. Complete the square when the coefficient of x² is not 1.
3. Solve quadratic equations by completing the square.
4. Find the turning point of a quadratic graph.
The Method
x² + bx + c = (x + b/2)² − (b/2)² + c
Halve the coefficient of x, square it, and subtract it.
Completing the Square
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Write x² + 6x + 5 in the form (x + p)² + q. |
1. Halve the coefficient of x
6 ÷ 2 = 3, so (x + 3)²
2. (x + 3)² expands to
x² + 6x + 9
3. Adjust the constant: we need 5, not 9
5 − 9 = −4
4. Write the result
(x + 3)² − 4
Answer: (x + 3)² − 4
A Negative Coefficient
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Write x² − 8x + 3 in the form (x + p)² + q. |
1. Halve −8
−4, so (x − 4)²
2. (x − 4)² expands to
x² − 8x + 16
3. Adjust the constant
3 − 16 = −13
Answer: (x − 4)² − 13
What the Form Tells You
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In y = (x + p)² + q the turning point is (−p, q). |
The graph of y equals x squared plus 6x plus 5 with its turning point at minus 3, minus 4 and roots at minus 5 and minus 1.
Reading the Completed Square Form
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y = (x + 3)² − 4 Turning point (−3, −4): the sign inside the bracket is reversed. |
Line of symmetry x = −3. |
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Minimum value −4, when x = −3. |
Never below the minimum (x + 3)² ≥ 0, so y ≥ −4. |
Solving by Completing the Square
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Solve x² + 6x − 2 = 0, leaving your answer in surd form. |
1. Complete the square
(x + 3)² − 9 − 2 = 0, so (x + 3)² − 11 = 0
2. Rearrange
(x + 3)² = 11
3. Take the square root of both sides
x + 3 = ± √11
4. Solve
x = −3 ± √11
Answer: x = −3 + √11 or x = −3 − √11
A Coefficient of x² Not 1
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Write 2x² + 12x + 7 in the form a(x + p)² + q. |
1. Take out the 2 from the x terms
2(x² + 6x) + 7
2. Complete the square inside
2[(x + 3)² − 9] + 7
3. Multiply out the 2
2(x + 3)² − 18 + 7
4. Simplify
2(x + 3)² − 11
Answer: 2(x + 3)² − 11
A Proof
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Show that x² + 4x + 5 is always positive. |
1. Complete the square
x² + 4x + 5 = (x + 2)² + 1
2. A square is never negative
(x + 2)² ≥ 0
3. Add 1
(x + 2)² + 1 ≥ 1
Answer: (x + 2)² + 1 ≥ 1 > 0, so the expression is always positive.
Key Terms
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Completing the square Rewriting a quadratic as a squared bracket plus a constant. |
Turning point The lowest or highest point of a quadratic graph. |
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Minimum value The smallest value a function can take. |
Line of symmetry The vertical line through the turning point. |
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Surd A root left in exact form, such as √11. |
Coefficient The number multiplying a variable. |
Your Task: Complete the Square Race
12 minutes
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Write each in the form (x + p)² + q, then state the turning point of its graph. (a) x² + 4x + 1 (b) x² − 10x + 30 (c) x² + 2x − 8 (d) x² − 3x. 1. Halve the coefficient of x. 2. Subtract its square. 3. Read off the turning point. |
A good answer shows: (a) (x + 2)² − 3, turning point (−2, −3). (b) (x − 5)² + 5, turning point (5, 5). (c) (x + 1)² − 9, turning point (−1, −9). (d) (x − 3/2)² − 9/4, turning point (3/2, −9/4).
Note: Ask students to check by expanding.
Can I...?
☐ Complete the square when a = 1.
☐ Complete the square when a ≠ 1.
☐ Find the turning point from the completed form.
☐ Solve by completing the square.
☐ Give answers as surds.
☐ Use the form to find the minimum value.
☐ Prove a quadratic is always positive.
☐ Check by expanding.
Summary
✓ x² + bx + c = (x + b/2)² + c − (b/2)².
