Difference between revisions of "Directory:Derek Elder/Programs/Quadratic Formula"
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| Line 11: | Line 11: | ||
int main() | int main() | ||
{ | { | ||
| − | |||
int A = 0; | int A = 0; | ||
int B = 0; | int B = 0; | ||
| Line 24: | Line 23: | ||
double discrim = sqrt((B * B) - 4 * A * C); | double discrim = sqrt((B * B) - 4 * A * C); | ||
| − | double imaginary_discrim = sqrt(-((B * B) - 4 * A * C)); | + | double imaginary_discrim = sqrt(-((B * B) - (4 * A * C))); |
| − | if (discrim >= 0) | + | if(discrim >= 0) |
{ | { | ||
cout<<"The value of the discriminant is: "<<discrim<<"\n"; | cout<<"The value of the discriminant is: "<<discrim<<"\n"; | ||
| Line 37: | Line 36: | ||
if(A != 0) | if(A != 0) | ||
{ | { | ||
| − | if(A <0) | + | if(A < 0) |
{ | { | ||
A *= -1; | A *= -1; | ||
| Line 45: | Line 44: | ||
{ | { | ||
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two real solutions are:"<<"\n"; | cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two real solutions are:"<<"\n"; | ||
| − | cout<<"X = "<<(-B + discrim)/(2.0 * A)<<"\n"; | + | cout<<"X = "<<((-B + discrim)/(2.0 * A))<<"\n"; |
| − | cout<<"X = "<<(-B - discrim)/(2.0 * A)<<"\n"; | + | cout<<"X = "<<((-B - discrim)/(2.0 * A))<<"\n"; |
} | } | ||
else if (discrim == 0) | else if (discrim == 0) | ||
{ | { | ||
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the one real solution is:"<<"\n"; | cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the one real solution is:"<<"\n"; | ||
| − | cout<<"X = "<<-B /(2.0 * A)<<"\n"; | + | cout<<"X = "<<(-B/(2.0 * A))<<"\n"; |
} | } | ||
else | else | ||
{ | { | ||
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two imaginary solutions are:"<<"\n"; | cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two imaginary solutions are:"<<"\n"; | ||
| − | cout<<"X = "<<-B / (2.0 * A)<<" + "<<imaginary_discrim /(2 * A)<<"i"<<"\n"; | + | cout<<"X = "<<(-B/(2.0 * A))<<" + "<<(imaginary_discrim/(2 * A))<<"i"<<"\n"; |
| − | cout<<"X = "<<-B / (2.0 * A)<<" - "<<imaginary_discrim /(2 * A)<<"i"<<"\n"; | + | cout<<"X = "<<(-B/(2.0 * A))<<" - "<<(imaginary_discrim/(2 * A))<<"i"<<"\n"; |
} | } | ||
Latest revision as of 17:32, 6 May 2008
#include <iostream>
#include <cmath>
using namespace std;
double sqrt(int discrim)
{
return sqrt(double(discrim));
}
int main()
{
int A = 0;
int B = 0;
int C = 0;
cout<<"Please enter a value for A: ";
cin>>A;
cout<<"Please enter a value for B: ";
cin>>B;
cout<<"Please enter a value for C: ";
cin>>C;
double discrim = sqrt((B * B) - 4 * A * C);
double imaginary_discrim = sqrt(-((B * B) - (4 * A * C)));
if(discrim >= 0)
{
cout<<"The value of the discriminant is: "<<discrim<<"\n";
}
else
{
cout<<"The value of the discriminant is: "<<imaginary_discrim<<"i"<<"\n";
}
if(A != 0)
{
if(A < 0)
{
A *= -1;
}
if (discrim > 0)
{
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two real solutions are:"<<"\n";
cout<<"X = "<<((-B + discrim)/(2.0 * A))<<"\n";
cout<<"X = "<<((-B - discrim)/(2.0 * A))<<"\n";
}
else if (discrim == 0)
{
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the one real solution is:"<<"\n";
cout<<"X = "<<(-B/(2.0 * A))<<"\n";
}
else
{
cout<<"When A = "<<A<<", B = "<<B<<", and C = "<<C<<", the two imaginary solutions are:"<<"\n";
cout<<"X = "<<(-B/(2.0 * A))<<" + "<<(imaginary_discrim/(2 * A))<<"i"<<"\n";
cout<<"X = "<<(-B/(2.0 * A))<<" - "<<(imaginary_discrim/(2 * A))<<"i"<<"\n";
}
}
else
cout<<"There is no solution for the equation when A = 0."<<"\n";
return 0;
}