Compute roots of quadratic with "if" statments

[noble3ad]

Hello all,

I'm trying to write a program that will compute the real roots of the quadratic equation ax^2 + bx + c = 0 given by:

x1 = (- b + sqrt (b^2 - 4ac)) / 2a and x2 = (- b - sqrt (b^2 - 4ac)) / 2a.

Based on the values of "a", "b", and "c" entered by a user, the roots will be calculated based on the following set of rules:

1. if a and b are zero, there are no solutions.
2. if a is zero, then there in one real root (-c / b).
3. if the discriminate (sqrt(b^2 - 4ac)) is negative, then there are no real roots.
4. for all other combinations, there are two real roots.

I figured I could accomplish this with a series of "if" statements....no else statements....

The code I came up with is as follows:

int main()
{
	float a;
	float b;
	float c;
	float x1;
	float x2;
	float one_root;
	
	printf("Please enter value for a:");
	scanf("%f", &a);
	
	printf("Please enter value for b:");
	scanf("%f", &b);

	printf("Please enter value for c:");
	scanf("%f", &c);
	
	if (a == 0 && b ==0)
		{
			printf("There are no solutions.");
		}
	if (a == 0)
		{
			one root = -c / b;
			printf("There is one root of %f.);
		}
	if (sqrt(pow (b ,2)) - (4 * a * c)) < 0)
		{
			printf("There are no real roots.");
		}
	if (a < 0 && a > 0 && b < 0 && b > 0)
		{
			x1 = (-b + sqrt ((pow (b, 2)) - (4 * a * c))) / (2 * a);
			x2 = (-b + sqrt ((pow (b, 2)) - (4 * a * c))) / (2 * a);
			printf("The two real roots are %f and %f.", x1, x2);
		}

	return 0;
}

I haven't had a chance to try it to see if it works because I can't get to a computer with software that will let me do it right now. What I really want to know right now is that it's ok to write the code just using "if" statements and leaving out the "else".

Thanks

[hammerhead]

Use else if statement. According to your code, if a=0 and b=0 then both the first and the second if statements will be executed. And the output will be
"There are no solutions"
"There is one solution x"

use this

if (a == 0 && b ==0)
		{
			printf("There are no solutions.");
		}
       else if (a == 0)
		{
			one root = -c / b;
			printf("There is one root of %f.);
		}
	else if (sqrt(pow (b ,2)) - (4 * a * c)) < 0)
		{
			printf("There are no real roots.");
		}
	else if (a < 0 && a > 0 && b < 0 && b > 0)
		{
			x1 = (-b + sqrt ((pow (b, 2)) - (4 * a * c))) / (2 * a);
			x2 = (-b + sqrt ((pow (b, 2)) - (4 * a * c))) / (2 * a);
			printf("The two real roots are %f and %f.", x1, x2);
		}

[noble3ad]

Thanks, hammer! Seems to be working.