Write a program to find all roots of a quadratic equation.
To understand this program, you need to have understanding of if else
statement and
math.sqrt
method.
Approach: First of all, you need to know what is quadratic equation. It is an equation which when arranged in equation form returns the result as zero as shown below.
ax2 + bx + c = 0
we can calculate the root of a quadratic equation by using the formula
x = (-b ± √(b2-4ac)) / (2a)
The sign ±
indicates that there will be two roots of a number.
root1 = (-b + √(b2-4ac)) / (2a) root1 = (-b - √(b2-4ac)) / (2a)
The equation b2-4ac
is called determinant of a quadratic equation. It denotes the nature of root.
if determinant > 0, roots are real and different if determinant == 0, roots are real and equal if determinant < 0, roots are complex complex and different
Here we go with the program
public class RootQuadratic { public static void main(String[] args) { double a = 2.3, b = 4, c = 5.6; double root1, root2; double determinant = b * b - 4 * a * c; // condition check for real and different roots if(determinant > 0) { root1 = (-b + Math.sqrt(determinant)) / (2 * a); root2 = (-b - Math.sqrt(determinant)) / (2 * a); System.out.format("root1 = %.2f and root2 = %.2f", root1 , root2); } // Condition check for real and equal roots else if(determinant == 0) { root1 = root2 = -b / (2 * a); System.out.format("root1 = root2 = %.2f;", root1); } // If roots are not real else { double realPart = -b / (2 *a); double imaginaryPart = Math.sqrt(-determinant) / (2 * a); System.out.format("root1 = %.2f+%.2fi and root2 = %.2f-%.2fi", realPart, imaginaryPart, realPart, imaginaryPart); } } }
Output
root1 = -0.87+1.30i and root2 = -0.87-1.30i