The Quadratic Formula — Explained
The quadratic formula x = (−b ± √(b²−4ac)) / 2a is one of the most important results in algebra. It solves every equation of the form ax² + bx + c = 0 — whether the roots are whole numbers, fractions, irrational numbers, or even complex numbers. Unlike factoring, the formula never fails.
This guide explains where the formula comes from, how to read the discriminant to predict the type of roots before solving, how to extract the vertex and axis of symmetry, and when to prefer factoring or completing the square instead. Eight fully worked examples progress from simple integers to complex conjugate roots.
The Quadratic Formula — What It Is
For any quadratic equation ax² + bx + c = 0 where a ≠ 0, the solutions are: x = (−b ± √(b² − 4ac)) / 2a.
The ± symbol means two solutions exist in general — one using the + sign and one using the − sign. These are called x₁ and x₂. The expression under the square root, D = b² − 4ac, is the discriminant and determines whether the roots are real or complex.
Three numbers fully define a quadratic: a (coefficient of x²), b (coefficient of x), and c (constant term). The formula works for any values of these coefficients, including fractions and decimals.
| Component | Symbol | Role in the Formula |
|---|---|---|
| Leading coefficient | a | Cannot be 0; if a=0 the equation is linear, not quadratic |
| Middle coefficient | b | Determines horizontal position of vertex: h = −b/2a |
| Constant term | c | The y-intercept of the parabola; f(0) = c |
| Discriminant | D = b²−4ac | Determines number and type of roots — the most informative single value |
| Vertex x-coordinate | h = −b/2a | Also the average of the two roots |
| Vertex y-coordinate | k = c − b²/4a | Maximum or minimum value of the parabola |
Derivation — Where Does the Formula Come From?
The quadratic formula is derived by completing the square on the general equation ax² + bx + c = 0. Here is the full derivation:
Step 1: Divide every term by a (valid since a ≠ 0): x² + (b/a)x + (c/a) = 0.
Step 2: Move the constant to the right side: x² + (b/a)x = −c/a.
Step 3: Complete the square. Add (b/2a)² to both sides: x² + (b/a)x + (b/2a)² = (b/2a)² − c/a.
Step 4: The left side is now a perfect square: (x + b/2a)² = b²/4a² − c/a = (b² − 4ac) / 4a².
Step 5: Take the square root of both sides: x + b/2a = ± √(b² − 4ac) / 2a.
Step 6: Subtract b/2a from both sides: x = −b/2a ± √(b² − 4ac) / 2a = (−b ± √(b²−4ac)) / 2a. This is the quadratic formula.
Understanding this derivation is valuable — it reveals that the formula is simply a formalized completing-the-square process, and it shows exactly why the vertex is at (−b/2a, k).
The Discriminant — Predicting Roots Before Solving
Before applying the full formula, compute D = b² − 4ac. The sign of D tells you everything about the nature of the roots:
D > 0: Two distinct real roots. The parabola crosses the x-axis at two points. If D is a perfect square, both roots are rational; if not, they are irrational surds.
D = 0: Exactly one real root (a repeated root). The parabola touches the x-axis at exactly one point — its vertex. The single root is x = −b/2a.
D < 0: No real roots — two complex conjugate roots of the form p ± qi. The parabola does not touch the x-axis at all. The roots are a ± bi where a = −b/2a and b = √|D| / 2|a|.
| Discriminant D | Root type | Number of x-intercepts | Parabola position |
|---|---|---|---|
| D > 0, perfect square | Two distinct rational roots | 2 | Crosses x-axis at two rational points |
| D > 0, not perfect square | Two distinct irrational roots | 2 | Crosses x-axis at two irrational points |
| D = 0 | One repeated real root | 1 (tangent) | Vertex touches x-axis exactly |
| D < 0 | Two complex conjugate roots | 0 | Entire parabola above (a>0) or below (a<0) x-axis |
8 Worked Examples — Integer to Complex Roots
Example 1 — Two rational roots: x² − 5x + 6 = 0. D = 25 − 24 = 1 > 0. x = (5 ± 1)/2 → x₁ = 3, x₂ = 2.
Example 2 — One repeated root: x² − 6x + 9 = 0. D = 36 − 36 = 0. x = 6/2 = 3 (double root).
