C++: Recursive Bisection Algorithm Keeps Returning 0

I am trying to implement a Bisection Root Finding Algorithm from scratch, and the equation I am passing it is x^2 - 42 = 0. My code keeps giving the output "The root of 42 is 0 within an error of 1e-05. ", meaning that the recursion stack of my bisection algo keeps returning 0 for some reason. The code is below:

#include <iostream>
#include <cmath>
using namespace std;

class Bisection
{
public:

  double a;
  double b;
  double c;
  double epsilon;

  void set_values(double, double, double);

  double f(double x)
  {
    return x*x - 42;
  }

  double solve(double a, double b, double epsilon)
  {
    c = (a+b)/2.0;
    if(abs(f(c)) < epsilon)
    {
      return c;
    }
    else
    {
      if (f(a) * f(c) < 0.0)
      {
        b = c;
        solve(a, b, epsilon);
      }
      else if (f(c) * f(b) < 0.0)
      {
         a = c;
         solve(a, b, epsilon);
      }
    }
    return 0;
  }
};

void Bisection::set_values(double left, double right, double error)
{
  a = left;
  b = right;
  epsilon = error;
}

int main()
{
  Bisection myObj;
  myObj.set_values(0.0, 10.0, 0.00001);
  //cout << myObj.f(7);
  cout << "The root of 42 is " << myObj.solve(myObj.a, myObj.b, myObj.epsilon) << " within     an error of " << myObj.epsilon << ".";
  return 0;
}

I had to include the "return 0;" line because my compiler was throwing back "non-void function does not return a value in all control paths" without it. How can I change my code to include a return value for all recursion paths but still yield the right answer? Is there an error in my fundamental design of the algorithm, or is it a simple fix?

Other problems with the code aside, you need to actually do something with the result of solve. Usually this means returning the result from solve directly until you reach the base case and return that:

double solve(double a, double b, double epsilon)
{
    c = (a+b)/2.0;
    if(abs(f(c)) < epsilon)
    {
      return c; //this is your base case
    }
    else
    {
      if (f(a) * f(c) < 0.0)
      {
        b = c;
        return solve(a, b, epsilon);
      }
      else if (f(c) * f(b) < 0.0)
      {
         a = c;
         return solve(a, b, epsilon);
      }
    }
}