寻找二次方程根的 Java 程序
函数的根是 x 截距。根据定义，位于 x 轴上的点的 y 坐标是 零 。因此，要求一个二次函数的根，我们设 f (x) = 0，求解方程，ax 2 + bx + c = 0。
二次方程的条件–
ax^2 + bx + c = 0
where
a, b, c are real numbers and cannot be zero ie, there value must be from {-∞ to -1} and {1 to ∞}
求二次方程根的数学公式–
roots = (-b ± √(b2-4ac)) / (2a)
± represents there are two roots.
二次方程的根是–
first = (-b + √(b2-4ac)) / (2a)
second = (-b - √(b2-4ac)) / (2a)
行列式的(b^2–4ac)告诉我们根的性质–
- If (b 2–4ac) > 0 , the roots are real and different.
- If (b 2–4ac) = = 0 , the roots are real and equal.
- If (b 2–4ac) < 0 , the roots are complex and different.
代码求一个二次方程的根:
Java
// Java program to find the roots of
// quadratic equation
public class Main {
public static void main(String[] args)
{
// value of the constants a, b, c
double a = 7.2, b = 5, c = 9;
// declared the two roots
double firstroot, secondroot;
// determinant (b^2 - 4ac)
double det = b * b - 4 * a * c;
// check if determinant is greater than 0
if (det > 0) {
// two real and distinct roots
firstroot = (-b + Math.sqrt(det)) / (2 * a);
secondroot = (-b - Math.sqrt(det)) / (2 * a);
System.out.format(
"First Root = %.2f and Second Root = %.2f",
firstroot, secondroot);
}
// check if determinant is equal to 0
else if (det == 0) {
// two real and equal roots
// determinant is equal to 0
// so -b + 0 == -b
firstroot = secondroot = -b / (2 * a);
System.out.format(
"First Root = Second Root = %.2f;",
firstroot);
}
// if determinant is less than zero
else {
// roots are complex number and distinct
double real = -b / (2 * a);
double imaginary = Math.sqrt(-det) / (2 * a);
System.out.printf("First Root = %.2f+%.2fi",
real, imaginary);
System.out.printf("\nSecond Root = %.2f-%.2fi",
real, imaginary);
}
}
}
输出
First Root = -0.35+1.06i
Second Root = -0.35-1.06i
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