In each of these questions, two equations I. and II are given. You have to solve both the equations and give answer
I. ` 5 x^(2) + 31 x + 48 = 0 `
II. ` 3 y ^(2) + 27 y + 42 = 0 `

Options

A. If ` x gt y `
B. If ` x ge y`
C. If ` x lt y `
D. if x = y or relationship between x and y cannot be established

Answer

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

### Step 1: Solve the first equation \( 5x^2 + 31x + 48 = 0 \)

1. **Identify coefficients**:
- \( a = 5 \)
- \( b = 31 \)
- \( c = 48 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 31^2 - 4 \cdot 5 \cdot 48 = 961 - 960 = 1 \]

4. **Substitute values into the quadratic formula**:
\[ x = \frac{-31 \pm \sqrt{1}}{2 \cdot 5} = \frac{-31 \pm 1}{10} \]

5. **Calculate the two possible values for \( x \)**:
- First value:
\[ x_1 = \frac{-31 + 1}{10} = \frac{-30}{10} = -3 \]
- Second value:
\[ x_2 = \frac{-31 - 1}{10} = \frac{-32}{10} = -\frac{16}{5} \]

### Step 2: Solve the second equation \( 3y^2 + 27y + 42 = 0 \)

1. **Identify coefficients**:
- \( a = 3 \)
- \( b = 27 \)
- \( c = 42 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 27^2 - 4 \cdot 3 \cdot 42 = 729 - 504 = 225 \]

4. **Substitute values into the quadratic formula**:
\[ y = \frac{-27 \pm \sqrt{225}}{2 \cdot 3} = \frac{-27 \pm 15}{6} \]

5. **Calculate the two possible values for \( y \)**:
- First value:
\[ y_1 = \frac{-27 + 15}{6} = \frac{-12}{6} = -2 \]
- Second value:
\[ y_2 = \frac{-27 - 15}{6} = \frac{-42}{6} = -7 \]

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

- The values obtained are:
- \( x_1 = -3 \)
- \( x_2 = -\frac{16}{5} \) (which is approximately -3.2)
- \( y_1 = -2 \)
- \( y_2 = -7 \)

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

1. **Comparing \( x_1 \) and \( y_1 \)**:
- \( -3 \) (from \( x \)) is less than \( -2 \) (from \( y \)), so \( x_1 < y_1 \).

2. **Comparing \( x_1 \) and \( y_2 \)**:
- \( -3 \) (from \( x \)) is greater than \( -7 \) (from \( y \)), so \( x_1 > y_2 \).

3. **Comparing \( x_2 \) and \( y_1 \)**:
- \( -\frac{16}{5} \) (which is approximately -3.2) is less than \( -2 \) (from \( y \)), so \( x_2 < y_1 \).

4. **Comparing \( x_2 \) and \( y_2 \)**:
- \( -\frac{16}{5} \) (approximately -3.2) is greater than \( -7 \) (from \( y \)), so \( x_2 > y_2 \).

### Conclusion

From the comparisons:
- \( x_1 < y_1 \)
- \( x_1 > y_2 \)
- \( x_2 < y_1 \)
- \( x_2 > y_2 \)

Thus, we can conclude that there is no consistent relationship that can be established between \( x \) and \( y \).

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