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
❓ 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?
✅ 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)
🎯 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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