A18: Solving Quadratic Equations
Solve quadratic equations algebraically by factorising; completing the square and quadratic formula (Higher); solve graphically
📋 Key Concepts
Quadratic equation: An equation of the form ax² + bx + c = 0 where a ≠ 0.
Methods:
- Factorising (all levels)
- Quadratic formula (Higher)
- Completing the square (Higher)
- Graphically (reading x-intercepts)
📝 Solving by Factorising
Method:
- Rearrange to ax² + bx + c = 0
- Factorise into two brackets
- Set each bracket equal to zero
- Solve for x
Example 1
Solve: x² + 5x + 6 = 0
Solution:
Factorise: (x + 2)(x + 3) = 0
x + 2 = 0 or x + 3 = 0
x = -2 or x = -3
Example 2
Solve: x² - 9 = 0
Solution:
Factorise (difference of two squares): (x + 3)(x - 3) = 0
x = 3 or x = -3
Example 3
Solve: x² - 7x + 10 = 0
Solution:
Factorise: (x - 5)(x - 2) = 0
x = 5 or x = 2
Example 4
Solve: 2x² + 7x + 3 = 0
Solution:
Factorise: (2x + 1)(x + 3) = 0
2x + 1 = 0 or x + 3 = 0
x = -½ or x = -3
📝 The Quadratic Formula (Higher)
Formula: For ax² + bx + c = 0:
x = -b ± √(b² - 4ac)⁄2a
Example 5
Solve: x² + 6x + 4 = 0 (give to 2 d.p.)
Solution:
a = 1, b = 6, c = 4
x = -6 ± √(36 - 16)⁄2
x = -6 ± √20⁄2
x = -6 ± 4.47⁄2
x = -6 + 4.47⁄2 or -6 - 4.47⁄2
x = -0.77 or x = -5.24
📝 Completing the Square (Higher)
Method: Rewrite x² + bx as (x + b⁄2)² - (b⁄2)²
Example 6
Solve x² + 8x + 5 = 0 by completing the square.
Solution:
x² + 8x = -5
(x + 4)² - 16 = -5
(x + 4)² = 11
x + 4 = ±√11
x = -4 + √11 or x = -4 - √11
x ≈ -0.68 or x ≈ -7.32
📝 Solving Graphically
Method: Draw the graph of y = ax² + bx + c. The solutions are the x-intercepts (where y = 0).
Example 7
The graph of y = x² - 4x + 3 crosses the x-axis at x = 1 and x = 3. Write the solutions.
Solution:
x² - 4x + 3 = 0 has solutions x = 1 and x = 3
📝 Number of Solutions
Discriminant: b² - 4ac determines the number of solutions:
- b² - 4ac > 0: Two distinct solutions
- b² - 4ac = 0: One solution (repeated root)
- b² - 4ac < 0: No real solutions
Example 8
How many solutions does x² + 4x + 5 = 0 have?
Solution:
b² - 4ac = 16 - 20 = -4 < 0
No real solutions
🧮 Worked Examples — Exam Style
Quadratics by factorising: find two numbers multiplying to c and adding to b. Formula when factorising fails.
Solve x² + 2x − 24 = 0.
Factors of −24 adding to +2: +6 and −4
(x + 6)(x − 4) = 0 → x = −6 or x = 4
M1 for the correct factorisation, A1 for both roots. BOTH roots required — one loses the A mark.
Solve x² − 5x + 1 = 0 using the formula (2 d.p.).
x = (5 ± √(25 − 4))/2 = (5 ± √21)/2
√21 ≈ 4.5826 → x ≈ 4.79 or x ≈ 0.21
M1 for correct substitution, A1 for both values. Write the formula once, substitute once, then evaluate both branches.
Exam wording to expect: "Solve … giving your answers to 2 decimal places", "by factorising", "show that the equation has no real solutions".
❓ Practice Questions
Q1: Solve: x² + 7x + 12 = 0
Q2: Solve: x² - 16 = 0
Q3: Solve: x² - 3x - 10 = 0
Q4: Solve using the formula: x² + 5x + 2 = 0 (to 2 d.p.)
Q5: Solve by completing the square: x² + 10x + 6 = 0
Q6: How many solutions does x² + 6x + 9 = 0 have?
