A quadratic equation has the form ax² + bx + c = 0. There are four methods to solve them — knowing which to use and when makes algebra much faster.
Standard Form
Every quadratic equation can be written as:
ax² + bx + c = 0
Where a ≠ 0 (if a = 0, it's a linear equation).
Examples:
- x² − 5x + 6 = 0 (a=1, b=−5, c=6)
- 2x² + 3x − 2 = 0 (a=2, b=3, c=−2)
- x² − 9 = 0 (a=1, b=0, c=−9)
Method 1: Factoring
Works best when the equation factors cleanly into integers. Fastest method when applicable.
Steps:
- Write in standard form
- Find two numbers that multiply to (a × c) and add to b
- Split the middle term and factor by grouping
- Set each factor equal to zero
Example: x² − 5x + 6 = 0
- Need two numbers: multiply to 6, add to −5 → −2 and −3
- Factor: (x − 2)(x − 3) = 0
- Solutions: x = 2 or x = 3
Example: 2x² + 5x + 3 = 0
- a × c = 6, need factors adding to 5 → 2 and 3
- Rewrite: 2x² + 2x + 3x + 3 = 0
- Factor: 2x(x + 1) + 3(x + 1) = 0
- Factor: (2x + 3)(x + 1) = 0
- Solutions: x = −3/2 or x = −1
When to use: When you can spot the factors quickly. If you don't find factors in 30 seconds, switch methods.
Method 2: The Quadratic Formula
Works for every quadratic equation. Use this when factoring is not obvious.
x = (−b ± √(b² − 4ac)) ÷ (2a)
Example: 2x² + 3x − 2 = 0 (a=2, b=3, c=−2)
- Discriminant: b² − 4ac = 9 − (4 × 2 × −2) = 9 + 16 = 25
- √25 = 5
- x = (−3 ± 5) ÷ 4
- x = (−3 + 5) ÷ 4 = 0.5 or x = (−3 − 5) ÷ 4 = −2
The Discriminant: How Many Solutions?
The expression b² − 4ac tells you the nature of solutions before you solve:
| Discriminant | Number of Solutions | Type |
|---|---|---|
| b² − 4ac > 0 | Two distinct real solutions | Real numbers |
| b² − 4ac = 0 | One repeated solution | Real, equal roots |
| b² − 4ac < 0 | No real solutions | Two complex/imaginary roots |
Example: x² + 2x + 5 = 0
- Discriminant = 4 − 20 = −16 → no real solutions
- Complex solutions: x = (−2 ± √(−16)) ÷ 2 = −1 ± 2i
Method 3: Completing the Square
Transforms the equation into (x + p)² = q form. Essential for understanding vertex form and deriving the quadratic formula.
Steps:
- Move constant to right side
- Divide through by a (if a ≠ 1)
- Add (b/2a)² to both sides
- Factor left side as a perfect square
- Take square root of both sides
Example: x² + 6x + 5 = 0
- x² + 6x = −5
- Add (6/2)² = 9 to both sides: x² + 6x + 9 = 4
- (x + 3)² = 4
- x + 3 = ±2
- x = −1 or x = −5
Method 4: Graphing
The solutions (roots) are the x-intercepts of the parabola y = ax² + bx + c.
- Two x-intercepts → two real solutions
- One x-intercept (vertex on x-axis) → one repeated solution
- No x-intercepts → no real solutions (complex roots)
When to use: For visual understanding or when using a graphing calculator. Not practical for exact answers.
Choosing the Right Method
| Situation | Best Method |
|---|---|
| Integer coefficients, looks factorable | Factoring first |
| Any quadratic, need exact answer | Quadratic formula |
| Understanding vertex/minimum/maximum | Completing the square |
| Visual understanding or approximation | Graphing |
| b² − 4ac < 0 | Quadratic formula (gives complex roots) |
Quick Reference: Common Patterns
Difference of squares: x² − k² = (x + k)(x − k) = 0 → x = ±k
Perfect square trinomial: x² + 2kx + k² = (x + k)² = 0 → x = −k (repeated)
No middle term: ax² + c = 0 → x = ±√(−c/a) (real only if c and a have opposite signs)
Sum and Product of Roots
For ax² + bx + c = 0 with roots r₁ and r₂:
r₁ + r₂ = −b/a
r₁ × r₂ = c/a
Example verification: x² − 5x + 6 = 0, roots 2 and 3:
- Sum: 2 + 3 = 5 = −(−5)/1 ✓
- Product: 2 × 3 = 6 = 6/1 ✓
Use our cubic equation solver for degree-3 equations, or apply the quadratic formula above for any standard quadratic.