If `(p^(2) + 2)x^(2) + 2p^(2)x + (p^(2) - 4) = 0` has roots of opposite sign, then find the range of p.

Options

A. (-2, 2]
B. [-2, 2]
C. (-2, 3)
D. (-2, 2)

Answer

To solve the problem, we need to determine the range of \( p \) for which the quadratic equation

\[ (p^2 + 2)x^2 + (2p^2)x + (p^2 - 4) = 0 \]

has roots of opposite signs.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = p^2 + 2 \)
- \( b = 2p^2 \)
- \( c = p^2 - 4 \)

2. **Condition for opposite sign roots**:
For the roots of the quadratic to be of opposite signs, the product of the roots must be negative. The product of the roots can be given by the formula \( \frac{c}{a} \). Therefore, we need:

\[ \frac{c}{a} < 0 \]

This translates to:

\[ \frac{p^2 - 4}{p^2 + 2} < 0 \]

3. **Analyze the inequality**:
The inequality \( \frac{p^2 - 4}{p^2 + 2} < 0 \) implies that the numerator and denominator must have opposite signs.

- The denominator \( p^2 + 2 \) is always positive since \( p^2 \) is non-negative and adding 2 keeps it positive.
- Therefore, we need the numerator \( p^2 - 4 \) to be negative:

\[ p^2 - 4 < 0 \]

4. **Solve the inequality**:
Rearranging the inequality gives:

\[ p^2 < 4 \]

Taking the square root of both sides results in:

\[ -2 < p < 2 \]

5. **Conclusion**:
The range of \( p \) for which the quadratic equation has roots of opposite signs is:

\[ p \in (-2, 2) \]

### Final Answer:
The range of \( p \) is \( (-2, 2) \).