Example 3 — Irrational roots: x² − 4x + 1 = 0. D = 16 − 4 = 12. x = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 ± √3 ≈ 3.732 or 0.268.
Example 4 — Complex roots: x² + 2x + 5 = 0. D = 4 − 20 = −16. x = (−2 ± √(−16))/2 = (−2 ± 4i)/2 = −1 ± 2i.
Example 5 — Non-monic (a ≠ 1): 2x² − 7x + 3 = 0. D = 49 − 24 = 25. x = (7 ± 5)/4 → x₁ = 3, x₂ = 1/2.
Example 6 — Negative leading coefficient: −3x² + 6x − 3 = 0. D = 36 − 36 = 0. x = −6/(−6) = 1 (double root).
Example 7 — Decimal coefficients: 0.5x² + 1.5x − 2 = 0. Multiply by 2: x² + 3x − 4 = 0. D = 9 + 16 = 25. x = (−3 ± 5)/2 → x₁ = 1, x₂ = −4.
Example 8 — Large discriminant: 3x² − 2x − 8 = 0. D = 4 + 96 = 100. x = (2 ± 10)/6 → x₁ = 2, x₂ = −4/3.
Vertex Form and the Quadratic Formula
Every quadratic ax² + bx + c can be rewritten in vertex form: a(x − h)² + k where (h, k) is the vertex.
The connection to the quadratic formula: h = −b/2a is exactly the midpoint of the two roots x₁ and x₂. This makes geometric sense — the axis of symmetry passes through the vertex, and the two roots are mirror images across it.
To find k: substitute h into the original equation. k = ah² + bh + c = c − b²/4a.
Vertex form is useful for graphing (the vertex is immediately visible), for optimization problems (finding maximum or minimum of a quadratic function), and for understanding the transformations applied to the basic parabola y = x².
When to Use the Quadratic Formula vs Factoring
The quadratic formula always works but is not always the fastest method. Use factoring when the roots are small integers or simple fractions — it's quicker. Use the formula when:
1. The discriminant is not a perfect square — factoring over the integers is impossible; you need the formula for exact irrational roots.
2. Coefficients are large, decimal, or fractional — guessing factors becomes impractical.
3. You need complex roots — factoring over the reals cannot produce complex numbers.
4. The problem requires vertex coordinates — the formula naturally gives h = −b/2a along the way.
Use completing the square when converting to vertex form is the primary goal, or when a=1 and b is even (makes the arithmetic cleaner).
| Method | Best when | Always works? | Gives vertex? |
|---|---|---|---|
| Factoring | Integer roots, simple trinomials | No (not all factor over ℤ) | No |
| Quadratic formula | Any quadratic, especially irrational/complex roots | Yes | Yes (h = −b/2a) |
| Completing the square | Finding vertex form, a=1 with even b | Yes | Yes (directly) |
| Graphing | Estimating roots visually | Approximate only | Yes |
Vieta's Formulas — Sum and Product of Roots
Without solving the equation, you can find the sum and product of the roots directly from the coefficients. These are known as Vieta's Formulas:
Sum of roots: x₁ + x₂ = −b/a.
Product of roots: x₁ × x₂ = c/a.
Example: x² − 5x + 6 = 0. Sum = 5/1 = 5 (check: 2+3=5 ✓). Product = 6/1 = 6 (check: 2×3=6 ✓).
Vieta's formulas are useful for checking your solutions, for constructing quadratics from given roots (if roots are r and s, the equation is x² − (r+s)x + rs = 0), and for solving problems where you know the sum or product of roots without needing the roots individually.
Common Mistakes When Applying the Quadratic Formula
Sign error on b: The formula uses −b, not b. If b = −5, then −b = +5. Forgetting this flip is the single most common error.
Forgetting the ± under the radical: Every discriminant D > 0 produces two roots. Missing one root is a serious error in problem-solving.
Dividing before computing the numerator: The entire numerator (−b ± √D) must be computed before dividing by 2a. Writing −b ± √D / 2a (without parentheses) is an order-of-operations error.
Computing b² incorrectly with negative b: If b = −3, then b² = (−3)² = 9, not −9. The discriminant is never negative because of b².
Abandoning the problem when D < 0: D < 0 means no real solutions, not "no solutions." The two complex conjugate roots x = −b/2a ± (√|D|/2a)i are valid answers.