Solving Quadratic Equations in Julia- A Step-by-Step Tutorial

What This Article Covers

Quadratic equations look like ax² + bx + c = 0. Julia handles them fast. This tutorial shows you exactly how to solve them with code you can copy and run.

If you want the quick version: use the quadratic formula directly or use Julia's built-in polynomial functions. Both methods work. I'll show you both.

Quick Refresher: The Quadratic Formula

You probably learned this in school. The solutions are:

x = (-b ± √(b² - 4ac)) / 2a

The part under the square root is the discriminant: b² - 4ac. It tells you what kind of solutions you'll get:

• Positive discriminant: two real solutions
• Zero discriminant: one real solution
• Negative discriminant: two complex solutions

Setting Up Julia

I'm assuming you have Julia installed. If not, download it from julialang.org. The installation takes about 5 minutes.

Start the REPL and you're ready to code. No extra packages needed for the basics—the standard library handles complex numbers natively.

Method 1: Direct Implementation

Here's the most straightforward approach. No tricks, no shortcuts—just the math written as code.

Basic Function

function solve_quadratic(a, b, c)
    discriminant = b^2 - 4*a*c
    
    if discriminant >= 0
        x1 = (-b + sqrt(discriminant)) / (2*a)
        x2 = (-b - sqrt(discriminant)) / (2*a)
        return (x1, x2)
    else
        # Handle complex roots
        real_part = -b / (2*a)
        imag_part = sqrt(-discriminant) / (2*a)
        x1 = complex(real_part, imag_part)
        x2 = complex(real_part, -imag_part)
        return (x1, x2)
    end
end

Testing It

# Example: 2x² + 5x - 3 = 0
solutions = solve_quadratic(2, 5, -3)
println("x₁ = $(solutions[1]), x₂ = $(solutions[2])")

This gives you x₁ = 0.5 and x₂ = -3.0. Verify: 2(0.5)² + 5(0.5) - 3 = 0 ✓

Method 2: Using Polynomials Package

Julia has a Polynomials package that handles this for you. Install it once, use it forever.

using Pkg
Pkg.add("Polynomials")

using Polynomials

# Create polynomial: 2x² + 5x - 3
p = Polynomial([2, 5, -3])

# Get roots directly
roots(p)

This returns the same answers. The package is useful when you're working with higher-degree polynomials too.

Method 3: Using the Roots Package

The Roots package specializes in finding zeros. It's overkill for simple quadratics but handles messy equations well.

Pkg.add("Roots")

using Roots

f(x) = 2x^2 + 5x - 3

# Find all roots
find_zeros(f, -10, 10)

This searches for roots in the interval [-10, 10]. It's slower than the direct formula but works when you don't know where the roots are.

Comparing the Methods

Common Mistakes to Avoid

• Dividing by zero: If a = 0, you don't have a quadratic equation. Add a check for it.
• Integer overflow: For huge coefficients, use Float64 explicitly: float(a)
• Precision issues: For high-precision work, use BigFloat: big(2)^2

Robust Version with Error Handling

function solve_quadratic_safe(a, b, c)
    if a == 0
        error("Not a quadratic equation (a = 0)")
    end
    
    discriminant = b^2 - 4*a*c
    denom = 2*a
    
    if discriminant == 0
        return (-b / denom, nothing)  # One solution
    elseif discriminant > 0
        return ((-b + sqrt(discriminant)) / denom, 
                (-b - sqrt(discriminant)) / denom)
    else
        real_part = -b / denom
        imag_part = sqrt(-discriminant) / denom
        return (complex(real_part, imag_part),
                complex(real_part, -imag_part))
    end
end

Real-World Example

Say you're calculating projectile motion. An object follows -4.9t² + 20t + 5 = 0. When does it hit the ground?

solutions = solve_quadratic(-4.9, 20, 5)
ground_time = filter(x -> x > 0 && imag(x) == 0, solutions)
println("Object hits ground at t = $(ground_time[1]) seconds")

Result: approximately 4.54 seconds. The negative solution (-0.22) gets discarded since negative time is meaningless.

Quick Start Checklist

• Open Julia REPL
• Copy the solve_quadratic function above
• Test with: solve_quadratic(1, -5, 6) → expect x = 2 and x = 3
• Verify manually: 1(2)² - 5(2) + 6 = 0 ✓

Bottom Line

For quadratic equations in Julia, the direct formula is your best bet. It's fast, accurate, and doesn't require dependencies. Install the Polynomials package only if you're also dealing with cubic or quartic equations.