Q7: Solve x² − 8x + 15 = 0 by factorising
Q8: Solve x² + 4x − 21 = 0
✅ Answers
- x = -3 or x = -4
- x = 4 or x = -4
- x = 5 or x = -2
- x = -0.44 or x = -4.56
- x = -5 + √19 or x = -5 - √19
- One solution (x = -3, repeated root)
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. (x + 3)(x + 4) = 0 → x = −3 or x = −4
Q2. x² = 16 → x = ±4
Q3. (x − 5)(x + 2) = 0 → x = 5 or x = −2
Q4. x = (−5 ± √17)/2 → √17 ≈ 4.1231 → x ≈ −0.44 or −4.56
Q5. (x + 5)² = 19 → x = −5 ± √19
Q6. (x + 3)² = 0 → one repeated solution, x = −3
Q7. (x − 3)(x − 5) = 0 → x = 3 or x = 5
Q8. (x + 7)(x − 3) = 0 → x = −7 or x = 3
📌 Key Facts to Memorise
- Factorise: two numbers with product ac (a = 1: product c) and sum b
- Formula: x = (−b ± √(b² − 4ac))/(2a)
- b² − 4ac < 0 → no real solutions · = 0 → one repeated root
- Difference of two squares: x² − k² = (x − k)(x + k) → roots ±k
- Check roots by substituting back into the ORIGINAL equation
🗺️ In the Exam
Solving quadratics: 3–5 marks — factorising on Paper 1, formula on Papers 2/3. Rounding instructions ("2 d.p.") signal the formula route.
Links to: quadratic graphs (A11), simultaneous equations (A19), completing the square (A11+).
🧭 Board Focus — OCR
Spec J560. Three papers of 100 marks each: Paper 1 non-calculator, Papers 2 and 3 calculator, at Foundation/Higher tiers. OCR marks problem-solving explicitly, and its papers tend to front-load multi-step worded contexts — practise extracting the mathematics from prose before calculating.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Always rearrange to ax² + bx + c = 0 first
- Try factorising first - it's often the quickest method
- Use the quadratic formula when factorising is difficult
- Check answers by substituting back
- Remember: b² - 4ac tells you how many solutions
🧠 Problem-Solving Strategies
Problem-Solving
Always rearrange to ax² + bx + c = 0 first. Try factorising (quickest), then use the quadratic formula if it won't factorise. Completing the square helps find the turning point. Check the discriminant (b² - 4ac) to know how many solutions to expect.
Multi-Step Problem
A ball is thrown upward from height 2 m with velocity 15 m/s. Its height after t seconds is h = -5t² + 15t + 2. (a) When does the ball hit the ground? (b) Find the maximum height.
Solution:
(a) -5t² + 15t + 2 = 0 → 5t² - 15t - 2 = 0 → t = 15 ± √(225+40)⁄10 = 15 ± √265⁄10. t ≈ 15 + 16.28⁄10 ≈ 3.13 s
(b) Complete the square: -5(t² - 3t) + 2 = -5(t - 1.5)² + 5(2.25) + 2 = -5(t - 1.5)² + 13.25. Max height = 13.25 m at t = 1.5 s
⚠️ Common Errors
Watch Out!
1. Wrong: x² = 9 so x = 3 Correct: x = 3 OR x = -3 (don't forget the negative solution)
2. Wrong: x² + 5x + 6 = 2 factorises as (x + 2)(x + 3) = 2 Correct: Must rearrange to = 0 first: x² + 5x + 4 = 0, then (x + 1)(x + 4) = 0
3. Wrong: In the quadratic formula, using c = 6 for x² + 5x = -6 Correct: Rearrange first: x² + 5x + 6 = 0, so c = 6 (positive, not -6)
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A rectangle has sides (x + 2) cm and (x - 1) cm. Its area is 20 cm². (a) Form and solve a quadratic equation. (b) Find the dimensions. (c) Verify that your answer is correct.
(a) (x + 2)(x - 1) = 20 → x² + x - 2 = 20 → x² + x - 22 = 0
x = -1 ± √(1 + 88)⁄2 = -1 ± √89⁄2. x must be positive, so x = -1 + 9.43⁄2 ≈ 4.22
(b) Length ≈ 6.22 cm, Width ≈ 3.22 cm
(c) 6.22 × 3.22 ≈ 20.03 ≈ 20 cm² ✓
Mark scheme: (a) 2 marks for equation and solving. (b) 2 marks for dimensions. (c) 2 marks for verification.
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A company's profit P in £1000s is modelled by P = -x² + 10x - 16, where x is the price in pounds.
(a) Find the prices that give zero profit.
(b) What price maximises profit, and what is the maximum?
(c) Explain why not all prices between the break-even points give positive profit.
Answers: (a) -x² + 10x - 16 = 0 → (x-2)(x-8) = 0 → x = £2 or x = £8. (b) Turning point at x = 5: P = -25 + 50 - 16 = 9 (£9000). (c) This is incorrect — actually all prices between £2 and £8 DO give positive profit since the parabola opens downward. The profit is negative only outside this range.
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