Question

C语言 x^3+ax^2+bx+c=0求解

Answer 1

用弦截法求方程x^3+ax^2+bx+c=0的根：(其中要求用户输入的x1和x2是用户估计的根的区间)
#include
<stdio.h>
#include
<math.h>
float
a,b,c;
/*求f(x)=x^3+ax^2+bx+c=0的函数值。*/
float
f(float
x)
{
float
y;
y=x*x*x+a*x*x+b*x+c;
return(y);
}
/*求(x1,f(x1))和(x2,f(x2))的连线与x轴的交点横坐标：x＝[x1×f(x2)－x2×f(x1)]
/
[f(x2)－f(x1)]*/
float
xpoint(float
x1,float
x2)
{
float
x;
x=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));
return(x);
}
/*求(x1,x2)区间内方程的实根*/
float
root(float
x1,float
x2)
{
float
x;
do
{
x=xpoint(x1,x2);
if(f(x)*f(x1)>0)
x1=x;
else
x2=x;
}
while(fabs(f(x))>=1e-4);
return(x);
}
main()
{
float
x1,x2,x;
printf("Please
input
a:
");
scanf("%f",&a);
printf("Please
input
b:
");
scanf("%f",&b);
printf("Please
input
c:
");
scanf("%f",&c);
do
{
printf("Please
input
x1
and
x2:
");
scanf("%f%f",&x1,&x2);
}while(f(x1)*f(x2)>=0);
x=root(x1,x2);
printf("A
root
of
equation
is
%.4f\n",x);
}
例如，a,b,c分别输入-5
16
-80，x1和x2分别输入1
10，结果为5

Answer 2

由条件有：A+B+C=1
16A-4B=45
Y'=3X^2+2AX+B
无极值
则
导数的判别式
4A^2-12B<0
则
4（A-6)^2<9
即
3/2<A<15/2