思路:
1.a=0,方程不是二次方程(是一元一次方程);
2.b2-4ac=0,有两个相等的实根;
3. b2-4ac>0,有两个不等的实更;
4. b2-4ac<0,有两个共轭复根(高中所学为无解,但现在应该以p+qi和p-qi的形式输 出复根,其中p=-b/2a,q=sqrt(b2-4ac)/2a)
Ps 共轭复根的两个实数解为:
X1 = -b/2a+i*sqrt(b2-4ac)/2a
X2 = -b/2a-i*sqrt(b2-4ac)/2a
扫描二维码关注公众号,回复:
6162461 查看本文章
![](/qrcode.jpg)
1 #include<stdio.h> 2 #include<math.h> 3 double a,b,c,p,q; 4 int main() 5 { 6 scanf("%lf %lf %lf",&a,&b,&c); 7 if((b*b-4*a*c)>0) 8 { 9 fun3(a,b,c); 10 } 11 else if((b*b-4*a*c)==0) 12 { 13 fun2(a,b,c); 14 } 15 else 16 { 17 fun1(a,b,c); 18 } 19 return 0; 20 } 21 22 int fun1(double a,double b,double c) //小于0 23 { 24 double m; 25 m=-(b*b-4*a*c); 26 p=-b/(2*a); 27 q=sqrt(m)/(2*a); 28 printf("x1=%.3lf+%.3lfi x2=%.3lf-%.3lfi\n",p,q,p,q); 29 return 0; 30 } 31 32 int fun2(double a,double b,double c) //等于0 33 { 34 p=-b/(2*a); 35 printf("x1=%.3lf x2=%.3lf\n",p,p); 36 return 0; 37 } 38 39 int fun3(double a,double b,double c) //大于0 40 { 41 p=-b/(2*a); 42 q=(sqrt(b*b-4*a*c))/(2*a); 43 printf("x1=%.3lf x2=%.3lf\n",p+q,p-q); 44 return 0; 45 }