In the following questions two equation numbered I and II are given You have to solve both equations and give answer
(i) `x^(2)=729`
(ii) ` y=sqrt(729)`

Options

A. if x > y
B. if x ≥ y
C. if x < y
D. if x ≤ y

Answer

To solve the given equations, we will work through each equation step by step.

### Step 1: Solve the first equation \( x^2 = 729 \)

1. To solve for \( x \), we take the square root of both sides:
\[ x = \pm \sqrt{729} \]

2. Next, we calculate \( \sqrt{729} \):
\[ \sqrt{729} = 27 \]

3. Therefore, the solutions for \( x \) are:
\[ x = 27 \quad \text{or} \quad x = -27 \]

### Step 2: Solve the second equation \( y = \sqrt{729} \)

1. We already calculated \( \sqrt{729} \) in the previous step:
\[ y = 27 \]

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

1. We have two possible values for \( x \): \( 27 \) and \( -27 \).
2. The value of \( y \) is \( 27 \).

3. Now we compare \( x \) and \( y \):
- For \( x = 27 \):
\[ x = y \quad (27 = 27) \]
- For \( x = -27 \):
\[ x < y \quad (-27 < 27) \]

### Conclusion

Since \( x \) can either be equal to \( y \) or less than \( y \), the final conclusion is:
\[ x \leq y \]

### Final Answer
The correct option is **4: \( x \leq y \)**.

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