If the set of all $ a \in \mathbb{R} \setminus \{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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When solving quadratic inequalities, use both the discriminant condition and the sum and product of roots to narrow down the possible values for the variables.
The roots are both positive. For the quadratic equation \( (1 - a)x^2 + 2(a - 3)x + 9 = 0 \), the discriminant condition \( D \geq 0 \) must be met.
\[4(a - 3)^2 - 4 \cdot (a - 3) \cdot 9 \geq 0\]
\[a^2 - 6a + 9 + 9a + 9 \geq 0\]
\[a^2 + 3a \geq 0\]
This simplifies to:
\[a(a + 3) \geq 0 \quad \text{(Equation (i))}\]
Solving for \( a \) yields:
\[a \in (-\infty, -3] \cup [0, \infty)\]
Next, the conditions for the sum and product of the roots are applied. The sum of the roots must be positive:
\[-\frac{b}{2a}>0\]
This translates to:
\[\frac{2(a - 3)}{2(a - 1)}>0\]
Which implies:
\[a \in (-\infty, 1) \quad \text{(Equation (ii))}\]
Combining both conditions gives the valid range for \( a \):
\[a \in (-\infty, -3] \cup [0, 1)\]
Substituting into the given equation:
\[2\alpha + \beta + \gamma = 7\]
The final value is:
\[2\alpha + \beta + \gamma = 7\]
