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To find the quadratic equation in standard form, \( y = ax^2 + bx + c \), given the roots \( x = 3 + 3\sqrt{3} \) and \( x = 3 - 3\sqrt{3} \), and the fact that the leading coefficient is \( a = 1 \), we can follow these steps:

### Step 1: Use the fact that the quadratic has given roots.

The general form for a quadratic with roots \( r_1 \) and \( r_2 \) is:
\[
y = a(x - r_1)(x - r_2)
\]
where \( r_1 = 3 + 3\sqrt{3} \) and \( r_2 = 3 - 3\sqrt{3} \).

Since the leading coefficient is 1, we have:
\[
y = (x - (3 + 3\sqrt{3}))(x - (3 - 3\sqrt{3}))
\]

### Step 2: Expand the expression.

This expression can be rewritten using the difference of squares:
\[
y = \left( (x - 3) - 3\sqrt{3} \right) \left( (x - 3) + 3\sqrt{3} \right)
\]
Now, apply the difference of squares formula:
\[
y = (x - 3)^2 - (3\sqrt{3})^2
\]

### Step 3: Simplify the expression.

First, calculate \( (3\sqrt{3})^2 \):
\[
(3\sqrt{3})^2 = 9 \times 3 = 27
\]

Now, expand \( (x - 3)^2 \):
\[
(x - 3)^2 = x^2 - 6x + 9
\]

Thus, the quadratic equation becomes:
\[
y = x^2 - 6x + 9 - 27
\]
Simplifying:
\[
y = x^2 - 6x - 18
\]

### Final Answer:
The quadratic equation in standard form is:
\[
y = x^2 - 6x - 18
\]

Would you like a more detailed explanation of any part of this process? 

Here are some related questions to further explore:
1. How do you derive the quadratic equation from roots in general?
2. What is the significance of the leading coefficient in quadratic equations?
3. How does the difference of squares formula work?
4. What is the relationship between the discriminant and the nature of the roots?
5. Can we solve the quadratic equation \( y = x^2 - 6x - 18 \) using the quadratic formula?

**Tip:** Always double-check your algebra when expanding binomials like \( (x - r_1)(x - r_2) \), as it's easy to make small mistakes!

Find the formula, in standard form y = ax^2 + bx + c, for a quadratic that has roots at x = 3 + 3√3 and x = 3 - 3√3, and has a leading coefficient of 1.

Learn how to find the quadratic equation in standard form with given roots x = 3 + 3√3 and x = 3 - 3√3. This problem involves using algebraic concepts, difference of squares, and root properties. Suitable for high school students in Grades 9-12.

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