For which interval of values of a will the eq | vedclass

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

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$(1, 2)$

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(C) For a quadratic equation $Ax^2 + Bx + C = 0$ to have roots of opposite signs,the product of the roots must be negative,i.e.,$\frac{C}{A} Here,$A = 3$ and $C = a^2 - 3a + 2$.So,$\frac{a^2 - 3a + 2}{3} This implies $a^2 - 3a + 2 Factoring the quadratic expression,we get $(a - 1)(a - 2) The inequality holds when $a$ lies between the roots of the quadratic expression $(a - 1)(a - 2) = 0$.Thus,$1 < a < 2$,which can be written as the interval $(1, 2)$.

For which interval of values of $a$ will the equation $3x^2 + 2(a^2 + 1)x + (a^2 - 3a + 2) = 0$ have roots of opposite signs?

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