C#程序求一元二次方程根
答案:1 悬赏:50
解决时间 2021-01-17 01:04
- 提问者网友：欲望失宠
- 2021-01-16 03:01
C#程序求一元二次方程根
最佳答案
- 二级知识专家网友：逐風
- 2021-01-16 04:33
public static void Main()
{
double a, b, c;
Console.Write("a=");
a = Convert.ToDouble(Console.ReadLine());
Console.Write("b=");
b = Convert.ToDouble(Console.ReadLine());
Console.Write("c=");
c = Convert.ToDouble(Console.ReadLine());
Console.WriteLine(a + "x^2+" + b + "x+" + c + "=0");
double d = b * b - 4 * a * c;
double x1, x2;
if (d > 0)
{
x1 = (-b - Math.Sqrt(d)) / 2.0 / a;
x2 = (-b + Math.Sqrt(d)) / 2.0 / a;
Console.WriteLine("方程有两个不同的实根\nx1={0}\nx2={1}", x1, x2);
}
else if (d == 0)
{
x1 = x2 = (-b) / 2.0 / a;
Console.WriteLine("方程有两个相同的实根\nx1={0}\nx2={1}", x1, x2);
}
else
{
double i = Math.Sqrt(-d) / 2.0/a;
x1 = x2 = -b / 2.0 / a;
Console.WriteLine("方程有两个复根\nx1={0}-{1}i\nx2={2}+{3}i", x1, i, x2, i);
}
}
{
double a, b, c;
Console.Write("a=");
a = Convert.ToDouble(Console.ReadLine());
Console.Write("b=");
b = Convert.ToDouble(Console.ReadLine());
Console.Write("c=");
c = Convert.ToDouble(Console.ReadLine());
Console.WriteLine(a + "x^2+" + b + "x+" + c + "=0");
double d = b * b - 4 * a * c;
double x1, x2;
if (d > 0)
{
x1 = (-b - Math.Sqrt(d)) / 2.0 / a;
x2 = (-b + Math.Sqrt(d)) / 2.0 / a;
Console.WriteLine("方程有两个不同的实根\nx1={0}\nx2={1}", x1, x2);
}
else if (d == 0)
{
x1 = x2 = (-b) / 2.0 / a;
Console.WriteLine("方程有两个相同的实根\nx1={0}\nx2={1}", x1, x2);
}
else
{
double i = Math.Sqrt(-d) / 2.0/a;
x1 = x2 = -b / 2.0 / a;
Console.WriteLine("方程有两个复根\nx1={0}-{1}i\nx2={2}+{3}i", x1, i, x2, i);
}
}
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