The value of a (a ge 3) for which the sum of | vedclass

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(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.From the given equation $x^2 - (a - 2)x + (a - 3) = 0$,we have:Sum of roots: $\alp

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$3$

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(A) Let the roots of the quadratic equation be $\alpha$ and $\beta$.From the given equation $x^2 - (a - 2)x + (a - 3) = 0$,we have:Sum of roots: $\alpha + \beta = a - 2$Product of roots: $\alpha \beta = a - 3$We need to minimize the sum of the cubes of the roots,$S = \alpha^3 + \beta^3$.Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$:$S = (a - 2)^3 - 3(a - 3)(a - 2)$$S = (a - 2) [(a - 2)^2 - 3(a - 3)]$$S = (a - 2) [a^2 - 4a + 4 - 3a + 9]$$S = (a - 2) (a^2 - 7a + 13)$$S = a^3 - 7a^2 + 13a - 2a^2 + 14a - 26$$S = a^3 - 9a^2 + 27a - 26$$S = (a - 3)^3 + 1$Since $a \ge 3$,the expression $(a - 3)^3 + 1$ is minimized when $(a - 3)^3 = 0$,which occurs at $a = 3$.

The value of $a (a \ge 3)$ for which the sum of the cubes of the roots of $x^2 - (a - 2)x + (a - 3) = 0$ assumes the least value is

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