In each question two equations numbered I and II are given . you have to solve both the equations and mark the appropriate option
I `3x^2+8x+4=0`
II `6y^2+7y+2=0`

Options

A. The relation between x and y cannot be established.
B. `xlty`
C. `xgty`
D. `xley`

Answer

To solve the equations given in the question, we will follow these steps:

### Step 1: Solve the first equation \(3x^2 + 8x + 4 = 0\)

1. Identify the coefficients:
- \(a = 3\), \(b = 8\), \(c = 4\).

2. Calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 8^2 - 4 \cdot 3 \cdot 4 = 64 - 48 = 16. \]

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

4. Use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-8 \pm \sqrt{16}}{2 \cdot 3} = \frac{-8 \pm 4}{6}. \]

5. Calculate the two values of \(x\):
- First root:
\[ x_1 = \frac{-8 + 4}{6} = \frac{-4}{6} = -\frac{2}{3} \approx -0.67. \]
- Second root:
\[ x_2 = \frac{-8 - 4}{6} = \frac{-12}{6} = -2. \]

### Step 2: Solve the second equation \(6y^2 + 7y + 2 = 0\)

1. Identify the coefficients:
- \(a = 6\), \(b = 7\), \(c = 2\).

2. Calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 6 \cdot 2 = 49 - 48 = 1. \]

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

4. Use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{1}}{2 \cdot 6} = \frac{-7 \pm 1}{12}. \]

5. Calculate the two values of \(y\):
- First root:
\[ y_1 = \frac{-7 + 1}{12} = \frac{-6}{12} = -\frac{1}{2} \approx -0.5. \]
- Second root:
\[ y_2 = \frac{-7 - 1}{12} = \frac{-8}{12} = -\frac{2}{3} \approx -0.67. \]

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

1. The values obtained are:
- For \(x\): \(x_1 = -\frac{2}{3} \approx -0.67\) and \(x_2 = -2\).
- For \(y\): \(y_1 = -\frac{1}{2} \approx -0.5\) and \(y_2 = -\frac{2}{3} \approx -0.67\).

2. Now, we compare:
- \(x_1 = -\frac{2}{3}\) and \(y_1 = -\frac{1}{2}\): Here, \(y_1\) is greater than \(x_1\) (since -0.5 is greater than -0.67).
- \(x_1 = -\frac{2}{3}\) and \(y_2 = -\frac{2}{3}\): They are equal.
- \(x_2 = -2\) and \(y_1 = -\frac{1}{2}\): Here, \(x_2\) is less than \(y_1\).
- \(x_2 = -2\) and \(y_2 = -\frac{2}{3}\): Here, \(x_2\) is less than \(y_2\).

### Conclusion

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

### Final Answer

The correct option is that \(x\) is either smaller than or equal to \(y\).

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