Trying to calc solution to cubic

Ok this is what ive done so far:

Code:

---

/*
**  Solve a cubic polynomial
**  By xxxxxxxxxxxxxxxx
*/

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

float a1,a2,a3,b,c,d,G,roots,S,x1,x2,x3, C, theta, P, Q;

int main(int argc, char* argv[])
{

       
        printf("Please enter the coefficient of x^2:");
                scanf("%f",&a1);
                printf("Please enter the coefficient of x:");
                scanf("%f",&a2);
                printf("Please enter the constant term:");
                scanf("%f",&a3);
               
    Q=((a1*a1-3*a2)/9);
        S = ((2*a1*a1*a1)-(9*a1*a2+(27*a3)))/54;
        theta=acos(S/sqrt(Q*Q*Q));
        P= pow ( (sqrt( (S*S)-(Q*Q*Q)+fabs(S) )), 1/3);

        C = ((Q*Q*Q)-(S*S));

        if (C<=0)
        {
                theta=acos(S/sqrt(Q*Q*Q));
                x1 = (((-2*sqrt(Q))*(cos(theta/3))) - a1)/3;
                x2 = ((-2*sqrt(Q))*(cos((theta+2*3.1412)))/3) - a1/3;
                x3 = ((-2*sqrt(Q))*(cos((theta+4*3.1412)))/3) - a1/3;
                printf("First root: %f.\n", x1);
                printf("Second root: %f.\n", x2);
                printf("Third root: %f.\n", x3);
        }
        else {
               
                        if(G>0) {
                               
                                x1 = -G*(P + (Q/P))-(a1/3);
                                printf("Only root: %f.\n", x1);
                       
                        } else {
                                x1 = G*(P + (Q/P))-(a1/3);
                       
                        }
                        printf("There is only one single root: %f.\n", x1);
          }

return(0);
}

---

However it doesnt work for all values? More importantly I get this also:

Please enter the coefficient of x^2:9
Please enter the coefficient of x:9
Please enter the constant term:9
There is only one single root: -1.#IND00.
Press any key to continue

I got that with 9,9,9. What does -1.#IND00 mean?

Any help would be greatly appreciated.

#INF are infinities, which basically correspond to overflow
#IND are indefinites, which basically correspond to underflow. They are essentially smaller than the smallest possible float, but still bigger than zero.
There is also
#NAN which is Not A Number.

The first obvious problem is 1/3 will be zero (by integer arithmetic rules), but I suspect you want 1.0/3.0.

Quote:

---

Originally Posted by Salem

#INF are infinities, which basically correspond to overflow
#IND are indefinites, which basically correspond to underflow. They are essentially smaller than the smallest possible float, but still bigger than zero.
There is also
#NAN which is Not A Number.

The first obvious problem is 1/3 will be zero (by integer arithmetic rules), but I suspect you want 1.0/3.0.

---

Thanks Salem, here is the updated code, but still doesn't work!

Code:

---

/*
**  Solve a cubic polynomial
**  By x
*/

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

float a1,a2,a3,b,c,d,G,roots,S,x1,x2,x3, C, theta, P, Q;

int main(int argc, char* argv[])
{

       
        printf("Please enter the coefficient of x^2:");
                scanf("%f",&a1);
                printf("Please enter the coefficient of x:");
                scanf("%f",&a2);
                printf("Please enter the constant term:");
                scanf("%f",&a3);
               
    Q=((a1*a1-3*a2)/9.0);
        S = ((2*a1*a1*a1)-(9*a1*a2+(27*a3)))/54.0;
        theta=acos(S/sqrt(Q*Q*Q));
        P= pow ( (sqrt( (S*S)-(Q*Q*Q)+fabs(S) )), 1/3);

        C = ((Q*Q*Q)-(S*S));

        if (C<=0)
        {
                theta=acos(S/sqrt(Q*Q*Q));
                x1 = (((-2*sqrt(Q))*(cos(theta/3))) - a1)/3.0;
                x2 = ((-2*sqrt(Q))*(cos((theta+2*3.1412)))/3.0) - a1/3.0;
                x3 = ((-2*sqrt(Q))*(cos((theta+4*3.1412)))/3.0) - a1/3.0;
                printf("First root: %f.\n", x1);
                printf("Second root: %f.\n", x2);
                printf("Third root: %f.\n", x3);
        }
        else {
               
                        if(G>0) {
                               
                                x1 = -G*(P + (Q/P))-(a1/3.0);
                                printf("Only root: %f.\n", x1);
                       
                        } else {
                                x1 = G*(P + (Q/P))-(a1/3.0);
                       
                        }
                        printf("There is only one single root: %f.\n", x1);
          }

return(0);
}

---

But I still get the same errors

Except the one which really needed to change is the same :rolleyes:

Sorry, you meant to change this line?

P= pow ( (sqrt( (S*S)-(Q*Q*Q)+fabs(S) )), 1.0/3.0);

I have changed it, still gets an error. Closes with no answer :(