Solve the given quadratic equations and mark the correct options based on your answer
I. ` x^(2) - 2 x = 15`
II. ` y^(2) + 5y + 4 = 0`

Options

A. ` x gt y `
B. ` x ge y `
C. ` x lt y `
D. `x = y` or no relation established

Answer

Let's solve the given quadratic equations step by step.

### Step 1: Solve the first equation \( x^2 - 2x = 15 \)

1. **Rearrange the equation:**
\[ x^2 - 2x - 15 = 0 \]

2. **Factor the quadratic equation:**
We need two numbers that multiply to \(-15\) (the constant term) and add to \(-2\) (the coefficient of \(x\)). The numbers are \(3\) and \(-5\).
\[ (x - 5)(x + 3) = 0 \]

3. **Set each factor to zero:**
\[ x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \]

4. **Solve for \(x\):**
\[ x = 5 \quad \text{or} \quad x = -3 \]

### Step 2: Solve the second equation \( y^2 + 5y + 4 = 0 \)

1. **Factor the quadratic equation:**
We need two numbers that multiply to \(4\) (the constant term) and add to \(5\) (the coefficient of \(y\)). The numbers are \(4\) and \(1\).
\[ (y + 4)(y + 1) = 0 \]

2. **Set each factor to zero:**
\[ y + 4 = 0 \quad \text{or} \quad y + 1 = 0 \]

3. **Solve for \(y\):**
\[ y = -4 \quad \text{or} \quad y = -1 \]

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

We have the following solutions:
- For \(x\): \(5\) and \(-3\)
- For \(y\): \(-4\) and \(-1\)

Now we will compare the values:
1. Compare \(x = 5\) with \(y = -4\):
\[ 5 > -4 \quad \text{(True)} \]

2. Compare \(x = 5\) with \(y = -1\):
\[ 5 > -1 \quad \text{(True)} \]

3. Compare \(x = -3\) with \(y = -4\):
\[ -3 > -4 \quad \text{(True)} \]

4. Compare \(x = -3\) with \(y = -1\):
\[ -3 < -1 \quad \text{(True)} \]

### Conclusion

From the comparisons:
- \(x = 5\) is greater than both values of \(y\).
- \(x = -3\) is greater than \(y = -4\) but less than \(y = -1\).

Since we have both greater and lesser relationships, we cannot establish a consistent relationship between \(x\) and \(y\).

Thus, the final answer is that no consistent relation can be established between \(x\) and \(y\).