In each of the following questions two equations numbered I and II are given. You have to solve both the equations and mark the appropriate option. Give answer
I `3x^2-8x+4=0`
II `6y^2-7y+2=0`

Options

A. if `xgty`
B. if `xlty`
C. if `xgey`
D. if `xley`

Answer

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

### Step 1: Solve the first equation \(3x^2 - 8x + 4 = 0\)

1. **Identify the coefficients**: Here, \(a = 3\), \(b = -8\), and \(c = 4\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-8)^2 - 4 \cdot 3 \cdot 4 = 64 - 48 = 16 \]
3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{8 \pm \sqrt{16}}{2 \cdot 3} = \frac{8 \pm 4}{6} \]
- First root:
\[ x_1 = \frac{8 + 4}{6} = \frac{12}{6} = 2 \]
- Second root:
\[ x_2 = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3} \]
4. **Conclusion for x**: The values of \(x\) are \(2\) and \(\frac{2}{3}\).

### Step 2: Solve the second equation \(6y^2 - 7y + 2 = 0\)

1. **Identify the coefficients**: Here, \(a = 6\), \(b = -7\), and \(c = 2\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 6 \cdot 2 = 49 - 48 = 1 \]
3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{1}}{2 \cdot 6} = \frac{7 \pm 1}{12} \]
- First root:
\[ y_1 = \frac{7 + 1}{12} = \frac{8}{12} = \frac{2}{3} \]
- Second root:
\[ y_2 = \frac{7 - 1}{12} = \frac{6}{12} = \frac{1}{2} \]
4. **Conclusion for y**: The values of \(y\) are \(\frac{2}{3}\) and \(\frac{1}{2}\).

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

- The values of \(x\) are \(2\) and \(\frac{2}{3}\).
- The values of \(y\) are \(\frac{2}{3}\) and \(\frac{1}{2}\).

### Step 4: Analyze the relationships

1. **For \(x = 2\)**:
- \(2 > \frac{2}{3}\)
- \(2 > \frac{1}{2}\)

2. **For \(x = \frac{2}{3}\)**:
- \(\frac{2}{3} = \frac{2}{3}\)
- \(\frac{2}{3} > \frac{1}{2}\)

### Conclusion

From the analysis:
- \(x\) can be greater than or equal to \(y\).
- Thus, the relationship can be summarized as \(x \geq y\).

### Final Answer
The correct option is that \(x\) is either greater than or equal to \(y\).

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