Solve the given quadratic equations and mark the correct option based on your answer
I.` 2 x^(2) - 19 x + 45 = 0`
II. ` 2 y^(2) - 9 y + 4 = 0 `

Options

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

Answer

To solve the given quadratic equations, we will follow these steps:

### Step 1: Solve the first quadratic equation \(2x^2 - 19x + 45 = 0\)

1. **Identify coefficients**: Here, \(a = 2\), \(b = -19\), and \(c = 45\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-19)^2 - 4 \cdot 2 \cdot 45 = 361 - 360 = 1 \]
3. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{19 \pm \sqrt{1}}{2 \cdot 2} = \frac{19 \pm 1}{4} \]
4. **Calculate the two possible values for \(x\)**:
- \(x_1 = \frac{20}{4} = 5\)
- \(x_2 = \frac{18}{4} = \frac{9}{2} = 4.5\)

### Step 2: Solve the second quadratic equation \(2y^2 - 9y + 4 = 0\)

1. **Identify coefficients**: Here, \(a = 2\), \(b = -9\), and \(c = 4\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-9)^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49 \]
3. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{9 \pm \sqrt{49}}{2 \cdot 2} = \frac{9 \pm 7}{4} \]
4. **Calculate the two possible values for \(y\)**:
- \(y_1 = \frac{16}{4} = 4\)
- \(y_2 = \frac{2}{4} = \frac{1}{2} = 0.5\)

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

- From the first equation, we have \(x_1 = 5\) and \(x_2 = 4.5\).
- From the second equation, we have \(y_1 = 4\) and \(y_2 = 0.5\).

Now we compare:
- \(5 > 4\) (i.e., \(x_1 > y_1\))
- \(4.5 > 4\) (i.e., \(x_2 > y_1\))
- \(5 > 0.5\) (i.e., \(x_1 > y_2\))
- \(4.5 > 0.5\) (i.e., \(x_2 > y_2\))

### Conclusion

Since in all comparisons \(x\) is greater than \(y\), we conclude that \(x > y\).

### Final Answer

The correct option is **A**: \(x > y\).

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