Quadratic Equation

Hi!
While solving this program I got a problem, that is-
ax^2 + bx + c = 0
and this will be solved by [-b +/- root(b^2 - 4ac)]/2a
now this equation will have two solutions, one is exat and another is imaginary. I'm having problem with imaginary portion, if thene is a negetive sign under the root i.r, root(b^2 - 4ac) the program will get error.
What to do?

 

Will you please give your program code, so that we can modify it for you?

 

you can do one thing, in the root(b^2 - 4ac) section you will get two thing, one is real and another is imaginary, while calculating the root means

Code:

d=b*b-4*a*c;
x=sqrt(d);

after this code add a If-Else i.e,

Code:

if(d>0)
x=sqrt(d);
else
{
d2=-1*d;
x=sqrt(d2);
}

Now while printing the value just add a "i" after the "%d" of imaginary part. i.e,

Code:

printf("%d i",&something);

Hope this will solve your problem!

 

Thanks Pankaj!
after coding again with your suggestion I'll confirm or inform whatever!

 

Here is the code of the complete solution of the problem.

Code:

#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float a,b,c,d,x1,x2,y,z;
clrscr();
printf("Enter the value of a: ");
scanf("%f",&a);
printf("Enter the value of b:");
scanf("%f",&b);
printf("Enter the value of c:");
scanf("%f",&c);
y=b*b-4*a*c;
if(y=>1)
{
d=sqrt(y);
x1=(-b+d)/2*a;
x2=(-b-d)/2*a;
printf("The solutions are\nX1= %f and\nX2= %f",x1,x2);
}
else
{
z=(-1)*y;
d=sqrt(z);
x1=(-b+d)/2*a;
x2=(-b-d)/2*a;
printf("The solutions are\nX1= %fi and\nX2= %fi",x1,x2);
}
getch();
}

thanks...