Solve the following quadratic equation and mark and answer as per instructions.
I. ` 2 x^(2) - x - 1 = 0`
II. ` 3 y^(2) - 5 y + 2 = 0`

Options

A. ` x le y `
B. ` x lt y`
C. x = y or no relation can be established
D. ` x ge y`

Answer

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

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

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

2. **Use the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ x = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ x = \frac{1 \pm \sqrt{9}}{4} \]
\[ x = \frac{1 \pm 3}{4} \]

3. **Calculate the two possible values for \(x\)**:
- \(x_1 = \frac{1 + 3}{4} = 1\)
- \(x_2 = \frac{1 - 3}{4} = -\frac{1}{2} = -0.5\)

### Step 2: Solve the second quadratic equation \(3y^2 - 5y + 2 = 0\)

1. **Identify the coefficients**:
- \(a = 3\), \(b = -5\), \(c = 2\).

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values:
\[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]
\[ y = \frac{5 \pm \sqrt{25 - 24}}{6} \]
\[ y = \frac{5 \pm \sqrt{1}}{6} \]
\[ y = \frac{5 \pm 1}{6} \]

3. **Calculate the two possible values for \(y\)**:
- \(y_1 = \frac{5 + 1}{6} = 1\)
- \(y_2 = \frac{5 - 1}{6} = \frac{4}{6} = \frac{2}{3} \approx 0.6667\)

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

We have:
- From the first equation: \(x_1 = 1\), \(x_2 = -0.5\)
- From the second equation: \(y_1 = 1\), \(y_2 \approx 0.6667\)

Now we compare:
1. \(x_1 = 1\) with \(y_1 = 1\): **Equal**
2. \(x_1 = 1\) with \(y_2 \approx 0.6667\): **Greater**
3. \(x_2 = -0.5\) with \(y_1 = 1\): **Smaller**
4. \(x_2 = -0.5\) with \(y_2 \approx 0.6667\): **Smaller**

### Conclusion

- We can establish that \(x_1 = y_1\) (both equal to 1).
- \(x_1 > y_2\) (1 is greater than approximately 0.6667).
- \(x_2 < y_1\) and \(x_2 < y_2\) (both comparisons show that -0.5 is less than both y-values).

Thus, the relationship between \(x\) and \(y\) is established, and we can conclude that there is no consistent inequality across all values.

### Final Answer

The values are:
- \(x = 1\) and \(x = -0.5\)
- \(y = 1\) and \(y \approx 0.6667\)

**Mark your answer accordingly based on the instructions provided.**

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