Couple Questions about a Quadratic Equation Program

[Aeon221]

Sorry about the length, but spoiler tags are apparently a no-go on this forum.

I've included a program I wrote (that works perfectly) that solves quadratics. It's functional, and has a sufficient UI, but what I'm looking for here is more along the lines of ways I can improve it.

For instance, I'd like to create a function that allows the user to save all the output they generate. I already know how to do that with push_back, a vector<double> and a ofstream to a .txt file, but I wanted it to maintain the contents of said file across sessions, and I'm not clear on how to do that. When I've tried using fstream with a .txt file on other programs I've written, it overwrites everything, which isn't all that helpful.

I'm also wondering if there's a shorter and more exact method for running square roots than the one I'm using?

And I'm sure this is obvious and I'm just a moron, but how on earth do you get the program to work on a computer that doesn't have a compiler?

Thanks very much!

#include <iostream>
#include <cstdlib>
#include <cmath>

using namespace std;

void alg(double a, double b, double c);
double sqrt(double m);
void menu();

int main()
{
	int x;
	do
	{
		cout << "Type 1 to use program, else type 0 to quit: ";
		cin >> x;
		if (x == 1)
		{
			menu ();
		}
		else;
	}while (x != 0);
}

void menu()
{
	cout << "\nQuadratic Equation program. Enter values for a, b, c to receive both roots" << endl;
	double a, b, c;
	cout << "\nDefine a: ";
	cin >> a;
	cout << "\nDefine b: ";
	cin >> b;
	cout << "\nDefine c: ";
	cin >> c;
	cout << endl;
	alg(a, b, c);
}

void alg (double a, double b, double c)
{
	double x1, x2, x12, xs1;
	
	x12= ((b*b)-(4*a*c));

	if (x12 <0)
	{
		cout << "(b^2 - 4ac) = " << x12 << endl;

		xs1= sqrt(-x12);
		cout << "The square root of (b^2 - 4ac) is: " << xs1 << "i." << endl;
		cout << "Therefore the roots of this equation are imaginary!" << endl;

		double f = 2*a;
		x1= (-b/(f));
		x2= (-b/(f));
		
		cout << "\nx = -b/2a + discriminant/2a is: " << x1 << " + (" << xs1 << "i / " << f << ")" << endl;
		cout << "\nx = -b/2a - discriminant/2a is: " << x2 << " + (" << xs1 << "i / " << f << ")" << endl;
	}

	else if (x12 >= 0)
	{
		cout << "(b^2 - 4ac) = " << x12 << endl;

		xs1= sqrt(x12);
		cout << "The square root of (b^2 - 4ac) is: " << xs1 << endl;
		cout << "Therefore the roots of this equation are real!" << endl;

		x1= ((-b + xs1)/(2*a));
		x2= ((-b - xs1)/(2*a));
		cout << "\nx= -b +... is: " << x1 << endl;
		cout << "\nx= -b -... is: " << x2 << endl;
	}
}

double sqroot(double m)
{
   double i=0;
   double x1,x2;
   double ((i*i) <= m);
   i+=0.1;
   x1=i;
   for(int j=0;j<10;j++)
   {
      x2=m;
      x2/=x1;
      x2+=x1;
      x2/=2;
      x1=x2;
   }
   return (x2);
}

As a side note -- which you may of course ignore -- I only started doing this coding stuff about three days ago (I've been reading the cplusplus.com pdf tutorial, Eckhart's free Thinking in C++ 2nd ed, and the cprogramming.com tutorial), mostly because I'm a bored and (currently) unemployed college grad. If you're up to it and you know a better (free and legal) source, feel free to provide a link here, or through a PM if it's rude to link to unrelated things in one of these threads.

[jsmith]

For starters, you aren't actually using your sqroot() function.

Secondly, obligatory link:
http://docs.sun.com/source/806-3568/ncg_goldberg.html

There is quite a lengthy discussion in there about the best way to compute the roots of a quadratic to avoid
floating point roundoff error.

EDIT: better link:
http://dlc.sun.com/pdf/800-7895/800-7895.pdf

[kempofighter]

When I've tried using fstream with a .txt file on other programs I've written, it overwrites everything, which isn't all that helpful.

Did you set the ate flag in the constructor or open call? Sounds like the file is being truncated when you open it.

[Aeon221]

First off, thanks to both of you. The info and docs are greatly appreciated. Now for some specifics!

@jsmith: Could you expand on how I'm not using my sqrt()? I've checked on problems I've solved by hand for both irrationals and imaginaries, and it returns the correct answer. And I just ran through a sqrt(4*4 - (4*4*4)) which returned x11 = -48 and xs1 = 6.9282i, which is (roughly) correct. Am I missing something, or are you saying that it's just doing a bad job of returning the decimal component? Because that it is, without a doubt.

@kempofighter: Oh. Yeah. Nope, I didn't! I figured it was something simple like that, but extensive googling provided no useful responses. Obviously if I had any brain in my head, I'd have, uh, checked here in the reference segment before asking.

[jsmith]

You named your function sqroot(), but alg() uses sqrt() instead. sqrt() is defined in the C runtime library.
Nowhere in your program do you call sqroot(). [You couldn't anyway, since it is not prototyped at the
top.]