Multiple choice

The values of $a$ for which the equation $3x^2 + 2(a^2 -3a + 2) = 0$ will have roots of opposite sign lie in the interval

  1. $(-\infty, 1)$
  2. $(-\infty, 0)$
  3. $(1, 2)$
  4. $\displaystyle{\left( \frac{3}{2}, 2 \right)}$
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C Correct answer
Explanation

For the quadratic equation 3x^2 + 2(a^2 - 3a + 2) = 0 to have roots of opposite signs, the product of the roots (c/a) must be negative. Here, c = 2(a^2 - 3a + 2) and a = 3. So, 2(a^2 - 3a + 2) / 3 < 0, which means a^2 - 3a + 2 < 0. Factoring gives (a-1)(a-2) < 0. This inequality holds when 1 < a < 2.

AI explanation

For the quadratic equation 3x^2 + 0x + 2(a^2 - 3a + 2) = 0 to have roots of opposite signs, the constant term must be negative. This requires 2(a^2 - 3a + 2) < 0. Factoring the quadratic expression gives 2(a - 1)(a - 2) < 0, which means a must lie between 1 and 2. Therefore, the interval for a is (1, 2).

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