I. ` 4x^(2) - 32 x + 63 = 0" "` II.` 2y^(2) - 11y + 15 = 0`

Options

A. ` x lt y`
B. `x gt y`
C. ` x le y`
D. `x ge y`

Explanation

To solve the given quadratic equations step by step, we will tackle each equation separately.

### Step 1: Solve the first equation \(4x^2 - 32x + 63 = 0\)

1. **Identify the coefficients**:
- \(a = 4\), \(b = -32\), \(c = 63\)

2. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-32)^2 - 4 \cdot 4 \cdot 63 = 1024 - 1008 = 16 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{32 \pm \sqrt{16}}{2 \cdot 4} = \frac{32 \pm 4}{8} \]

5. **Calculate the two possible values of \(x\)**:
- \(x_1 = \frac{32 + 4}{8} = \frac{36}{8} = 4.5\)
- \(x_2 = \frac{32 - 4}{8} = \frac{28}{8} = 3.5\)

### Step 2: Solve the second equation \(2y^2 - 11y + 15 = 0\)

1. **Identify the coefficients**:
- \(a = 2\), \(b = -11\), \(c = 15\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = (-11)^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{11 \pm \sqrt{1}}{2 \cdot 2} = \frac{11 \pm 1}{4} \]

5. **Calculate the two possible values of \(y\)**:
- \(y_1 = \frac{11 + 1}{4} = \frac{12}{4} = 3\)
- \(y_2 = \frac{11 - 1}{4} = \frac{10}{4} = 2.5\)

### Step 3: Compare the values of \(x\) and \(y\)

- The values of \(x\) are \(4.5\) and \(3.5\).
- The values of \(y\) are \(3\) and \(2.5\).

### Step 4: Determine the relationship between \(x\) and \(y\)

- The maximum value of \(x\) is \(4.5\) and the maximum value of \(y\) is \(3\).
- Therefore, \(x > y\).

### Final Conclusion

The relationship between the values of \(x\) and \(y\) is \(x > y\).

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