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(A) For the roots of the quadratic equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ to be positive,we require: $1$. Discriminant $D \geq 0$: $D = [2(a-3)]^2 - 4(

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$7$

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(A) For the roots of the quadratic equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ to be positive,we require: $1$. Discriminant $D \geq 0$: $D = [2(a-3)]^2 - 4(1-a)(9) \geq 0$ $4(a^2 - 6a + 9) - 36(1-a) \geq 0$ $a^2 - 6a + 9 - 9 + 9a \geq 0$ $a^2 + 3a \geq 0 \implies a(a+3) \geq 0$ $a \in (-\infty, -3] \cup [0, \infty)$ $2$. Sum of roots $> 0$: $-\frac{b}{a} = -\frac{2(a-3)}{1-a} = \frac{2(a-3)}{a-1} > 0$ $a \in (-\infty, 1) \cup (3, \infty)$ $3$. Product of roots $> 0$: $\frac{c}{a} = \frac{9}{1-a} > 0 \implies 1-a > 0 \implies a Taking the intersection of all conditions: $a \in (-\infty, -3] \cup [0, 1)$ Comparing with $(-\infty, -\alpha] \cup [\beta, \gamma)$,we get $\alpha = 3, \beta = 0, \gamma = 1$. Thus,$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 7$.

If the set of all $a \in R - \{1\}$,for which the roots of the equation $(1-a)x^2 + 2(a-3)x + 9 = 0$ are positive,is $(-\infty, -\alpha] \cup [\beta, \gamma)$,then $2\alpha + \beta + \gamma$ is equal to . . . . . . .

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