In each of the following questions, two equations (I) and (II) are given. Solve the equations and mark the correct option
I. ` 2 x^(2) + 11 x + 12 = 0`
II. ` 8 y^(2) - 22 y - 21 = 0`

Options

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

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation \( 2x^2 + 11x + 12 = 0 \)

1. **Factor the quadratic equation**:
We need to factor the equation \( 2x^2 + 11x + 12 \). We look for two numbers that multiply to \( 2 \times 12 = 24 \) and add to \( 11 \). The numbers \( 8 \) and \( 3 \) work because \( 8 + 3 = 11 \) and \( 8 \times 3 = 24 \).

So, we can rewrite the equation as:
\[ 2x^2 + 8x + 3x + 12 = 0 \]
Grouping the terms:
\[ (2x^2 + 8x) + (3x + 12) = 0 \]
Factoring out common terms:
\[ 2x(x + 4) + 3(x + 4) = 0 \]
Now factor out \( (x + 4) \):
\[ (x + 4)(2x + 3) = 0 \]

2. **Set each factor to zero**:
\[ x + 4 = 0 \quad \Rightarrow \quad x = -4 \]
\[ 2x + 3 = 0 \quad \Rightarrow \quad x = -\frac{3}{2} \]

### Step 2: Solve the second equation \( 8y^2 - 22y - 21 = 0 \)

1. **Factor the quadratic equation**:
We need to factor \( 8y^2 - 22y - 21 \). We look for two numbers that multiply to \( 8 \times (-21) = -168 \) and add to \( -22 \). The numbers \( -28 \) and \( 6 \) work because \( -28 + 6 = -22 \) and \( -28 \times 6 = -168 \).

So, we can rewrite the equation as:
\[ 8y^2 - 28y + 6y - 21 = 0 \]
Grouping the terms:
\[ (8y^2 - 28y) + (6y - 21) = 0 \]
Factoring out common terms:
\[ 4y(2y - 7) + 3(2y - 7) = 0 \]
Now factor out \( (2y - 7) \):
\[ (2y - 7)(4y + 3) = 0 \]

2. **Set each factor to zero**:
\[ 2y - 7 = 0 \quad \Rightarrow \quad y = \frac{7}{2} \]
\[ 4y + 3 = 0 \quad \Rightarrow \quad y = -\frac{3}{4} \]

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

We have the following values:
- For \( x \): \( -4 \) and \( -\frac{3}{2} \)
- For \( y \): \( \frac{7}{2} \) and \( -\frac{3}{4} \)

Now we will compare the values:
1. Compare \( x = -4 \) with \( y = \frac{7}{2} \) and \( y = -\frac{3}{4} \):
- \( -4 < \frac{7}{2} \) (True)
- \( -4 < -\frac{3}{4} \) (True)

2. Compare \( x = -\frac{3}{2} \) with \( y = \frac{7}{2} \) and \( y = -\frac{3}{4} \):
- \( -\frac{3}{2} < \frac{7}{2} \) (True)
- \( -\frac{3}{2} > -\frac{3}{4} \) (False)

### Conclusion

From the comparisons, we can conclude that in all cases, \( x \) is less than \( y \). Therefore, the final relation is:
\[ x < y \]