Solving quadratic equations
Write it as ax² + bx + c = 0, then solve by factorisation, completing the square or the formula. A quadratic has at most two roots.
Three methods and when to use them
| Method | How | Use it when |
|---|---|---|
| Factorisation | Split the middle term, factorise, set each factor equal to 0 | ac has an easy factor pair that adds up to b |
| Completing the square | Make the x terms into a perfect square (x + p)², then take square roots | The question asks for it |
| Formula | Put a, b and c into x = [−b ± √(b² − 4ac)]/(2a) | Factors are hard to find, or the roots have surds such as √13 |
Worked example 1: by factorisation
Solve 2x² − 7x + 3 = 0.
- a × c = 2 × 3 = 6. Find two numbers with product 6 and sum −7: they are −6 and −1.
- Split the middle term: 2x² − 6x − x + 3 = 0.
- Group: 2x(x − 3) − 1(x − 3) = 0, so (2x − 1)(x − 3) = 0.
- 2x − 1 = 0 gives x = 1/2. x − 3 = 0 gives x = 3.
- Check x = 3: 2(9) − 21 + 3 = 0. True.
x = 1/2 or x = 3
Worked example 2: by completing the square
Solve 2x² + 5x − 3 = 0 by completing the square.
- Divide by 2 so that x² has coefficient 1: x² + (5/2)x − 3/2 = 0.
- Move the constant to the right: x² + (5/2)x = 3/2.
- Half of 5/2 is 5/4. Add (5/4)² = 25/16 to both sides.
- (x + 5/4)² = 3/2 + 25/16 = 24/16 + 25/16 = 49/16.
- Take square roots: x + 5/4 = ± 7/4.
- x = −5/4 + 7/4 = 1/2, or x = −5/4 − 7/4 = −3.
x = 1/2 or x = −3
Worked example 3: by the formula
Solve 3x² − 5x + 1 = 0 using the quadratic formula.
- Formula: x = [−b ± √(b² − 4ac)]/(2a).
- Here a = 3, b = −5 and c = 1.
- b² − 4ac = (−5)² − 4(3)(1) = 25 − 12 = 13.
- x = [5 ± √13]/6.
- With √13 ≈ 3.606, x ≈ 1.434 or x ≈ 0.232 (rounded to 3 decimal places).
x = (5 + √13)/6 or x = (5 − √13)/6
Worked example 4: a bus journey word problem
A bus covers a 300 km journey from Chennai at a uniform speed. If its speed were 10 km/h more, it would take 1 hour less. Find its speed.
- Let the speed be x km/h. Formula: time = distance ÷ speed, so the time is 300/x hours.
- At (x + 10) km/h the time is 300/(x + 10) hours.
- The difference is 1 hour: 300/x − 300/(x + 10) = 1.
- Multiply by x(x + 10): 300(x + 10) − 300x = x(x + 10), so 3000 = x² + 10x.
- x² + 10x − 3000 = 0. Two numbers with product −3000 and sum 10 are 60 and −50, so (x + 60)(x − 50) = 0.
- x = 50 or x = −60. A speed cannot be negative, so reject −60.
- Check: 300/50 = 6 hours and 300/60 = 5 hours. The difference is 1 hour.
The speed of the bus is 50 km/h
Planning a rectangular terrace garden or a kolam of a fixed area leads to a quadratic. Problems on speed, age and area all become ax² + bx + c = 0.
1 marks: “Which number must be added to x² + 8x to make it a perfect square?” 16, since (8/2)² = 16 and x² + 8x + 16 = (x + 4)².
2 marks: “Solve x² − 2x − 15 = 0 by factorisation.” Numbers with product −15 and sum −2 are −5 and 3. (x − 5)(x + 3) = 0, so x = 5 or x = −3.
5 marks: “The length of a rectangular garden is 3 m more than its breadth. Its area is 108 m². Find its length and breadth.” Let the breadth be x m. x(x + 3) = 108, so x² + 3x − 108 = 0, and (x + 12)(x − 9) = 0. Reject −12. Breadth 9 m, length 12 m.
Factorise if you can, use the formula if you can't. Always reject a root that makes no sense, like a negative speed or age.