If roots of the equation $3 x ^ { 2 } + 2 \left( a ^ { 2 } + 1 \right) x + \left( a ^ { 2 } - 3 a + 2 \right) = 0$ are of opposite signs, then a lies in the interval

For the quadratic equation 3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0 to have roots of opposite signs, the product of the roots must be strictly negative. Using Vieta's formulas, the product of the roots is c/a, which gives (a^2 - 3a + 2) / 3 < 0. For this fraction to be negative, the numerator a^2 - 3a + 2 must be less than zero, so we factor it to get (a - 1)(a - 2) < 0. This inequality holds true when a lies strictly between 1 and 2.

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