Why do I keep getting zero and nan for these variables? - c++

Question

Here is my code, for n and m, I consistently get 0 for n and nan for m, and I do not know why. I have tried everything I could, but n always shows 0 whilst m is always nan.
#include <iostream>
#include <iomanip>
#include <math.h>
#include <complex>
using namespace std;
int main()
{
float a, b, c, root1, root2, n, m ;
cout << "Please input values for a, b and c to represent the variables in equation : ax^2 + bx + c" << endl; // Dont know how to raise 2 instead of using ^2
cin >> a >> b >> c;
const float d = (b * b) - (4 * a * c);
cout << "Determinant is equal to " << d << endl;
cout << setprecision(3); // to set number of decimal places
n = b / 2 * a;
m = sqrt(d) / 2 * a;
if (d > 0) {
root1 = (-b + sqrt(d)) / (2.0 * a); // positive root
root2 = (-b - sqrt(d)) / (2.0 * a); // negative root
cout << "Determinant is greater than zero, therefore the equation has two real roots, which are " << root1 << " and " << root2 << endl;
}
else if (d < 0) {
cout << "Determinant is less than zero, therefore there would be two imaginary roots: " << n << " + " << m << "i ";
cout << "and " << n << " - " << m << "i" << endl;
}
}

Answer 1

you have to put parenthesis.
n = b / (2 * a);
m = sqrt(d) / (2 * a);

Answer 2

If the d is negative, you should first convert it to a non-negative number before feed it into sqrt(). Remember C/C++ does not equal to math language.
My quick fix:
#include <complex>
#include <iomanip>
#include <iostream>
#include <math.h>
using namespace std;
int main() {
float a, b, c, root1, root2, n, m;
cout << "Please input values for a, b and c to represent the variables in equation : ax^2 + bx + c " << endl; // Dont know how to raise 2 instead of using ^2
cin >> a >> b >> c;
const float d = (b * b) - (4 * a * c);
cout << "Determinant is equal to " << d << endl;
cout << setprecision(3); // to set number of decimal places
n = b / (2 * a);
if (d > 0) {
m = sqrt(d) / (2 * a);
root1 = (-b + sqrt(d)) / (2.0 * a); // positive root
root2 = (-b - sqrt(d)) / (2.0 * a); // negative root
cout << "Determinant is greater than zero, therefore the equation has two "
"real roots, which are "
<< root1 << " and " << root2 << endl;
} else if (d < 0) {
m = sqrt(-d) / (2 * a);
cout << "Determinant is less than zero, therefore there would be two "
"imaginary roots: "
<< n << " + " << m << "*i ";
cout << "and " << n << " - " << m << "*i" << endl;
}
}
A reference to sqrt() function:
http://www.cplusplus.com/reference/cmath/sqrt/

Answer 3

You will need to make some changes to get the correct answers for all cases First, you need to include the case of zero determinant which produces two identical real roots. You will need to calculate m with -d in the third case, and you have to make sure to change its sign into positive sign if necessary. For more aesthetic results, you can include special cases to omit n if it equals 0, and m if it equals 1.
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
float a, b, c, root1, root2, n, m;
cout << "Please input values for a, b and c to represent the variables in equation : ax^2 + bx + c" << endl;
cin >> a >> b >> c;
const float d = b * b - 4 * a * c;
cout << "Determinant is equal to " << d << endl;
cout << setprecision(3); // to set number of decimal places
if (d > 0)
{
root1 = (-b + sqrt(d)) / 2 / a;
root2 = (-b - sqrt(d)) / 2 / a;
cout << "Determinant is greater than zero, therefore the equation has two real roots, which are " << root1 << " and " << root2 << endl;
}
else if (d == 0)
{
root1 = -b / 2 / a;
cout << "Determinant is zero, therefore the equation has two identical real roots, which both are " << root1 << endl;
}
else
{
n = -b / 2 / a;
m = sqrt(-d) / 2 / a;
if (m < 0)
m *= -1;
if (n == 0)
{
if (m == 1)
cout << "Determinant is less than zero, therefore there would be two imaginary roots: i and -i" << endl;
else
cout << "Determinant is less than zero, therefore there would be two imaginary roots: " << m << "i and -" << m << "i" << endl;
}
else
{
if (m == 1)
cout << "Determinant is less than zero, therefore there would be two imaginary roots: " << n << " + " << "i and " << n << " - " << "i" << endl;
else
cout << "Determinant is less than zero, therefore there would be two imaginary roots: " << n << " + " << m << "i and " << n << " - " << m << "i" << endl;
}
}
}