Quadratic Equation Solver
How do you solve a quadratic equation?
With the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. The discriminant Δ = b² - 4ac decides the outcome: Δ>0 two real solutions, Δ=0 one repeated solution, Δ<0 no real solution.
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Quadratic Equation Solver
Solve ax² + bx + c = 0 with formula and steps
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Quadratic Equations: Theory and Solution
A quadratic equation (or second-degree equation) is an algebraic equation in which the highest degree of the unknown is 2. It is one of the pillars of algebra and appears in countless applications: from physics to trajectory calculations, from economics to engineering.
Standard Form
Every quadratic equation can be written in standard form:
ax² + bx + c = 0
where a, b, and c are real coefficients and a ≠ 0 (if a were zero, the equation would be linear).
The Discriminant
The key to solving a quadratic equation is the discriminant (denoted by the Greek letter delta):
Δ = b² – 4ac
The value of the discriminant determines the number and nature of the solutions:
| Value of Δ | Real Solutions | Description |
|---|---|---|
| Δ > 0 | 2 distinct | The parabola intersects the x-axis at two points |
| Δ = 0 | 1 (repeated) | The parabola is tangent to the x-axis |
| Δ < 0 | 0 (2 complex) | The parabola does not intersect the x-axis |
The Quadratic Formula
The solutions (roots) of the equation are obtained using the quadratic formula:
x = (–b ± √Δ) / (2a)
which, expanding the discriminant, becomes:
x = (–b ± √(b² – 4ac)) / (2a)
Examples for Each Case
Case 1: Δ > 0 — Two Distinct Real Solutions
Solve x² – 5x + 6 = 0 (a=1, b=–5, c=6)
- Δ = (–5)² – 4(1)(6) = 25 – 24 = 1
- x₁ = (5 + 1) / 2 = 3
- x₂ = (5 – 1) / 2 = 2
Case 2: Δ = 0 — Repeated Solution
Solve x² – 6x + 9 = 0 (a=1, b=–6, c=9)
- Δ = (–6)² – 4(1)(9) = 36 – 36 = 0
- x = 6 / 2 = 3 (repeated solution)
Case 3: Δ < 0 — No Real Solutions
Solve x² + 2x + 5 = 0 (a=1, b=2, c=5)
- Δ = (2)² – 4(1)(5) = 4 – 20 = –16
- No real solutions exist
- Complex solutions: x = (–2 ± 4i) / 2 → x₁ = –1 + 2i, x₂ = –1 – 2i
Vertex of the Parabola
The equation ax² + bx + c = 0 is associated with the parabola y = ax² + bx + c. The vertex of the parabola has coordinates:
V = (–b/(2a), –Δ/(4a))
The vertex represents the minimum point (if a > 0) or the maximum point (if a < 0) of the parabola. For the equation x² – 5x + 6 = 0:
- xᵥ = 5 / 2 = 2.5
- yᵥ = –1 / 4 = –0.25
- Vertex: V(2.5, –0.25)
Relationships Between Roots and Coefficients (Vieta's Formulas)
If x₁ and x₂ are the roots of the equation, the following relationships hold:
- x₁ + x₂ = –b/a (sum of roots)
- x₁ × x₂ = c/a (product of roots)
These formulas allow you to quickly verify the solutions found and are useful for constructing equations from desired roots.
Special Cases
| Type | Condition | Quick Method |
|---|---|---|
| Pure | b = 0 → ax² + c = 0 | x = ±√(–c/a) |
| Incomplete | c = 0 → ax² + bx = 0 | x(ax + b) = 0 → x=0, x=–b/a |
| Trivial | b = 0 and c = 0 | x = 0 (repeated) |
How to Use the Calculator
Enter the coefficients a, b, and c of the equation in standard form. The solver will calculate the discriminant, the solutions (real or complex), the factored form (if Δ ≥ 0), the coordinates of the parabola's vertex, the axis of symmetry, and the concavity. Decimal and negative values are accepted for the coefficients.