Quantity I : Value of x
` ( x + 3) ^(2)= ( x - 3) ^(2) + x^(2)`
Quantity II : value of y
` y^(2) - 29 y + 204 = 0`

Options

A. A)Quantity I ` gt` Quantity II
B. B)Quantity I ` lt` Quantity II
C. C)Quantity I ` ge` Quantity II
D. D)Quantity I ` le` Quantity II

Answer

To solve the problem, we need to evaluate two quantities: Quantity I involves solving an equation for \( x \), and Quantity II involves solving a quadratic equation for \( y \).

### Step-by-Step Solution:

**Quantity I: Solve for \( x \)**

1. **Start with the given equation:**
\[ (x + 3)^2 = (x - 3)^2 + x^2 \]

2. **Expand both sides:**
- Left side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
- Right side:
\[ (x - 3)^2 + x^2 = (x^2 - 6x + 9) + x^2 = 2x^2 - 6x + 9 \]

3. **Set the expanded forms equal to each other:**
\[ x^2 + 6x + 9 = 2x^2 - 6x + 9 \]

4. **Rearrange the equation:**
\[ x^2 + 6x + 9 - 2x^2 + 6x - 9 = 0 \]
Simplifying gives:
\[ -x^2 + 12x = 0 \]

5. **Factor the equation:**
\[ x(-x + 12) = 0 \]

6. **Solve for \( x \):**
- \( x = 0 \)
- \( x - 12 = 0 \) gives \( x = 12 \)

So, the possible values of \( x \) are \( 0 \) and \( 12 \).

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**Quantity II: Solve for \( y \)**

1. **Start with the quadratic equation:**
\[ y^2 - 29y + 204 = 0 \]

2. **Use the quadratic formula:**
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -29, c = 204 \).

3. **Calculate the discriminant:**
\[ b^2 - 4ac = (-29)^2 - 4 \cdot 1 \cdot 204 = 841 - 816 = 25 \]

4. **Substitute back into the formula:**
\[ y = \frac{29 \pm \sqrt{25}}{2} = \frac{29 \pm 5}{2} \]

5. **Calculate the two possible values for \( y \):**
- \( y = \frac{34}{2} = 17 \)
- \( y = \frac{24}{2} = 12 \)

So, the possible values of \( y \) are \( 17 \) and \( 12 \).

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### Comparison of Quantities:

- From Quantity I, we have \( x = 0 \) or \( x = 12 \).
- From Quantity II, we have \( y = 12 \) or \( y = 17 \).

Now, we compare the values:
1. If \( x = 0 \), then \( 0 < 12 \) and \( 0 < 17 \).
2. If \( x = 12 \), then \( 12 = 12 \) and \( 12 < 17 \).

Thus, we find that:
- \( x \leq y \)

### Final Conclusion:
The relationship between Quantity I and Quantity II is:
\[ \text{Quantity I} \leq \text{Quantity II} \]