In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option.
I.`2x^2 +7x +5=0`
II. `3y^2+5y+2=0`

Options

A. `x lt y`
B. `x gt y`
C. `x le y`
D. `x ge y`

Answer

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

### Step 1: Solve the first equation \(2x^2 + 7x + 5 = 0\)

1. **Identify the coefficients**:
- Here, \(a = 2\), \(b = 7\), and \(c = 5\).

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot 5 = 49 - 40 = 9 \]

4. **Calculate the roots**:
\[ x = \frac{-7 \pm \sqrt{9}}{2 \cdot 2} = \frac{-7 \pm 3}{4} \]
- For \(x_1\):
\[ x_1 = \frac{-7 + 3}{4} = \frac{-4}{4} = -1 \]
- For \(x_2\):
\[ x_2 = \frac{-7 - 3}{4} = \frac{-10}{4} = -\frac{5}{2} \]

5. **Roots of the first equation**:
\[ x = -1 \quad \text{and} \quad x = -\frac{5}{2} \]

### Step 2: Solve the second equation \(3y^2 + 5y + 2 = 0\)

1. **Identify the coefficients**:
- Here, \(a = 3\), \(b = 5\), and \(c = 2\).

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1 \]

4. **Calculate the roots**:
\[ y = \frac{-5 \pm \sqrt{1}}{2 \cdot 3} = \frac{-5 \pm 1}{6} \]
- For \(y_1\):
\[ y_1 = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3} \]
- For \(y_2\):
\[ y_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1 \]

5. **Roots of the second equation**:
\[ y = -\frac{2}{3} \quad \text{and} \quad y = -1 \]

### Step 3: Compare the roots

- We have the roots:
- For \(x\): \(-1\) and \(-\frac{5}{2}\)
- For \(y\): \(-1\) and \(-\frac{2}{3}\)

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

- The values of \(x\) and \(y\) are:
- \(x_1 = -1\) and \(x_2 = -\frac{5}{2}\)
- \(y_1 = -\frac{2}{3}\) and \(y_2 = -1\)

- Comparing the values:
- Since \(-1 = -1\), we have \(x_1 = y_2\).
- Since \(-\frac{5}{2} \approx -2.5\) and \(-\frac{2}{3} \approx -0.67\), we have \(x_2 < y_1\).

### Conclusion

From the comparison, we can conclude:
\[ x \leq y \]
Thus, the correct answer is that \(x\) is less than or equal to \(y\).