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#include(stdio.h) //note use '<' & '>' in place of '(' & ')'
#include(conio.h)
#include(math.h)
int main( )
{
int a , b, c, d;
float x1, x2;
ab:
printf("Enter the value of a, b, c:");
scanf("%d%d%d", &a, &b, &c);
if(a==0)
{
printf("\nIt is a Linear Equation");
prinft("\nRe-Enter the data again.");
goto ab;
}
d=(b*b)-(4*a*c);
if(d==0)
{
printf("\nRoots are real and equal");
x1= x2=(-b)/(2*a);
printf("\nRoots are: %f \t %f ", x1, x2);
}
else
{
if(d<0)
printf("\nRoots are imaginary");
else
{
x1=((-b)+sqrt(d))/(2*a);
x2=((-b)-sqrt(d))/(2*a);
printf("\nRoots are real");
printf("\nRoots are : %f \t %f", x1, x2);
}
}
getch( );
}
Output:-
Enter the values of a, b,c : 5
2
4
Roots are imaginary.
Wednesday, July 29, 2009
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ReplyDeleteCann't we find the imaginary roots also?
ReplyDeletewhat about the roots ??
ReplyDeletecant we get them?
yes.. we can find them using the formula
ReplyDeletex1=-b/(2.0*a);
x2=(sqrt(-d))/(2*a);
printf("The roots are imaginary");
printf("\n The roots are\n");
printf("\n%0.2f+i%0.2f\n",x1,x2);
printf("\n%0.2f+%0.2f\n",x1,x2);
printf("\n%0.2f+i%0.2f\n",x1,x2); //should be +ve
ReplyDeleteprintf("\n%0.2f-%0.2f\n",x1,x2); //Should be -ve
very nice
ReplyDelete