In the following questions two equations numbered I and II are given. You have to solve both the equations and answer
`4x^2 + 9x+5=0`
`3y^2 + 5y+2=0`

Options

A. if `x gt y`
B. if `x ge y`
C. if `xlt y`
D. if `x le y`

Answer

To solve the given equations step by step, we will follow the quadratic formula method for both equations.

### Step 1: Solve the first equation \(4x^2 + 9x + 5 = 0\)

1. Identify coefficients:
- \(a = 4\)
- \(b = 9\)
- \(c = 5\)

2. Calculate the discriminant (\(D\)):
\[ D = b^2 - 4ac = 9^2 - 4 \cdot 4 \cdot 5 = 81 - 80 = 1 \]

3. Since \(D > 0\), we will have two real and distinct roots.

4. Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-9 \pm \sqrt{1}}{2 \cdot 4} \]
\[ x = \frac{-9 \pm 1}{8} \]

5. Calculate the two values of \(x\):
- For \(x_1\):
\[ x_1 = \frac{-9 + 1}{8} = \frac{-8}{8} = -1 \]
- For \(x_2\):
\[ x_2 = \frac{-9 - 1}{8} = \frac{-10}{8} = -\frac{5}{4} = -1.25 \]

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

1. Identify coefficients:
- \(a = 3\)
- \(b = 5\)
- \(c = 2\)

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

3. Since \(D > 0\), we will have two real and distinct roots.

4. Use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{1}}{2 \cdot 3} \]
\[ y = \frac{-5 \pm 1}{6} \]

5. Calculate the two values of \(y\):
- For \(y_1\):
\[ y_1 = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3} \approx -0.67 \]
- For \(y_2\):
\[ y_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1 \]

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

We have the following values:
- \(x_1 = -1\), \(x_2 = -1.25\)
- \(y_1 \approx -0.67\), \(y_2 = -1\)

Now we can compare:
1. \(x_1 = -1\) and \(y_1 \approx -0.67\): Here, \(x_1 < y_1\).
2. \(x_1 = -1\) and \(y_2 = -1\): Here, \(x_1 = y_2\).
3. \(x_2 = -1.25\) and \(y_1 \approx -0.67\): Here, \(x_2 < y_1\).
4. \(x_2 = -1.25\) and \(y_2 = -1\): Here, \(x_2 < y_2\).

### Conclusion

From the comparisons, we can conclude that:
- \(x\) is less than or equal to \(y\).

Thus, the final answer is that \(x\) is smaller than or equal to \(y\).