Solve the given quadratic equations and mark the correct option based on your answer
I. ` 4 x^(2) - 7 x + 3 = 0`
II. ` 7 y ^(2) - 17 y + 6 = 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 quadratic equations given in the question, we will follow these steps:

### Step 1: Solve the first equation \( 4x^2 - 7x + 3 = 0 \)

1. **Identify coefficients**: Here, \( a = 4 \), \( b = -7 \), and \( c = 3 \).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 4 \cdot 3 = 49 - 48 = 1 \]
3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{7 \pm \sqrt{1}}{2 \cdot 4} = \frac{7 \pm 1}{8} \]
- First root:
\[ x_1 = \frac{7 + 1}{8} = \frac{8}{8} = 1 \]
- Second root:
\[ x_2 = \frac{7 - 1}{8} = \frac{6}{8} = \frac{3}{4} \]

### Step 2: Solve the second equation \( 7y^2 - 17y + 6 = 0 \)

1. **Identify coefficients**: Here, \( a = 7 \), \( b = -17 \), and \( c = 6 \).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-17)^2 - 4 \cdot 7 \cdot 6 = 289 - 168 = 121 \]
3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{17 \pm \sqrt{121}}{2 \cdot 7} = \frac{17 \pm 11}{14} \]
- First root:
\[ y_1 = \frac{17 + 11}{14} = \frac{28}{14} = 2 \]
- Second root:
\[ y_2 = \frac{17 - 11}{14} = \frac{6}{14} = \frac{3}{7} \]

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

- The roots of the first equation are \( x_1 = 1 \) and \( x_2 = \frac{3}{4} \).
- The roots of the second equation are \( y_1 = 2 \) and \( y_2 = \frac{3}{7} \).

Now we compare the values:

1. Compare \( x_1 = 1 \) with \( y_1 = 2 \):
- \( 1 < 2 \) (So, \( x_1 < y_1 \))

2. Compare \( x_1 = 1 \) with \( y_2 = \frac{3}{7} \):
- \( 1 > \frac{3}{7} \) (So, \( x_1 > y_2 \))

3. Compare \( x_2 = \frac{3}{4} \) with \( y_1 = 2 \):
- \( \frac{3}{4} < 2 \) (So, \( x_2 < y_1 \))

4. Compare \( x_2 = \frac{3}{4} \) with \( y_2 = \frac{3}{7} \):
- \( \frac{3}{4} > \frac{3}{7} \) (So, \( x_2 > y_2 \))

### Conclusion

From the comparisons:
- \( x_1 < y_1 \)
- \( x_1 > y_2 \)
- \( x_2 < y_1 \)
- \( x_2 > y_2 \)

Since we have both \( x < y \) and \( x > y \) for different pairs, we cannot establish a consistent relationship between \( x \) and \( y \). Therefore, the answer is that no relation can be established between \( x \) and \( y \).

### Final Answer
**Option D: No relation can be established between \( x \) and \( y \)**.

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