✓ Turning point of y = (x + p)² + q is (−p, q).
✓ Solve by rearranging to (x + p)² = k.
✓ (x + p)² ≥ 0 helps with proofs.
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EXAM FOCUS Write x² − 8x + 3 in the form (x + a)² + b. (2 marks) Halve the coefficient of x (keeping its sign) and square it. Then adjust the constant. Always check by expanding. |
Exam Practice: Completing the Square
Answer all questions. Show your working. · 30 minutes
▸ Question 1 · 2 marks · Non-calculator. Write x² + 6x + 5 in the form (x + a)² + b.
▸ Question 2 · 2 marks · Non-calculator. Write x² − 8x + 3 in the form (x + a)² + b.
▸ Question 3 · 3 marks · Non-calculator. Solve x² + 6x − 2 = 0. Give your answers in the form p ± √q.
▸ Question 4 · 3 marks · Non-calculator. The graph of y = x² − 4x + 7 has a minimum point. Find the coordinates of the minimum point.
▸ Question 5 · 3 marks · Non-calculator. Write 2x² + 12x + 7 in the form a(x + p)² + q.
▸ Question 6 · 3 marks · Show that. Show that x² + 4x + 5 is always positive for all values of x.
Question 1 · 2 marks · Non-calculator
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“Write x² + 6x + 5 in the form (x + a)² + b.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 2 marks, so plan before writing.
Question 1 · mark scheme
2 marks available. Award a mark for each point made.
▸ (x + 3)². M1
▸ (x + 3)² − 4. A1
▸ Model answer. (x + 3)² − 9 + 5 = (x + 3)² − 4.
Question 2 · 2 marks · Non-calculator
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“Write x² − 8x + 3 in the form (x + a)² + b.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 2 marks, so plan before writing.
Question 2 · mark scheme
2 marks available. Award a mark for each point made.
▸ (x − 4)². M1
▸ (x − 4)² − 13. A1
▸ Model answer. (x − 4)² − 16 + 3 = (x − 4)² − 13.
Question 3 · 3 marks · Non-calculator
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“Solve x² + 6x − 2 = 0. Give your answers in the form p ± √q.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 3 · mark scheme
3 marks available. Award a mark for each point made.
▸ (x + 3)² − 9 − 2. M1
▸ (x + 3)² = 11. M1
▸ −3 ± √11. A1
▸ Model answer. (x + 3)² − 11 = 0, so (x + 3)² = 11 and x = −3 ± √11.
Question 4 · 3 marks · Non-calculator
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“The graph of y = x² − 4x + 7 has a minimum point. Find the coordinates of the minimum point.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 4 · mark scheme
3 marks available. Award a mark for each point made.
▸ (x − 2)². M1
▸ (x − 2)² + 3. M1
▸ (2, 3). A1
▸ Model answer. x² − 4x + 7 = (x − 2)² + 3, so the minimum point is (2, 3).
Question 5 · 3 marks · Non-calculator
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“Write 2x² + 12x + 7 in the form a(x + p)² + q.” |
HOW TO ANSWER IT Command word: Non-calculator. Worth 3 marks, so plan before writing.
Question 5 · mark scheme
3 marks available. Award a mark for each point made.
▸ 2(x² + 6x). M1
▸ 2(x + 3)² − 18 + 7. M1
▸ 2(x + 3)² − 11. A1
▸ Model answer. 2(x² + 6x) + 7 = 2[(x + 3)² − 9] + 7 = 2(x + 3)² − 11.
Question 6 · 3 marks · Show that
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“Show that x² + 4x + 5 is always positive for all values of x.” |
HOW TO ANSWER IT Command word: Show that. Worth 3 marks, so plan before writing.
Question 6 · mark scheme
3 marks available. Award a mark for each point made.
▸ (x + 2)² + 1. M1
▸ (x + 2)² ≥ 0. M1
▸ Conclusion. C1
▸ Model answer. x² + 4x + 5 = (x + 2)² + 1. A square is never negative, so (x + 2)² ≥ 0 and the expression is at least 1, which is positive.