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# Java Program to Find the Roots of a Quadratic Equation

Last modified: October 6, 2022

## 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*.

### 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:

```
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:**

```
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:**

`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:**

```
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.