(i) ` 12x^(2) + 11x + 2 = 0`
(ii) ` 25y^(2) + 15y+2 = 0`

Options

A. ` x ge y`
B. x = y or no relation can be established between x & y.
C. ` x lt y`
D. `x le y`

Explanation

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

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

We start with the quadratic equation:

\[ 12x^2 + 11x + 2 = 0 \]

To factor this, we need two numbers that multiply to \( 12 \times 2 = 24 \) and add up to \( 11 \). The numbers that satisfy this condition are \( 3 \) and \( 8 \).

Now we can rewrite the middle term:

\[ 12x^2 + 8x + 3x + 2 = 0 \]

Next, we group the terms:

\[ (12x^2 + 8x) + (3x + 2) = 0 \]

Factoring out the common terms in each group:

\[ 4x(3x + 2) + 1(3x + 2) = 0 \]

Now we can factor out \( (3x + 2) \):

\[ (3x + 2)(4x + 1) = 0 \]

Setting each factor to zero gives us:

1. \( 3x + 2 = 0 \) → \( x = -\frac{2}{3} \)
2. \( 4x + 1 = 0 \) → \( x = -\frac{1}{4} \)

### Step 2: Solve the second equation \( 25y^2 + 15y + 2 = 0 \)

Now we move to the second quadratic equation:

\[ 25y^2 + 15y + 2 = 0 \]

We need two numbers that multiply to \( 25 \times 2 = 50 \) and add up to \( 15 \). The numbers that satisfy this condition are \( 5 \) and \( 10 \).

Rewriting the middle term:

\[ 25y^2 + 5y + 10y + 2 = 0 \]

Grouping the terms:

\[ (25y^2 + 5y) + (10y + 2) = 0 \]

Factoring out the common terms:

\[ 5y(5y + 1) + 2(5y + 1) = 0 \]

Factoring out \( (5y + 1) \):

\[ (5y + 1)(5y + 2) = 0 \]

Setting each factor to zero gives us:

1. \( 5y + 1 = 0 \) → \( y = -\frac{1}{5} \)
2. \( 5y + 2 = 0 \) → \( y = -\frac{2}{5} \)

### Summary of Solutions

The solutions to the equations are:

- For \( x \): \( x = -\frac{2}{3} \) or \( x = -\frac{1}{4} \)
- For \( y \): \( y = -\frac{1}{5} \) or \( y = -\frac{2}{5} \)