Simplifying radicals

[The Mighty Boosh]

I seem to have trouble for either finding a descriptive piece of code, or finding a description of how to simplify radicals. I plan to use this as a separate function in my expanded quadratic equation program. The function I use is just perfect the way it is, except for when I get to a number dealing with 15, since 15 is not prime nor is it made of any perfect squares.
Here is the function:

double getPerfSquareDivisor(double squareRoot)
{
	double divisor = 1;
	double exponent;
	double num = 0;
	char prime = ' ';
	if (squareRoot < 0)
	{
		squareRoot *= -1;
	}

	cout << "squareRoot:  " << squareRoot << endl;
	prime = getIsPrime(squareRoot);
	cout << "prime(getPerfSquareDivisor):  " << prime << endl;
	if (prime == 'N')
	{
		for (double exponBase = 1; exponBase * exponBase <= squareRoot && prime == 'N'; exponBase += 1)
		{
			exponent = pow(exponBase, 2);
			cout << "Exponent is:  " << exponent << endl;
			num = squareRoot / exponent;
			cout << "Num is:  " << num << endl;
			outsideRad = exponent;
			prime = getIsPrime(num);
			cout << "primein for loop(getIsPrime):  " << prime << endl;
		}
		divisor = num;
		cout << "Divisor:  " << divisor << endl;
	}
	else
	{
		cout << "squareRoot:  " << squareRoot << endl;
		divisor = squareRoot;
		cout << "Divisor(Prime):  " << divisor << endl;
	}
	//End if
	return divisor;
}

Most of the couts are not going to be in the final product, I am just currently using them to somewhat debug, which has not really been working out. I know this is not the correct way to do this, but I'm just showing that I have at least tried to figure it out on my own. I am also aware that most ways of calculating this correctly use the modulus operator, which I tend not to use because of my favoritism of the double data type, and my limited knowledge of it.
Thanks! Your input is appreciated!