In the following questions two equation numbered I and II are given You have to solve both equations and . . . . .
(i) `x^(2)-4=0`
(ii) `y^(2)+6y+9=0`

Answer

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

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

1. Start with the equation:
\[ x^2 - 4 = 0 \]

2. Rearranging gives:
\[ x^2 = 4 \]

3. Taking the square root of both sides:
\[ x = \pm 2 \]

Thus, the solutions for \( x \) are:
\[ x = 2 \quad \text{and} \quad x = -2 \]

### Step 2: Solve the second equation \( y^2 + 6y + 9 = 0 \)

1. Start with the equation:
\[ y^2 + 6y + 9 = 0 \]

2. This can be factored as:
\[ (y + 3)(y + 3) = 0 \]
or
\[ (y + 3)^2 = 0 \]

3. Therefore, solving gives:
\[ y + 3 = 0 \implies y = -3 \]

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

1. The values we found are:
- \( x = 2 \) or \( x = -2 \)
- \( y = -3 \)

2. Now we compare:
- For \( x = 2 \): \( 2 > -3 \)
- For \( x = -2 \): \( -2 > -3 \)

In both cases, \( x \) is greater than \( y \).

### Conclusion

Thus, the relationship established is:
\[ x > y \]

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