In each of the following questions, two equations (I) and (II) are given . Solve the equation and mark the correct option :
I. ` x = 3 sqrt( 4049)`
II. ` y^(2) = 256`

Options

A. If ` x gt y `
B. If ` x ge y `
C. if ` x lt y `
D. x = y or relation cat't be established

Answer

To solve the given equations and determine the relationship between \( x \) and \( y \), we will follow these steps:

### Step 1: Solve for \( x \)
The first equation is:
\[ x = 3 \sqrt{4049} \]

To find \( x \), we first need to calculate \( \sqrt{4049} \).

### Step 2: Calculate \( \sqrt{4049} \)
Since \( 4049 \) is not a perfect square, we can approximate it.

The perfect squares around \( 4049 \) are \( 63^2 = 3969 \) and \( 64^2 = 4096 \). Thus, \( \sqrt{4049} \) is slightly less than \( 64 \).

Calculating it more precisely:
\[ \sqrt{4049} \approx 63.6 \]

Now we can find \( x \):
\[ x = 3 \times 63.6 \approx 190.8 \]

### Step 3: Solve for \( y \)
The second equation is:
\[ y^2 = 256 \]

To find \( y \), we take the square root of both sides:
\[ y = \pm \sqrt{256} \]

Calculating \( \sqrt{256} \):
\[ \sqrt{256} = 16 \]
Thus, we have:
\[ y = 16 \quad \text{or} \quad y = -16 \]

### Step 4: Compare \( x \) and \( y \)
Now we have:
- \( x \approx 190.8 \)
- \( y = 16 \) or \( y = -16 \)

We will compare \( x \) with both values of \( y \):
1. Comparing \( x \) with \( y = 16 \):
\[ 190.8 > 16 \]
2. Comparing \( x \) with \( y = -16 \):
\[ 190.8 > -16 \]

### Conclusion
From the comparisons, we can conclude:
- \( x \) is greater than both values of \( y \).

Thus, the relationship established is that \( x > y \) for \( y = 16 \) and \( x > y \) for \( y = -16 \).

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

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