How to print the imaginary root of quadratic equation in c? - c

Question

I need to print the imaginary root in the quadratic equation.
but when I execute my code the result shows me that the imaginary root is 0.00000i.
even I use to <complex.h> also same.
Can everybody help me to check the code that I bold?
//C program to find the root of the quadratic equation
#include<stdio.h>
#include<math.h>
#include<complex.h>
int main()
{
double a, b, c, x, x1, x2, disc, xr, ximg1, ximg2;
printf("Please enter the value of quadratic equation, a: ");
scanf("%lf", &a);
printf("Please enter the value of quadratic equation, b: ");
scanf("%lf", &b);
printf("Please enter the value of quadratic equation, c: ");
scanf("%lf", &c);
if( a == 0 )
{
x = -(c/b);
printf("\nThis is not a quadratic equation.\n");
printf("x = %.3lf", x);
}
else{
disc = (b*b) - 4*a*c;
if( disc == 0 ){
x1 = -b / 2 * a;
x2 = -b / 2 * a;
printf("x1 = %lf, x2 = %lf", x1, x2);
}
else if(disc > 0){
x1 = ( -b + sqrt( disc ) ) / 2 * a;
x2 = ( -b - sqrt( disc ) ) / 2 * a;
printf("x1 = %.1lf, x2 = %.1lf", x1, x2);
}
else{
ximg1 = sqrt( disc ) / 2 * a;
ximg2 = - sqrt( disc ) / 2 * a;
xr = - b / ( 2 * a );
**printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi", crealf(xr), cimagf(ximg1), cimagf(ximg2));**
}
}
return 0;
}
The output are shown as below:
Please enter the value of quadratic equation, a: 4
Please enter the value of quadratic equation, b: 1
Please enter the value of quadratic equation, c: 3
xr = -0.125000, ximg1 = 0.000000i, ximg2 = 0.000000i
Process returned 0 (0x0) execution time : 3.914 s
Press any key to continue.

Answer 1

You should print the roots as complex numbers using the computed real and imaginary parts. No need for <complex.h> nor complex types:
double xr = - b / ( 2 * a );
double ximg1 = -sqrt( -disc ) / (2 * a);
double ximg2 = -ximg1;
printf("x1 = %lf%+lfi, x2 = %lf%+lfi\n", xr, ximg1, xr, ximg2);

Answer 2

As you have used complex header file you just need to use the csqrt() instead of sqrt(),
//C program to find the root of the quadratic equation
#include<stdio.h>
#include<math.h>
#include<complex.h>
int main()
{
double a, b, c, x, x1, x2, disc, xr, ximg1, ximg2;
printf("Please enter the value of quadratic equation, a: ");
scanf("%lf", &a);
printf("Please enter the value of quadratic equation, b: ");
scanf("%lf", &b);
printf("Please enter the value of quadratic equation, c: ");
scanf("%lf", &c);
if( a == 0 )
{
x = -(c/b);
printf("\nThis is not a quadratic equation.\n");
printf("x = %.3lf", x);
}
else{
disc = (b*b) - 4*a*c;
if( disc == 0 ){
x1 = -b / 2 * a;
x2 = -b / 2 * a;
printf("x1 = %lf, x2 = %lf", x1, x2);
}
else if(disc > 0){
x1 = ( -b + csqrt( disc ) ) / 2 * a;//changed the function here
x2 = ( -b - csqrt( disc ) ) / 2 * a;//changed the function here
printf("x1 = %.1lf, x2 = %.1lf", x1, x2);
}
else{
ximg1 = sqrt( disc ) / 2 * a;
ximg2 = - sqrt( disc ) / 2 * a;
xr = - b / ( 2 * a );
**printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi", crealf(xr), cimagf(ximg1), cimagf(ximg2));**
}
}
return 0;
}

Answer 3

Complex types not needed
When code reaches the below, disc < 0. Find the square root of the negation.
// ximg1 = sqrt( disc ) / 2 * a;
ximg1 = sqrt( -disc ) / 2 * a; // Use -disc
// ximg2 = - sqrt( disc ) / 2 * a;
ximg2 = -ximg1; // Simply negate ximg1
xr = - b / ( 2 * a );
printf("xr = %lf, ximg1 = %lfi, ximg2 = %lfi",
// crealf(xr), cimagf(ximg1), cimagf(ximg2));
xr, ximg1, ximg2);
Tip: use "%e", it is more informative.
printf("xr = %le, ximg1 = %lei, ximg2 = %lei",
xr, ximg1, ximg2);
Bug
Instead of / 2 * a;, in various places, certainly you want / (2 * a);.