C++ Bisection Algorithm for a Quadratic Equation - c++

Question

I had a prior snag with this problem that was resolved but I felt that since the nature of my new issue is not related to successful compiling, rather to actual logic of the code, that it would be acceptable to make a new topic. Here is my code so far:
#include "assign4.h"
#include <iostream>
using namespace std;
int main(int argc, char * argv[]){
solution s;
double root;
cout << "Enter interval endpoints: ";
cin >> s.xLeft >> s.xRight;
cout << "Enter tolerance: ";
cin >> s.epsilon;
root = s.bisect (s.xLeft, s.xRight, s.epsilon, &solution::f, s.error);
if (!(s.error))
cout << "Root found at " << root << "\nValue of f(x) at root is: " << s.f(root) << "\n";
else {
cout << "The solution of a quadratic equation with coefficients: " << endl;
// cout << "a = " << a << ", b = " << b << ", c = " << c << endl;
cout << "has not been found." << endl;
}
return 0;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#include "assign4.h"
#include <iostream>
#include <math.h>
using namespace std;
double solution::bisect (double xLeft, double xRight, double epsilon, double (solution::*f)(double), bool& error) {
double xMid;
double fLeft, fRight;
double fMid;
fLeft = (this->*f)(xLeft);
fRight = (this->*f)(xRight);
error = (fLeft * fRight) < 0;
if (error)
return -999.0;
for (double i = 0; i < 20; i++) {
xMid = (xLeft + (xLeft + 1.0)) / 2.0;
fMid = (this->*f)(xMid);
if (fLeft * fMid > 0.0) {
xLeft = xMid + 0.5;
xRight = xLeft + 1.0;
fLeft = fMid;
}
else if (fLeft * fMid < 0.0){
xRight = xMid;
fRight = fMid;
}
else {
return xMid;
}
cout << "New Interval is [" << xLeft << ", " << xRight << "]" << endl;
}
return (xLeft + xRight) / 2.0;
}
double solution::f (double x) {
return ((1 * pow(x,2.0)) + (5 * x) + 2);
}
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#ifndef ASSIGN4_H
#define ASSIGN4_H
class solution {
public:
double xLeft, xRight;
double epsilon;
bool error;
double bisect(double, double, double, double (solution::*f)(double), bool&);
double f(double);
};
#endif // ASSIGN4_H
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
My goal with this assignment is to find any roots should they exist. My issue is that every bisection example I have found only talks about how to find a single root at once. The interval I am to use is [-10.0, 10.0] and eventually I am going to receive coefficients of the equation passed via an array encapsulated in a structure, but for now I have hard coded the coefficients.
So my problem is that I can currently get with 0.2 of the first root for the equation I have hard coded (x^2 + 5x + 2) but I am uncertain of how to step past that root and keep searching for another root up until the end of my interval. I am also unsure of how to accurately hit the root and not be marginally off.
Apologies for the wall of text and any help is appreciated! :)

Answer 1

You can input the interval such that it includes both the roots. Then call bisect twice -- once for the lower root and once for the higher root.
int main(int argc, char * argv[]){
solution s;
double root;
cout << "Enter interval endpoints: ";
cin >> s.xLeft >> s.xRight;
cout << "Enter tolerance: ";
cin >> s.epsilon;
double xMid = 0.5*(s.xLeft + s.xRight);
root = s.bisect (s.xLeft, xMid, s.epsilon, &solution::f, s.error);
if (!(s.error))
cout << "Root found at " << root << "\nValue of f(x) at root is: " << s.f(root) << endl;
else
{
cout << "The solution of a quadratic equation with coefficients: " << endl;
cout << "has not been found." << endl;
}
root = s.bisect (xMid, s.xRight, s.epsilon, &solution::f, s.error);
if (!(s.error))
cout << "Root found at " << root << "\nValue of f(x) at root is: " << s.f(root) << endl;
else
{
cout << "The solution of a quadratic equation with coefficients: " << endl;
cout << "has not been found." << endl;
}
return 0;
}