In each of the following questions, two equations (I)and (II) are given. Solve the equations and mark the correct options
I. ` 9 x^(2) + 11 x + 2 = 0 `
II. ` 8 y^(2) + 6y + 1 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. If x = y or no relation can be established between x and y

Answer

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

### Step 1: Solve the first equation \(9x^2 + 11x + 2 = 0\)

1. **Identify the coefficients**:
- \(a = 9\)
- \(b = 11\)
- \(c = 2\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 11^2 - 4 \cdot 9 \cdot 2 = 121 - 72 = 49 \]

3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{49}}{2 \cdot 9} = \frac{-11 \pm 7}{18} \]

4. **Calculate the two possible values for \(x\)**:
- First root:
\[ x_1 = \frac{-11 + 7}{18} = \frac{-4}{18} = -\frac{2}{9} \approx -0.222 \]
- Second root:
\[ x_2 = \frac{-11 - 7}{18} = \frac{-18}{18} = -1 \]

### Step 2: Solve the second equation \(8y^2 + 6y + 1 = 0\)

1. **Identify the coefficients**:
- \(a = 8\)
- \(b = 6\)
- \(c = 1\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 6^2 - 4 \cdot 8 \cdot 1 = 36 - 32 = 4 \]

3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-6 \pm \sqrt{4}}{2 \cdot 8} = \frac{-6 \pm 2}{16} \]

4. **Calculate the two possible values for \(y\)**:
- First root:
\[ y_1 = \frac{-6 + 2}{16} = \frac{-4}{16} = -\frac{1}{4} = -0.25 \]
- Second root:
\[ y_2 = \frac{-6 - 2}{16} = \frac{-8}{16} = -\frac{1}{2} = -0.5 \]

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

Now we have the values:
- \(x_1 = -\frac{2}{9} \approx -0.222\) and \(x_2 = -1\)
- \(y_1 = -\frac{1}{4} = -0.25\) and \(y_2 = -\frac{1}{2} = -0.5\)

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

1. **Compare \(x_1\) and \(y_1\)**:
- \(-0.222\) (from \(x_1\)) is greater than \(-0.25\) (from \(y_1\)), so \(x_1 > y_1\).

2. **Compare \(x_1\) and \(y_2\)**:
- \(-0.222\) (from \(x_1\)) is greater than \(-0.5\) (from \(y_2\)), so \(x_1 > y_2\).

3. **Compare \(x_2\) and \(y_1\)**:
- \(-1\) (from \(x_2\)) is less than \(-0.25\) (from \(y_1\)), so \(x_2 < y_1\).

4. **Compare \(x_2\) and \(y_2\)**:
- \(-1\) (from \(x_2\)) is less than \(-0.5\) (from \(y_2\)), so \(x_2 < y_2\).

### Conclusion

From the comparisons, we can conclude:
- \(x_1 > y_1\)
- \(x_1 > y_2\)
- \(x_2 < y_1\)
- \(x_2 < y_2\)

Thus, there is no consistent relationship between \(x\) and \(y\) across all values. Therefore, the final answer is that no relation can be established between \(x\) and \(y\).