I.` 4x^(2) - 48x + 143 = 0`
` II. 4y^(2) - 52y+165 = 0`

Options

A. If ` x gt y`
B. If ` y gt x`
C. if `x ge y`
D. If x = y or a relationship between x and y cannot be established.

Explanation

To solve the equations \( I. \ 4x^2 - 48x + 143 = 0 \) and \( II. \ 4y^2 - 52y + 165 = 0 \), we will use the quadratic formula, also known as the Shridhara-Charler formula. The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 1: Solve for \( x \)

1. Identify coefficients for the first equation:
- \( a = 4 \)
- \( b = -48 \)
- \( c = 143 \)

2. Plug values into the quadratic formula:
\[ x = \frac{-(-48) \pm \sqrt{(-48)^2 - 4 \cdot 4 \cdot 143}}{2 \cdot 4} \]

3. Calculate \( b^2 - 4ac \):
\[ (-48)^2 = 2304 \]
\[ 4 \cdot 4 \cdot 143 = 2288 \]
\[ b^2 - 4ac = 2304 - 2288 = 16 \]

4. Substitute back into the formula:
\[ x = \frac{48 \pm \sqrt{16}}{8} \]
\[ x = \frac{48 \pm 4}{8} \]

5. Calculate the two possible values for \( x \):
- \( x_1 = \frac{48 + 4}{8} = \frac{52}{8} = 6.5 \)
- \( x_2 = \frac{48 - 4}{8} = \frac{44}{8} = 5.5 \)

### Step 2: Solve for \( y \)

1. Identify coefficients for the second equation:
- \( a = 4 \)
- \( b = -52 \)
- \( c = 165 \)

2. Plug values into the quadratic formula:
\[ y = \frac{-(-52) \pm \sqrt{(-52)^2 - 4 \cdot 4 \cdot 165}}{2 \cdot 4} \]

3. Calculate \( b^2 - 4ac \):
\[ (-52)^2 = 2704 \]
\[ 4 \cdot 4 \cdot 165 = 2640 \]
\[ b^2 - 4ac = 2704 - 2640 = 64 \]

4. Substitute back into the formula:
\[ y = \frac{52 \pm \sqrt{64}}{8} \]
\[ y = \frac{52 \pm 8}{8} \]

5. Calculate the two possible values for \( y \):
- \( y_1 = \frac{52 + 8}{8} = \frac{60}{8} = 7.5 \)
- \( y_2 = \frac{52 - 8}{8} = \frac{44}{8} = 5.5 \)

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

- Values of \( x \): \( 6.5, 5.5 \)
- Values of \( y \): \( 7.5, 5.5 \)

### Conclusion

- The values \( y_1 = 7.5 \) is greater than \( x_1 = 6.5 \).
- The values \( y_2 = 5.5 \) is equal to \( x_2 = 5.5 \).

Thus, we can conclude that \( y \) is greater than or equal to \( x \).