If the roots of the equation $3x^{2}+2x+p(p-1)=0$ are of opposite signs, then :

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Accepted Answer

To determine why Option 2 ($p \in (0,1)$) is the correct answer, we need to analyze the condition given in the problem: the roots of the quadratic equation $3x^2 + 2x + p(p-1) = 0$ are of opposite signs. ### Explanation: 1. **Condition for Roots to be of Opposite Signs:** - For a quadratic equation $ax^2 + bx + c = 0$, the roots are of opposite signs if and only if the product of the roots is negative. The product of the roots, given by Vieta's formulas, is $\frac{c}{a}$. - In our equation, $a = 3$ and $c = p(p-1)$. Therefore, the product of the roots is $\frac{p(p-1)}{3}$. - For the roots to be of opposite signs, $\frac{p(p-1)}{3} < 0$. This implies $p(p-1) < 0$. 2. **Solving the Inequality $p(p-1) < 0$:** - The expression $p(p-1)$ is a quadratic expression in $p$. - The roots of the equation $p(p-1) = 0$ are $p = 0$ and $p = 1$. - The sign of the quadratic expression $p(p-1)$ changes at these roots. To determine the intervals where $p(p-1) < 0$, we test intervals around the roots: - For $p \in (-\infty, 0)$, $p(p-1) > 0$. - For $p \in (0, 1)$, $p(p-1) < 0$. - For $p \in (1, \infty)$, $p(p-1) > 0$. 3. **Conclusion:** - The inequality $p(p-1) < 0$ holds true only in the interval $p \in (0, 1)$. - Therefore, the correct option is Option 2: $p \in (0, 1)$. ### Why Other Options are Incorrect: 1. **Option 1: $p \in (-\infty, 0)$** - In this interval, $p(p-1) > 0$, which means the product of the roots is positive, not negative. Thus, the roots cannot be of opposite signs. 2. **Option 3: $p \in (1, \infty)$** - In this interval, $p(p-1) > 0$, which again means the product of the roots is positive. Hence, the roots cannot be of opposite signs. 3. **Option 4: $p \in (4, \infty)$** - This is a subset of Option 3 and similarly results in $p(p-1) > 0$, meaning the product of the roots is positive. Therefore, the roots cannot be of opposite signs. By following the logical steps and analysis above, we confirm that Option 2 is indeed the correct answer, as it is the only interval where the product of the roots is negative, satisfying the condition that the roots are of opposite signs.