The value of a for which the quadratic equati | vedclass

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(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots with opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} < 0$.He

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$(1, 2)$

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(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots with opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} Here,$A = 3$ and $C = a^2 - 3a + 2$.Thus,$\frac{a^2 - 3a + 2}{3} Factoring the quadratic expression,we get $(a - 1)(a - 2) This inequality holds when $1 Since the discriminant $D = B^2 - 4AC = [2(a^2 + 1)]^2 - 4(3)(a^2 - 3a + 2) = 4(a^4 + 2a^2 + 1) - 12(a^2 - 3a + 2) = 4(a^4 - a^2 + 9a - 5)$ is always positive for $a \in (1, 2)$,the roots are guaranteed to be real.Therefore,the value of $a$ lies in $(1, 2)$.

The value of $a$ for which the quadratic equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ possesses roots with opposite signs,lies in

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