Multiple choice

The set of values of $\displaystyle p$ for which the roots of the equation $\displaystyle 3x^{2} + 2x + p\left ( p-1 \right ) = 0$ have opposite signs is

  1. $\displaystyle \left ( -1, -\infty \right )$
  2. $\displaystyle \left ( -\infty, \infty \right )$
  3. $\displaystyle \left ( 2, 6 \right )$
  4. $\displaystyle \left ( 0,  1 \right )$
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D Correct answer
Explanation

For roots to have opposite signs, the product of the roots (c/a) must be negative. Here, p(p-1)/3 < 0, which implies 0 < p < 1.

AI explanation

For the roots of the quadratic equation 3x squared + 2x + p(p - 1) = 0 to have opposite signs, the product of the roots must be less than zero. Using the relation between roots and coefficients, the product of the roots is p(p - 1) divided by 3. This implies p(p - 1) < 0, which gives the interval (0, 1).

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