Question

Write a C program, to solve the quadratic equation a x² + b x + c = 0 of given coefficients a, b and c. When running the program, it prompts for the input of coefficients a,b,c as floating numbers. After inputting three floating numbers, it computes and prints out the solutions, then prompts for another round of input. Your program will quit when getting input 0,0,0. Your program should handle four situations: (1) a=0, not a quadratic equation; (2) b² - 4ac = 0, the equation has two equal roots; (3) b² - 4ac < 0, the equation has complex roots; (4) b² - 4ac > 0, the equation has distinct real roots. Your program should be robust for invalid inputs.

output:

Please enter the coefficients a,b,c:
1,2,1
The equation has two equal real roots:
-1.000000
Please enter the coefficients a,b,c:
1,2,2
The equation has two complex roots:
-1.000000+1.000000i -1.000000-1.000000i
Please enter the coefficients a,b,c:
2,6,1
The equation has two distinct real roots:
-0.177124 -2.822875
Please enter the coefficients a,b,c:
a,b,c
Invalid input
Please enter the coefficients a,b,c:
1,2
Invalid input
Please enter the coefficients a,b,c:
0,1,2
not a quadratic equation
Please enter the coefficients a,b,c:
0,0,0
Goodbye

Answer 1

code:

#include<stdio.h>
int main()
{
float a,b,c;
while(1)
{

printf("Please enter the coefficients a,b,c:");
int k=scanf("%f%f%f",&a,&b,&c);
fflush(stdin);
if(k!=3)
{
printf("Invalid input\n");
continue;
}
if(a==0&&b==0&&c==0)
{
printf("Goodbye\n");
break;
}
if(a==0)
{
printf("not a quadratic equation\n");
continue;
}
float x=b*b-4*a*c;
if(x==0)
{
printf("The equation has two equal real roots:\n");
printf("%f\n",-b/(2.0*a));
}
else if(x>0)
{
printf("The equation has two distinct real roots:\n");
printf("%f %f\n",(-b+sqrt(x))/(2.0*a),(-b-sqrt(x))/(2.0*a));
}
else if(x<0)
{
printf("The equation has two complex roots:\n");
printf("%f+%fi ",-b/(2.0*a),sqrt(-x)/(2.0*a));
printf("%f%fi\n",-b/(2.0*a),-sqrt(-x)/(2.0*a));
}
}
}