Question

c编程：求一元二次方程的实根（考虑所有可能的情况）

RT，要求简单，初学者看得懂，无函数应用

Answer 1

#include"stdio.h" #include"math.h" void main { float a,b,c,disc,x1,x2,re,im;//实根，虚根 printf("请输入a,b,c的值:\n"); scanf("%f,%f,%f",&a,&b,&c); printf("the equation"); if(fabs(a)<=le-6)//判断a是否为0 printf("is not a quadratic\n"); else { disc=b*b-4*a*c; if(fabs(disc<=le-6) printf("has two equal roots:%8.4f\n,"-b/(2*a)); else if(disc>le-6) { x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); printf("has distinct real roots:%8.4f and %8.4f \n",x1,x2); } else { re=-b/(2*a); im=sqrt(-disc)/(2*a); printf("has complex roots:\n"; printf("%8.4f+%8.4fi\n",re,im); printf("%8.4f-%8.4fi\n",re,im); } } }

Answer 2

#include"stdio.h" #include"math.h" void main() { float a,b,c,d; printf("请输入a,b,c的值:\n"); scanf("%f%f%f",&a,&b,&c); d=b*b-4*a*c; if(d<0) { printf("NO ANSWER!!!\n"); } if(d==0) { printf("answer=%.2f\n",-b/2/a); } if(d>0) { printf("answer1=%.2f\n",(-b+sqrt(d))/2/a); printf("answer2=%.2f\n",(-b-sqrt(d))/2/a); } }