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📐 Quadratic Equation Solver

Solve ax^2 + bx + c = 0

x^2 -5x +6 = 0

a ≠ 0

Discriminant Δ = b2 - 4ac
1
Δ > 0, two distinct real roots
x1
3
x2
2
Vertex
(2.5, -0.25)
Axis of Symmetry
x = 2.5
Vieta's Formulas Verification
x1 + x2 = -b/a = 5
x1 · x2 = c/a = 6

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Quadratic Equation Solver Guide

The Quadratic Equation Solver quickly solves standard quadratic equations of the form ax^2 + bx + c = 0. Enter the three coefficients a (quadratic), b (linear), and c (constant), and the tool automatically computes the discriminant Δ=b^2-4ac, determines the nature of the roots based on the discriminant, and provides exact solutions.

This tool not only gives the two roots (real or complex), but also displays the discriminant value, vertex coordinates, axis of symmetry, and Vieta's formulas verification. This information is very helpful for understanding the geometric meaning of the equation (parabola properties). It's a great helper for middle school students learning quadratic functions and college students reviewing algebra fundamentals.

Core Formulas and Methods

【Quadratic Formula (Universal Formula)】
x = (-b ± sqrt(b^2-4ac)) / (2a)

【Discriminant】Δ = b^2 - 4ac
Δ > 0: Two distinct real roots
Δ = 0: One repeated real root
Δ < 0: Two conjugate complex roots

【Vieta's Formulas】
x1 + x2 = -b/a  (sum of roots)
x1 · x2 = c/a    (product of roots)

【Vertex Formula】Axis of symmetry x = -b/(2a), Vertex at (-b/(2a), f(-b/(2a)))

Practical Examples

📋
【Example 1】x^2 - 5x + 6 = 0 a=1, b=-5, c=6 Δ = (-5)^2 - 4×1×6 = 25-24 = 1 > 0 x1 = (5+1)/2 = 3, x2 = (5-1)/2 = 2 Verification: 3+2=5=-(-5)/1 ✓, 3×2=6=6/1 ✓ 【Example 2】x^2 + 2x + 1 = 0 (perfect square) a=1, b=2, c=1 Δ = 4-4 = 0 → repeated root x = -2/2 = -1 (double root) Factored form: (x+1)^2 = 0 【Example 3】x^2 + x + 1 = 0 (no real roots) a=1, b=1, c=1 Δ = 1-4 = -3 < 0 → complex roots x = (-1 ± sqrt(-3))/2 = -0.5 ± 0.866i

Important Notes

💡
- The quadratic coefficient a cannot be 0, otherwise it reduces to a linear equation (this tool handles this case automatically) - The discriminant determines the type of roots — this is the key to understanding the nature of solutions - Complex roots always appear as conjugate pairs (same real part, opposite imaginary parts) - Vieta's formulas allow you to understand root properties without computing actual roots - Three solving methods — factoring, completing the square, and the quadratic formula — each have their applicable scenarios

Application Scenarios

  • - Physics: Projectile trajectory equations, maximum height and range calculations
  • - Geometry optimization: Area maximization or perimeter minimization problems
  • - Engineering design: Bridge arch structures, parabolic antenna design parameters
  • - Economics: Profit maximization and break-even analysis
  • - Signal processing: Characteristic equations of second-order systems (e.g., RLC circuits)

Comparison of Three Solving Methods

There are three common methods for solving quadratic equations: (1) Factoring — suitable when factors are easily identifiable, such as x^2-5x+6=(x-2)(x-3)=0, fastest but relies on observation; (2) Completing the square — transforms the equation into (x+p)^2=q form, good for understanding geometric meaning; (3) Quadratic formula — universal and mechanically reliable, which is the method used by this tool.

Recommended strategy: Try factoring first (cross-multiplication), then use the formula if that doesn't work. For programming implementations, the quadratic formula is the most stable choice, but be aware of numerical stability issues when |b| is much larger than sqrt(Δ) (can be improved with rationalization).

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