For the two given questions I and II
`3p^2+ 2p – 1 = 0`
`2q^2 + 7q + 6 = 0`

Options

A. if p is greater than q.
B. if p is smaller than q
C. if p is equal to q
D. if p is either equal to or smaller than q

Answer

To solve the equations \(3p^2 + 2p - 1 = 0\) and \(2q^2 + 7q + 6 = 0\), we will find the roots of each equation step by step.

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

To factor the quadratic equation, we look for two numbers that multiply to \(3 \times -1 = -3\) and add to \(2\). The numbers \(3\) and \(-1\) fit this requirement.

Rewriting the equation:
\[ 3p^2 + 3p - p - 1 = 0 \]

### Step 2: Group the terms
Group the first two terms and the last two terms:
\[ (3p^2 + 3p) + (-p - 1) = 0 \]

### Step 3: Factor by grouping
Factor out the common terms:
\[ 3p(p + 1) - 1(p + 1) = 0 \]
Now, factor out \((p + 1)\):
\[ (3p - 1)(p + 1) = 0 \]

### Step 4: Find the roots for \(p\)
Set each factor to zero:
1. \(3p - 1 = 0 \Rightarrow p = \frac{1}{3}\)
2. \(p + 1 = 0 \Rightarrow p = -1\)

Thus, the roots for \(p\) are:
\[ p = \frac{1}{3}, \quad p = -1 \]

### Step 5: Solve the second equation \(2q^2 + 7q + 6 = 0\)

We need to find two numbers that multiply to \(2 \times 6 = 12\) and add to \(7\). The numbers \(3\) and \(4\) fit this requirement.

Rewriting the equation:
\[ 2q^2 + 3q + 4q + 6 = 0 \]

### Step 6: Group the terms
Group the first two terms and the last two terms:
\[ (2q^2 + 3q) + (4q + 6) = 0 \]

### Step 7: Factor by grouping
Factor out the common terms:
\[ q(2q + 3) + 2(2q + 3) = 0 \]
Now, factor out \((2q + 3)\):
\[ (2q + 3)(q + 2) = 0 \]

### Step 8: Find the roots for \(q\)
Set each factor to zero:
1. \(2q + 3 = 0 \Rightarrow q = -\frac{3}{2}\)
2. \(q + 2 = 0 \Rightarrow q = -2\)

Thus, the roots for \(q\) are:
\[ q = -\frac{3}{2}, \quad q = -2 \]

### Step 9: Compare the values of \(p\) and \(q\)
The roots for \(p\) are \(\frac{1}{3}\) and \(-1\), and the roots for \(q\) are \(-\frac{3}{2}\) and \(-2\).

### Step 10: Determine which is greater
Comparing the maximum values:
- The maximum value of \(p\) is \(\frac{1}{3}\).
- The maximum value of \(q\) is \(-\frac{3}{2}\).

Since \(\frac{1}{3} > -\frac{3}{2}\), we conclude that:
\[ p > q \]

### Final Result
Thus, the condition \(p > q\) holds true.