In the following questions two equations numbered I and II are given. You have to solve both the equations and answer
`x^2-20x+96=0`
`y^2=64`

Options

A. if `x gt y`
B. if `x ge y`
C. if `xlt y`
D. if `x le y`

Answer

To solve the given equations step by step, we will start with each equation separately:

### Step 1: Solve the first equation \( x^2 - 20x + 96 = 0 \)

1. **Identify the coefficients**: The equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -20 \), and \( c = 96 \).

2. **Factor the quadratic**: We need to find two numbers that multiply to \( c = 96 \) and add up to \( b = -20 \). The numbers are \( -12 \) and \( -8 \) because:
- \( -12 \times -8 = 96 \)
- \( -12 + (-8) = -20 \)

3. **Write the factored form**: The equation can be factored as:
\[ (x - 12)(x - 8) = 0 \]

4. **Set each factor to zero**:
\[ x - 12 = 0 \quad \text{or} \quad x - 8 = 0 \]

5. **Solve for \( x \)**:
\[ x = 12 \quad \text{or} \quad x = 8 \]

### Step 2: Solve the second equation \( y^2 = 64 \)

1. **Take the square root of both sides**:
\[ y = \pm \sqrt{64} \]

2. **Calculate the square root**:
\[ y = \pm 8 \]
This gives us two possible values for \( y \):
\[ y = 8 \quad \text{or} \quad y = -8 \]

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

Now we have the values:
- \( x = 8 \) or \( x = 12 \)
- \( y = 8 \) or \( y = -8 \)

We will compare these values:

1. **Case 1**: \( x = 8 \) and \( y = 8 \)
- Here, \( x = y \).

2. **Case 2**: \( x = 8 \) and \( y = -8 \)
- Here, \( x > y \).

3. **Case 3**: \( x = 12 \) and \( y = 8 \)
- Here, \( x > y \).

4. **Case 4**: \( x = 12 \) and \( y = -8 \)
- Here, \( x > y \).

### Conclusion

From the comparisons, we can conclude that:
- In all cases, \( x \) is greater than or equal to \( y \).

Thus, the final answer is:
- \( x \geq y \)