`14p^2+27p+9=0`
`56q^2-3q-9=0`

Options

A. `pltq`
B. `pgtq`
C. `pleqq`
D. p = q

Answer

To solve the equations \(14p^2 + 27p + 9 = 0\) and \(56q^2 - 3q - 9 = 0\), we will find the values of \(p\) and \(q\) and then compare them.

### Step 1: Solve for \(p\) in the equation \(14p^2 + 27p + 9 = 0\)

1. **Factor the quadratic equation**:
We can rewrite the equation as:
\[ 14p^2 + 21p + 6p + 9 = 0 \]
Now, group the terms:
\[ (14p^2 + 21p) + (6p + 9) = 0 \]
Factor out common terms:
\[ 7p(2p + 3) + 3(2p + 3) = 0 \]
Now, factor out \((2p + 3)\):
\[ (7p + 3)(2p + 3) = 0 \]

2. **Find the values of \(p\)**:
Set each factor to zero:
\[ 7p + 3 = 0 \quad \Rightarrow \quad p = -\frac{3}{7} \]
\[ 2p + 3 = 0 \quad \Rightarrow \quad p = -\frac{3}{2} \]
Thus, the values of \(p\) are:
\[ p_1 = -\frac{3}{7}, \quad p_2 = -\frac{3}{2} \]

### Step 2: Solve for \(q\) in the equation \(56q^2 - 3q - 9 = 0\)

1. **Factor the quadratic equation**:
Rewrite the equation as:
\[ 56q^2 + 21q - 24q - 9 = 0 \]
Group the terms:
\[ (56q^2 + 21q) + (-24q - 9) = 0 \]
Factor out common terms:
\[ 7q(8q + 3) - 3(8q + 3) = 0 \]
Factor out \((8q + 3)\):
\[ (7q - 3)(8q + 3) = 0 \]

2. **Find the values of \(q\)**:
Set each factor to zero:
\[ 7q - 3 = 0 \quad \Rightarrow \quad q = \frac{3}{7} \]
\[ 8q + 3 = 0 \quad \Rightarrow \quad q = -\frac{3}{8} \]
Thus, the values of \(q\) are:
\[ q_1 = \frac{3}{7}, \quad q_2 = -\frac{3}{8} \]

### Step 3: Compare the values of \(p\) and \(q\)

Now we will compare the values of \(p\) and \(q\):

- For \(p_1 = -\frac{3}{7}\) and \(p_2 = -\frac{3}{2}\):
- \(p_1 = -0.428\) and \(p_2 = -1.5\)

- For \(q_1 = \frac{3}{7}\) and \(q_2 = -\frac{3}{8}\):
- \(q_1 = 0.428\) and \(q_2 = -0.375\)

### Conclusion

1. Comparing \(p_1\) and \(q_1\):
\[ -\frac{3}{7} < \frac{3}{7} \quad \Rightarrow \quad p_1 < q_1 \]

2. Comparing \(p_2\) and \(q_2\):
\[ -\frac{3}{2} < -\frac{3}{8} \quad \Rightarrow \quad p_2 < q_2 \]

Thus, in both cases, we find that \(p < q\).

### Final Answer
The relation between \(p\) and \(q\) is:
\[ p < q \]