In the following questions, two equations numbered I and II are given. You have to solve both the equations and
I. `2x^2+5x+2=0`
II. `2y^2+11y+15=0`

Options

A. `xlty`
B. `xgty`
C. `x le y`
D. `x ge y`

Answer

To solve the given equations step by step, we will tackle each quadratic equation separately.

### Step 1: Solve the first equation \(2x^2 + 5x + 2 = 0\)

1. **Identify the coefficients**:
- \(a = 2\)
- \(b = 5\)
- \(c = 2\)

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-5 \pm \sqrt{9}}{2 \cdot 2} = \frac{-5 \pm 3}{4} \]

5. **Calculate the two possible values for \(x\)**:
- First solution:
\[ x_1 = \frac{-5 + 3}{4} = \frac{-2}{4} = -\frac{1}{2} \]
- Second solution:
\[ x_2 = \frac{-5 - 3}{4} = \frac{-8}{4} = -2 \]

### Step 2: Solve the second equation \(2y^2 + 11y + 15 = 0\)

1. **Identify the coefficients**:
- \(a = 2\)
- \(b = 11\)
- \(c = 15\)

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 11^2 - 4 \cdot 2 \cdot 15 = 121 - 120 = 1 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-11 \pm \sqrt{1}}{2 \cdot 2} = \frac{-11 \pm 1}{4} \]

5. **Calculate the two possible values for \(y\)**:
- First solution:
\[ y_1 = \frac{-11 + 1}{4} = \frac{-10}{4} = -\frac{5}{2} \]
- Second solution:
\[ y_2 = \frac{-11 - 1}{4} = \frac{-12}{4} = -3 \]

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

The solutions we found are:
- For \(x\): \(-\frac{1}{2}\) and \(-2\)
- For \(y\): \(-\frac{5}{2}\) and \(-3\)

### Step 4: Determine the relationship between \(x\) and \(y\)

- The greater value of \(x\) is \(-\frac{1}{2}\).
- The greater value of \(y\) is \(-\frac{5}{2}\).

Since \(-\frac{1}{2} > -\frac{5}{2}\), we conclude that \(x\) is greater than \(y\).

### Final Answer:
Thus, the relationship is \(x > y\).

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