C program sqrt not working

Question

I also entered include<math.h> but it still doesnt work. People are saying to enter -Im but im new to this where do I put -Im and how do I fix this.

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

int main()
{
    float a=0, b=0, c=0, root1=0, root2=0;

    printf("Enter the value of a,b and c to determine the roots\n");
    scanf("%f%f%f",&a,&b,&c);

    root1=(-b+sqrt(b*b-4*a*c))/(2*a);
    root1=(-b-sqrt(b*b-4*a*c))/(2*a);

    printf("The first roots of the quadratic equation are\nFirst root=%.1f\nSecond root=%.1f",root1,root2);


    return 0;
}

It would help if you told us how it "doesn't work". Are you getting an error? What error? — Jack Aidley, Oct 24 '13 at 15:50
You should assign the second root to `root2`, not overwrite `root1`. You should check the input succeeded. You should probably check that `b*b` is not smaller than `4*a*c`. — Jonathan Leffler, Oct 24 '13 at 15:52
You can try compiling your program like this "gcc prog.c -o prog -lm" — Charlie Burns, Oct 24 '13 at 15:52
What input are you giving it? What output do you get? What output did you expect? — abelenky, Oct 24 '13 at 15:56

Sebastien · Answer 1 · 2013-10-24T16:08:44.767

Two things: first you copy pasted "root1" twice so you will lose the "plus" value and root2 will be zero. Second, for the benefits of others, the problem is most probably at compile time and the googled answer is there:

http://www.cs.cf.ac.uk/Dave/C/node17.html

And you should test for imaginary values:

    if(b*b-4*a*c < 0){
      printf("error: complex solution unsupported, see http://en.wikipedia.org/wiki/Square_root\n");
      exit(1);
    }

score 0 · Answer 2 · answered Oct 24 '13 at 16:00

You have a copy-paste bug here:

root1=(-b+sqrt(b*b-4*a*c))/(2*a);
root1=(-b-sqrt(b*b-4*a*c))/(2*a);

should be:

root1=(-b+sqrt(b*b-4*a*c))/(2*a);
root2=(-b-sqrt(b*b-4*a*c))/(2*a);

Also you may need to link with the math library, e.g.

$ gcc -Wall foo.c -o foo -lm

C program sqrt not working

2 Answers2