Question

求方程ax^2+bx+c=0的根,用3个函数分别求当b^2-4ac大于0,等于0,和小于0时的亘并输出,两个数由键盘输入

Answer 1

#include<stdio.h>
#include<math.h>

void main()
{
void big(float a,float b,float c);
void equality(float a,float b,float c);
void small(float a,float b,float c);
float a,b,c,d;
printf("请输入三个数用，隔开：");
scanf("%f,%f,%f",&a,&b,&c);
d=b*b-4*a*c;
if(d>0) big(a,b,c);
if(d==0) equality(a,b,c);
if(d<0) small(a,b,c);
}
void big(float a,float b,float c)
{
float x1,x2;
x1=(-b+sqrt(b*b-4*a*c))/(2*a);
x2=(-b-sqrt(b*b-4*a*c))/(2*a);
printf("方程的根：%f或%f\n",x1,x2);
}
void equality(float a,float b,float c)
{
float x;
x=-b/(2*a);
printf("方程的根：%f\n",x);
}
void small(float a,float b,float c)
{
printf("无解\n");
}

Answer 2

10 input "输入方程的系数a,b,c=o";a,b,c
if a=0 then "一元二次方程的系数a不能为0”：goto 10
y=b*b-4*a*c
if y>=0 then
PRINT "x1=";(-b+sqr(y))/(2*a),"x2=";(-b-sqr(y))/(2*a)
else
PRINT "此一元二次方程无实数解"
end