自定义函数求一元二次方程

解题思路:

注意事项:

参考代码:

#include<iostream>

#include<math.h>

#include<iomanip>

using namespace std;

void f1(double a, double b, double c)

{

    int x1 = 0, x2 = 0;

    x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);

    x2 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a);

    cout << "x1=" << setprecision(3) << x1  <<" "<< "x2=" << setprecision(3) << x2 << endl;

}

void f2(double a, double b, double c)

{

    int x1 = 0, x2 = 0;

    cout << "x1=" << setprecision(3) << -b/(2*a)<<"+"<<setprecision(3)<<sqrt(-(b * b - 4 * a * c))/(2*a) << "i "<< "x2=" << setprecision(3) << -b / (2 *     a) << "-" <<setprecision(3)<< sqrt(-(b * b - 4 * a * c)) / (2 * a) << "i " << endl;

}

int main()

{

    double a, b, c;

    cin >> a >> b >> c;

    double t = b * b - 4 * a * c;

    if (t >= 0)

    f1(a, b, c);

    else

    f2(a, b, c);

    return 0;

}