Program in C to find root using bisection method?

Answer 1

#include

#include

#include

#define fn(x) (x*x*x-3*x*x-x+9)

void main()

{

clrscr();

float x,a,b,y,y1,y2;

cout<<"enter the initial value a&b"<

cin>>a>>b;

y1=fn(a);

y2=fn(b);

if(y1*y2>0)

{

cout<<"invalid interval"<

}

do

{

x=(a+b)/2;

y=fn(x);

if(y1*y<0)

{

b=x;

y2=y;

}

else

{

a=x;

y1=y;

}

}

while(fabs(b-a)>.001);

cout<<"the root of the equation ="<

getch();

}

Answer 2

1. # include
2. # include
3. # include
4. # include
5. # define EPS .00001
6. double F(double x);
7. double root(double fVal1,double fVal2);
8. void main()
9. {
10. double fVal1,fVal2;
11. clrscr();
12. printf("enter the values ");
13. scanf("%f %f",&fVal1,&fVal2);
14. printf("the roots of the euation is %1.9f\n",root(fVal1,fVal2));
15. getch();
16. }
17. double root(double fVal1,double fVal2)
18. {
19. int nCtr=0;
20. double fTest1,fTest2,fTest3,fVal,fNum;
21. fTest1=F(fVal1);
22. fTest2=F(fVal2);
23. if((fTest1*fTest2)>0)
24. {
25. printf("both are of equal sign");
26. exit(0);
27. }
28. else
29. {
30. do
31. {
32. nCtr++;
33. fVal1=(fVal1+fVal2)/2;
34. fTest3=F(fVal);
35. if(fTest1*fTest3>0)
36. {
37. fTest1=fTest3;
38. fVal1=fVal;
39. }
40. else
41. {
42. fTest2=fTest3;
43. fVal2=fVal;
44. }
45. fNum=fabs (fVal2-fVal1);
46. } while(fNum>EPS);
47. }
48. return fabs(fVal2)>fabs(fVal1)?fVal2:fVal1;
49. }
50. double F(double x)
51. {
52. return (x*x*x)+(3*x)-5;
53. }

Sep 3 '06 #4

Answer 3

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<iostream.h>

void main()

{

float a,b,c,x1,x2,disc;

clrscr();

printf("Enter the co-efficients\n");

scanf("%f%f%f",&a,&b,&c);

disc=b*b-4*a*c;

if(disc>0)

{

x1=(-b+sqrt(disc))/(2*a);

x2=(-b-sqrt(disc))/(2*a);

printf("The roots are distinct\n");

}

if(disc==0)

{

x1=x2=-b/(2*a);

printf("The roots are equal\n");

printf("x1=%f\nx2=%f\n",x1,x2);

}

else

x1=-b/(2*a);

x2=sqrt(fabs(disc))/(2*a);

printf("The roots are complex\n");

printf("The first root=%f+i%f\n",x1,x2);

printf("The second root=%f-i%f\n",x1,x2);

getch();

}

Answer 4

main()

{

float a,b,c,x1,x2,x,series;

double d;

printf("enter a,b,c and x1(pos) & x2(neg)");

scanf("%f%f%f%f%f", &a, &b, &c, &x1, &x2);

read:

x = (x1 + x2) / 2;

series = a * x * x + b * x + c;

d = fabs(series);

if (d > 0.0001)

{

if (x * x1 < 0)

x = x2;

else

x = x1;

goto read;

}

else

{

printf("ans=%f", x);

}

return 0;

}