I.` 3p^(2)+10p+8 = 0`
II.` 2q^(2) + 3q+1 = 0`

Options

A. If x = y or no relation can be established
B. If`x gt y`
C. If ` x lt y`
D. If`x ge y`

Explanation

To solve the equations and find the relationship between \( p \) and \( q \), we will follow these steps:

### Step 1: Solve the first equation \( 3p^2 + 10p + 8 = 0 \)

To solve this quadratic equation, we can use the factorization method. We need to find two numbers that multiply to \( 3 \times 8 = 24 \) and add up to \( 10 \).

The numbers that satisfy this are \( 6 \) and \( 4 \).

Now we can rewrite the equation:

\[ 3p^2 + 6p + 4p + 8 = 0 \]

Next, we group the terms:

\[ (3p^2 + 6p) + (4p + 8) = 0 \]

Factoring out the common terms:

\[ 3p(p + 2) + 4(p + 2) = 0 \]

Now we can factor by grouping:

\[ (3p + 4)(p + 2) = 0 \]

Setting each factor to zero gives us:

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

### Step 2: Solve the second equation \( 2q^2 + 3q + 1 = 0 \)

Similarly, we will factor this quadratic equation. We need to find two numbers that multiply to \( 2 \times 1 = 2 \) and add up to \( 3 \).

The numbers that satisfy this are \( 2 \) and \( 1 \).

Now we can rewrite the equation:

\[ 2q^2 + 2q + q + 1 = 0 \]

Next, we group the terms:

\[ (2q^2 + 2q) + (q + 1) = 0 \]

Factoring out the common terms:

\[ 2q(q + 1) + 1(q + 1) = 0 \]

Now we can factor by grouping:

\[ (2q + 1)(q + 1) = 0 \]

Setting each factor to zero gives us:

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

### Step 3: Determine the relationship between \( p \) and \( q \)

Now we have the values:

- For \( p \): \( -\frac{4}{3} \) and \( -2 \)
- For \( q \): \( -\frac{1}{2} \) and \( -1 \)

We can compare the values:

1. \( -\frac{4}{3} \) (approximately -1.33) is less than \( -1 \)
2. \( -2 \) is less than both \( -\frac{1}{2} \) and \( -1 \)

Thus, we can conclude that:

\[ p < q \]

### Final Answer

The relationship between \( p \) and \( q \) is:

\[ p < q \]