Quadratic Equations
In one line
A quadratic equation has x² as its highest power, has at most two answers (roots), and you can always find them with the quadratic formula.
Standard form
ax² + bx + c = 0, where a ≠ 0
- a, b and c are real numbers. If a = 0, it isn't quadratic any more.
- The roots are the values of x that make the equation true.
- Examples: x² − 5x + 6 = 0, 2x² + 3x = 0, x² = 49.
Method 1: factorisation (splitting the middle term)
Find two numbers that multiply to a × c and add to b.
Example: x² − 5x + 6 = 0a × c = 6, b = −5. The numbers −2 and −3 multiply to 6 and add to −5.
x² − 2x − 3x + 6 = 0 → x(x − 2) − 3(x − 2) = 0 → (x − 2)(x − 3) = 0.
So x = 2 or x = 3.
x² − 2x − 3x + 6 = 0 → x(x − 2) − 3(x − 2) = 0 → (x − 2)(x − 3) = 0.
So x = 2 or x = 3.
Method 2: the quadratic formula
x = (−b ± √(b² − 4ac)) ÷ 2a
Example: 2x² + x − 6 = 0a = 2, b = 1, c = −6. b² − 4ac = 1 + 48 = 49, √49 = 7.
x = (−1 + 7) ÷ 4 = 3/2, or x = (−1 − 7) ÷ 4 = −2.
x = (−1 + 7) ÷ 4 = 3/2, or x = (−1 − 7) ÷ 4 = −2.
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The discriminant: how many roots?
D = b² − 4ac
| Value of D | Nature of roots | Example |
|---|---|---|
| D > 0 | Two different real roots | x² − 5x + 6 = 0 (D = 1) |
| D = 0 | Two equal real roots (x = −b/2a) | x² − 6x + 9 = 0 (D = 0, x = 3) |
| D < 0 | No real roots | x² + x + 1 = 0 (D = −3) |
Quick check on any answer: the sum of the roots is −b/a and the product is c/a. For x² − 5x + 6 = 0: 2 + 3 = 5 and 2 × 3 = 6. ✓
Word problem
A rectangle's length is 3 m more than its width, and its area is 40 m². Find the width.Let width = x. Then x(x + 3) = 40 → x² + 3x − 40 = 0 → (x + 8)(x − 5) = 0.
Width can't be negative, so width = 5 m and length = 8 m.
Width can't be negative, so width = 5 m and length = 8 m.
Common mistakes
- Not moving everything to one side first. Write the equation as … = 0 before finding a, b and c.
- Getting the sign of b wrong. In x² − 5x + 6, b is −5, so −b is +5.
- Dividing both sides by x. In 2x² + 3x = 0 this loses the root x = 0. Factor instead: x(2x + 3) = 0.
- Keeping answers that make no sense. In word problems, reject negative lengths, ages or speeds.
Quick check
Revise in 30 seconds
| Form | ax² + bx + c = 0, a ≠ 0 |
|---|---|
| Formula | x = (−b ± √(b² − 4ac)) / 2a |
| Discriminant | D = b² − 4ac: >0 two real · =0 equal · <0 none |
| Sum / product | −b/a and c/a |
| Factorising | Find two numbers: multiply to ac, add to b |