Java Program to Find the Roots of a Quadratic Equation – 查找一元二次方程根的Java程序

最后修改: 2022年 9月 6日

1. Overview


In this article, we’ll see how to compute the solutions of a quadratic equation in Java. We’ll start by defining what a quadratic equation is, and then we’ll compute its solutions whether we work in the real or the complex number system.


2. The Solutions of a Quadratic Equation


Given real numbers a ≠ 0, b and c, let’s consider the following quadratic equation: ax² + bx + c = 0.

给定实数a≠0,b和c,让我们考虑下面的二次方程。ax² + bx + c = 0

2.1. The Roots of a Polynomial


The solutions of this equation are also called the roots of the polynomial ax² + bx + c. Thus, let’s define a Polynom class. We’ll throw an IllegalArgumentException if the a coefficient is equal to 0:

这个方程的解也被称为多项式ax² + bx + c的根。因此,让我们定义一个Polynom类。如果a系数等于0,我们将抛出一个IllegalArgumentException

public class Polynom {

    private double a;
    private double b;
    private double c;

    public Polynom(double a, double b, double c) {
        if (a==0) {
            throw new IllegalArgumentException("a can not be equal to 0");
        this.a = a;
        this.b = b;
        this.c = c;

    // getters and setters

We’ll solve this equation in the real number system: for this, we’ll look for some Double solutions.


2.2. Complex Number System


We’ll also show how to solve this equation in the complex number system. There is no default representation of a complex number in Java, so we’ll create our own. Let’s give it a static method ofReal to easily convert real numbers. This will be helpful in the following steps:


public class Complex {

    private double realPart;
    private double imaginaryPart;

    public Complex(double realPart, double imaginaryPart) {
        this.realPart = realPart;
        this.imaginaryPart = imaginaryPart;

    public static Complex ofReal(double realPart) {
        return new Complex(realPart, 0);

    // getters and setters

3. Calculate the Discriminant


The quantity Δ = b² – 4ac is called the discriminant of the quadratic equation. To calculate b squared in java, we have two solutions:


  • multiply b by itself
  • use Math.pow to raise it to the power of 2

Let’s stick with the first method and add a getDiscriminant method to the Polynom class:


public double getDiscriminant() {
    return b*b - 4*a*c;

4. Get the Solutions


Depending on the value of the discriminant, we’re able to know how many solutions exist and compute them.


4.1. With a Strictly Positive Discriminant


If the discriminant is strictly positive, the equation has two real solutions, (-b – √Δ) / 2a and (-b + √Δ) / 2a:

如果判别式是严格的正数,那么该方程有两个实数解,即(-b – √Δ) / 2a和(-b + √Δ) / 2a:

Double solution1 = (-polynom.getB() - Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA());
Double solution2 = (-polynom.getB() + Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA());

If we work in the complex number system, we then just need to make the conversion:


Complex solution1 = Complex.ofReal((-polynom.getB() - Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA()));
Complex solution2 = Complex.ofReal((-polynom.getB() + Math.sqrt(polynom.getDiscriminant())) / (2 * polynom.getA()));

4.2. With a Discriminant Equal to Zero


If the discriminant is equal to zero, the equation has a unique real solution -b / 2a:

如果判别式等于零,该方程有一个唯一的实数解-b / 2a:

Double solution = (double) -polynom.getB() / (2 * polynom.getA());

Similarly, if we work in a complex number system, we’ll transform the solution in the following way:


Complex solution = Complex.ofReal(-polynom.getB() / (2 * polynom.getA()));

4.3. With a Strictly Negative Discriminant


If the discriminant is strictly negative, the equation has no solution in the real number system. However, it can be solved in the complex number system: the solutions are (-b – i√-Δ) / 2a and its conjugate (-b + i√-Δ) / 2a:

如果判别式是严格的负数,该方程在实数系统中没有解。然而,它可以在复数系统中得到解决:解是(-b – i√-Δ) / 2a及其共轭物(-b + i√-Δ) / 2a:

Complex solution1 = new Complex(-polynom.getB() / (2* polynom.getA()), -Math.sqrt(-polynom.getDiscriminant()) / 2* polynom.getA());
Complex solution2 = new Complex(-polynom.getB() / (2* polynom.getA()), Math.sqrt(-polynom.getDiscriminant()) / 2* polynom.getA());

4.4. Gather the Results


To sum up, let’s build a method that will fill in a List with the solutions of the equation when they exist. In the real number system, this method looks like this:


public static List<Double> getPolynomRoots(Polynom polynom) {
    List<Double> roots = new ArrayList<>();
    double discriminant = polynom.getDiscriminant();
    if (discriminant > 0) {
        roots.add((-polynom.getB() - Math.sqrt(discriminant)) / (2 * polynom.getA()));
        roots.add((-polynom.getB() + Math.sqrt(discriminant)) / (2 * polynom.getA()));
    } else if (discriminant == 0) {
        roots.add(-polynom.getB() / (2 * polynom.getA()));
    return roots;

If we work in a complex number system, we’ll rather write:


public static List<Complex> getPolynomRoots(Polynom polynom) {
    List<Complex> roots = new ArrayList<>();
    double discriminant = polynom.getDiscriminant();
    if (discriminant > 0) {
        roots.add(Complex.ofReal((-polynom.getB() - Math.sqrt(discriminant)) / (2 * polynom.getA())));
        roots.add(Complex.ofReal((-polynom.getB() + Math.sqrt(discriminant)) / (2 * polynom.getA())));
    } else if (discriminant == 0) {
        roots.add(Complex.ofReal(-polynom.getB() / (2 * polynom.getA())));
    } else {
        roots.add(new Complex(-polynom.getB() / (2* polynom.getA()), -Math.sqrt(-discriminant) / 2* polynom.getA()));
        roots.add(new Complex(-polynom.getB() / (2* polynom.getA()), Math.sqrt(-discriminant) / 2* polynom.getA()));
    return roots;

5. Conclusion


In this tutorial, we’ve seen how to solve a quadratic equation in Java, whether we work with real or complex numbers.


As always, the code can be found over on GitHub.