We are given the roots of the quadratic equation as $3 - \sqrt{3}$ and $3 + \sqrt{3}$. To find the quadratic equation with these roots, we can use the fact that if the roots of a quadratic equation are $p$ and $q$, the equation can be written as:

$(x - p)(x - q) = 0$

Step 1: Identify the roots

The given roots are $p = 3 - \sqrt{3}$ and $q = 3 + \sqrt{3}$.

Step 2: Write the equation

The quadratic equation is:

$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$

Step 3: Simplify the expression

Now expand the product:

$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = (x - 3 + \sqrt{3})(x - 3 - \sqrt{3})$

This expression is a difference of squares, so:

$= (x - 3)^2 - (\sqrt{3})^2$

$= (x - 3)^2 - 3$

Now expand $(x - 3)^2$:

$= x^2 - 6x + 9 - 3$

$= x^2 - 6x + 6$

Final Answer:

The quadratic equation is:

$x^2 - 6x + 6 = 0$

Would you like further explanation or help with another problem? Here are some related questions for practice:

1. How would you find the quadratic equation if the roots were $2 + \sqrt{5}$ and $2 - \sqrt{5}$?
2. What is the relationship between the sum and product of the roots of a quadratic equation?
3. How can you use the quadratic formula to find the roots of a quadratic equation?
4. What happens when the discriminant of a quadratic equation is negative?
5. How can you express the roots of a quadratic equation in terms of its coefficients?

Tip: The sum of the roots of a quadratic equation $ax^2 + bx + c = 0$ is $-\frac{b}{a}$, and the product is $\frac{c}{a}$.