What Is a Quadratic Equation?
A quadratic equation is a polynomial equation of degree 2, meaning the highest power of the variable is 2. The standard form is:
ax² + bx + c = 0
Where a, b, and c are constants and a ≠ 0. Every quadratic equation can have zero, one, or two real solutions (also called roots or zeros).
The Quadratic Formula
The quadratic formula gives you the solutions (x values) of any quadratic equation in standard form:
x = (−b ± √(b² − 4ac)) / (2a)
The ± symbol means there are typically two solutions: one using + and one using −.
Where Does the Formula Come From?
The quadratic formula is derived by completing the square on the general form ax² + bx + c = 0. Here are the key steps:
- Divide both sides by a: x² + (b/a)x + (c/a) = 0
- Move c/a to the right: x² + (b/a)x = −c/a
- Complete the square by adding (b/2a)² to both sides
- Factor the left side as a perfect square: (x + b/2a)² = (b² − 4ac) / 4a²
- Take the square root of both sides and solve for x
This derivation proves the formula works for any values of a, b, and c — making it one of the most powerful tools in algebra.
The Discriminant: Predicting the Number of Solutions
The expression under the square root sign — b² − 4ac — is called the discriminant (often written as Δ or D).
| Discriminant Value | Number of Solutions | Type of Solutions |
|---|---|---|
| Δ > 0 | 2 | Two distinct real solutions |
| Δ = 0 | 1 | One repeated real solution (double root) |
| Δ < 0 | 0 real | Two complex (imaginary) solutions |
Worked Examples
Example 1: Two Real Solutions (Δ > 0)
Solve: x² − 5x + 6 = 0
Here, a = 1, b = −5, c = 6
- Discriminant: (−5)² − 4(1)(6) = 25 − 24 = 1 (positive, so two solutions)
- x = (−(−5) ± √1) / (2 × 1)
- x = (5 ± 1) / 2
- x₁ = (5 + 1) / 2 = 6/2 = 3
- x₂ = (5 − 1) / 2 = 4/2 = 2
Solutions: x = 3 and x = 2
Verification: (3)² − 5(3) + 6 = 9 − 15 + 6 = 0 ✓ and (2)² − 5(2) + 6 = 4 − 10 + 6 = 0 ✓
Example 2: One Real Solution (Δ = 0)
Solve: x² − 6x + 9 = 0
Here, a = 1, b = −6, c = 9
- Discriminant: (−6)² − 4(1)(9) = 36 − 36 = 0
- x = (6 ± √0) / 2 = 6/2 = 3
Solution: x = 3 (double root)
This means x² − 6x + 9 = (x − 3)².
Example 3: No Real Solutions (Δ < 0)
Solve: x² + 2x + 5 = 0
Here, a = 1, b = 2, c = 5
- Discriminant: (2)² − 4(1)(5) = 4 − 20 = −16 (negative)
- This equation has no real solutions — only complex solutions: x = −1 ± 2i
Tips for Using the Quadratic Formula
- Always rearrange to standard form first. Make sure the equation equals zero before identifying a, b, and c.
- Watch the sign of b. If the equation is x² − 5x + 6 = 0, then b = −5 (not +5). The formula uses −b, which would be −(−5) = +5.
- Check your discriminant first. Computing the discriminant before the full formula tells you immediately how many real solutions to expect.
- Simplify the radical if possible. √12 can be simplified to 2√3, which often leads to cleaner solutions.