In the following questions two equation numbered I and II are given You have to solve both equations and give answer
(i) `2x^(2)+11x+12=0`
(ii) `5y^(2)+27y+10=0`

Options

A. if x > y
B. if x ≥ y
C. if x < y
D. The relationshiup cant be eastsblished

Answer

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

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

1. **Identify the coefficients**:
- \( a = 2 \)
- \( b = 11 \)
- \( c = 12 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 11^2 - 4 \cdot 2 \cdot 12 = 121 - 96 = 25 \]

4. **Calculate the roots**:
\[ x = \frac{-11 \pm \sqrt{25}}{2 \cdot 2} = \frac{-11 \pm 5}{4} \]
This gives us two possible values for \( x \):
- \( x_1 = \frac{-11 + 5}{4} = \frac{-6}{4} = -\frac{3}{2} \)
- \( x_2 = \frac{-11 - 5}{4} = \frac{-16}{4} = -4 \)

### Step 2: Solve the second equation \( 5y^2 + 27y + 10 = 0 \)

1. **Identify the coefficients**:
- \( a = 5 \)
- \( b = 27 \)
- \( c = 10 \)

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

3. **Calculate the discriminant**:
\[ b^2 - 4ac = 27^2 - 4 \cdot 5 \cdot 10 = 729 - 200 = 529 \]

4. **Calculate the roots**:
\[ y = \frac{-27 \pm \sqrt{529}}{2 \cdot 5} = \frac{-27 \pm 23}{10} \]
This gives us two possible values for \( y \):
- \( y_1 = \frac{-27 + 23}{10} = \frac{-4}{10} = -\frac{2}{5} \)
- \( y_2 = \frac{-27 - 23}{10} = \frac{-50}{10} = -5 \)

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

We have the following values:
- From the first equation: \( x_1 = -\frac{3}{2} \) and \( x_2 = -4 \)
- From the second equation: \( y_1 = -\frac{2}{5} \) and \( y_2 = -5 \)

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

1. **Compare \( x_1 \) and \( y_1 \)**:
- \( -\frac{3}{2} \) (which is -1.5) is less than \( -\frac{2}{5} \) (which is -0.4).
- Thus, \( x_1 < y_1 \).

2. **Compare \( x_1 \) and \( y_2 \)**:
- \( -\frac{3}{2} < -5 \) is true.
- Thus, \( x_1 < y_2 \).

3. **Compare \( x_2 \) and \( y_1 \)**:
- \( -4 < -\frac{2}{5} \) is true.
- Thus, \( x_2 < y_1 \).

4. **Compare \( x_2 \) and \( y_2 \)**:
- \( -4 > -5 \) is true.
- Thus, \( x_2 > y_2 \).

### Conclusion:
Since we have \( x_1 < y_1 \) and \( x_2 > y_2 \), we cannot establish a consistent relationship between \( x \) and \( y \) across all values.

### Final Answer:
The relationship can't be established.

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