求方程ax2+bx+c=0的根,用3個函數分別求當b2-4ac大于0,等于0,和小于0時的根并輸出結果。從主函數a,b,c牲/span>

問題補充說明:用c語言編程,急啊

求方程ax2+bx+c=0的根,用3個函數分別求當b2-4ac大于0,等于0,和小于0時的根并輸出結果。從主函數a,b,c牲/span>


#include<cstdio>
#include<c360問答math>
//b^2-4ac==0
voidfun1(double&a,double&b,double&c,double&d){
doubleans=-b/(2*a);
printf("b^2-4ac==0,x1=x2=%lf.\n",ans)越勢醫倒濃位越者;
}
//b^2-4ac>0
voidfun2(double&a,double&b,double&c,double&d){
doubleans1,ans2;
ans1=(-b幫隊推龍+sqrt(d))/(2*a);
ans2=(-b-sqrt(d))/(2*a);
pri目苦期啊按么器ntf("b^2-4ac>0審盟科具在散李,x1=%lf,x2=%lf.\n",ans1,ans2);
}
//b^2-4ac<0
voidfun3(double&a,double&b,double&c,double&d){
doublereal,imar;
real=-b/(2*a);
imar=sqrt(-d)/(2*a);
printf("損b^2-4ac<0,x1=%lf+%lfi,x2=%lf-%lfi.\n",real,imar,real,imar);
}
intmain(){
doublea,b,c,d;
printf("pleaseinputa,b,c.\n");
while(scanf(混終新放討紀趕"%lf%lf%lf",&a,&b,&c)!=EOF){
d=b*b-4*a*c;
if流委展案(d==0)fun1(a,b,c,d);
elseif(d>0)fun2(a,b,c,d);
elsefun3(a,b,c,d);
printf("p至晚量列限土環亞光永治leaseinputa,b,c.\n");
}
}
【求方程ax2+bx+c=0的根,用3個函數分別求當b2-4ac大于0,等于0,和小于0時的根并輸出結果。從主函數a,b,c牲/span>】

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