The set of values of p for which the roots of the equation `3x^(2)+2x+p(p-1)=0` are of opposite signs is :

Options

A. `(-oo,0)`
B. `(0,1)`
C. `(1,oo)`
D. `(0,oo)`

Answer

To solve the problem, we need to find the set of values of \( p \) for which the roots of the equation

\[ 3x^2 + 2x + p(p-1) = 0 \]

are of opposite signs.

### Step 1: Understanding the Condition for Opposite Signs
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be of opposite signs, the product of the roots must be negative. If we denote the roots by \( \alpha \) and \( \beta \), then:

\[ \alpha \beta < 0 \]

### Step 2: Finding the Product of the Roots
Using Vieta's formulas, the product of the roots \( \alpha \) and \( \beta \) can be expressed as:

\[ \alpha \beta = \frac{c}{a} \]

For our equation, \( a = 3 \) and \( c = p(p-1) \). Thus, we have:

\[ \alpha \beta = \frac{p(p-1)}{3} \]

### Step 3: Setting Up the Inequality
To satisfy the condition for opposite signs, we need:

\[ \frac{p(p-1)}{3} < 0 \]

Multiplying both sides by 3 (which is positive and does not change the inequality):

\[ p(p-1) < 0 \]

### Step 4: Analyzing the Inequality
The inequality \( p(p-1) < 0 \) implies that the product of \( p \) and \( (p-1) \) is negative. This occurs when one of the factors is positive and the other is negative.

1. \( p < 0 \) and \( p - 1 > 0 \) (not possible since if \( p < 0 \), \( p - 1 < -1 < 0 \))
2. \( p > 0 \) and \( p - 1 < 0 \) (which simplifies to \( 0 < p < 1 \))

### Step 5: Conclusion
Thus, the set of values of \( p \) for which the roots of the equation are of opposite signs is:

\[ p \in (0, 1) \]

### Final Answer
The set of values of \( p \) is \( (0, 1) \).

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