In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer
I. ` x^(2) + 20 x + 100 = 0 `
II. ` y^(2) + 13 y + 30 = 0 `

Options

A. If `x= y` or no relation can be established
B. If ` x gt y `
C. If ` x lt y `
D. If `xle y `

Answer

To solve the given equations and find the relationship between \( x \) and \( y \), we will proceed step by step.

### Step 1: Solve the first equation \( I. \, x^2 + 20x + 100 = 0 \)

1. Recognize that the equation can be factored as a perfect square:
\[ x^2 + 20x + 100 = (x + 10)^2 = 0 \]
2. Set the factored equation to zero:
\[ (x + 10)^2 = 0 \]
3. Solve for \( x \):
\[ x + 10 = 0 \implies x = -10 \]

### Step 2: Solve the second equation \( II. \, y^2 + 13y + 30 = 0 \)

1. Factor the quadratic equation:
\[ y^2 + 13y + 30 = (y + 10)(y + 3) = 0 \]
2. Set each factor to zero:
\[ y + 10 = 0 \implies y = -10 \]
\[ y + 3 = 0 \implies y = -3 \]

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

1. From the first equation, we found \( x = -10 \).
2. From the second equation, we found two values for \( y \): \( y = -10 \) and \( y = -3 \).
3. Now, we compare:
- When \( y = -10 \):
\[ x = y \quad \text{(since both are -10)} \]
- When \( y = -3 \):
\[ x < y \quad \text{(since -10 < -3)} \]

### Step 4: Conclusion

Combining these results, we conclude:
- \( x \) can be equal to \( y \) when \( y = -10 \).
- \( x \) is less than \( y \) when \( y = -3 \).

Thus, the final relationship is:
\[ x \leq y \]

### Final Answer
The appropriate answer is \( x \leq y \).

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