In each of questions two eqautions numbered Iand II are given you have to solve both equations and give answer
I.`x^2+6x+9=0`
II.`y^2-y-20=0`

Options

A. If `xlty`
B. `if xley`
C. if `xgty`
D. if relation between x and y can not be established

Answer

To solve the given equations step by step, we will follow the process of factoring and finding the roots for both equations.

### Step 1: Solve the first equation \( x^2 + 6x + 9 = 0 \)

1. **Identify the equation**: The equation is \( x^2 + 6x + 9 = 0 \).
2. **Factor the equation**: We notice that this can be factored as:
\[ (x + 3)(x + 3) = 0 \]
or
\[ (x + 3)^2 = 0 \]
3. **Set each factor to zero**:
\[ x + 3 = 0 \]
4. **Solve for \( x \)**:
\[ x = -3 \]

### Step 2: Solve the second equation \( y^2 - y - 20 = 0 \)

1. **Identify the equation**: The equation is \( y^2 - y - 20 = 0 \).
2. **Factor the equation**: We need to find two numbers that multiply to \(-20\) and add to \(-1\). The numbers \(-5\) and \(4\) work:
\[ (y - 5)(y + 4) = 0 \]
3. **Set each factor to zero**:
\[ y - 5 = 0 \quad \text{or} \quad y + 4 = 0 \]
4. **Solve for \( y \)**:
\[ y = 5 \quad \text{or} \quad y = -4 \]

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

1. **Values obtained**: We have \( x = -3 \) and \( y = 5 \) or \( y = -4 \).
2. **Establish the relationship**:
- When \( y = 5 \): \( -3 < 5 \)
- When \( y = -4 \): \( -3 > -4 \)

### Conclusion

Since \( x \) can be greater than \( y \) in one case and less than \( y \) in another, we cannot establish a consistent relationship between \( x \) and \( y \).

### Final Answer

The relation between \( x \) and \( y \) cannot be established.

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