Key Concepts and Formulas

• Discriminant of a Quadratic: For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is $D = B^2 - 4AC$. The roots are real if and only if $D \ge 0$.
• Sum and Product of Roots: For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $-\frac{B}{A}$ and the product of the roots is $\frac{C}{A}$.
• Conditions for Positive Roots: For a quadratic to have two positive roots, the following conditions must be met: $D \ge 0$, $-\frac{B}{A} > 0$, and $\frac{C}{A} > 0$.

Step 1: Identify the Coefficients

We are given the equation $(1-a)x^2 + 2(a-3)x + 9 = 0$. Comparing this to the general quadratic form $Ax^2 + Bx + C = 0$, we identify the coefficients: $A = 1-a$ $B = 2(a-3)$ $C = 9$

Step 2: Apply the Discriminant Condition ($D \ge 0$)

For the roots to be real, the discriminant must be non-negative: $D = B^2 - 4AC \ge 0$ Substitute the coefficients: $[2(a-3)]^2 - 4(1-a)(9) \ge 0$ $4(a^2 - 6a + 9) - 36(1-a) \ge 0$ Divide by 4: $a^2 - 6a + 9 - 9(1-a) \ge 0$ $a^2 - 6a + 9 - 9 + 9a \ge 0$ $a^2 + 3a \ge 0$ $a(a+3) \ge 0$

The critical points are $a=0$ and $a=-3$. Testing the intervals $(-\infty, -3)$, $(-3, 0)$, and $(0, \infty)$, we find the inequality holds for $a \le -3$ or $a \ge 0$. Therefore: $a \in (-\infty, -3] \cup [0, \infty) \quad (1)$

Step 3: Apply the Sum of Roots Condition ($-\frac{B}{A} > 0$)

For the roots to be positive, their sum must be positive: $-\frac{B}{A} > 0$ $-\frac{2(a-3)}{1-a} > 0$ $\frac{2(a-3)}{a-1} > 0$ $\frac{a-3}{a-1} > 0$

The critical points are $a=1$ and $a=3$. Testing the intervals $(-\infty, 1)$, $(1, 3)$, and $(3, \infty)$, we find the inequality holds for $a < 1$ or $a > 3$. Therefore: $a \in (-\infty, 1) \cup (3, \infty) \quad (2)$

Step 4: Apply the Product of Roots Condition ($\frac{C}{A} > 0$)

For the roots to be positive, their product must be positive: $\frac{C}{A} > 0$ $\frac{9}{1-a} > 0$ Since the numerator is positive, the denominator must also be positive: $1-a > 0$ $a < 1$ Therefore: $a \in (-\infty, 1) \quad (3)$

Step 5: Find the Intersection of the Solution Sets

We need to find the intersection of the intervals from (1), (2), and (3): $( (-\infty, -3] \cup [0, \infty) ) \cap ( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1)$

First, find the intersection of $( (-\infty, 1) \cup (3, \infty) )$ and $(-\infty, 1)$: $( (-\infty, 1) \cup (3, \infty) ) \cap (-\infty, 1) = (-\infty, 1)$

Now, intersect this result with $(-\infty, -3] \cup [0, \infty)$: $(-\infty, -3] \cup [0, \infty) ) \cap (-\infty, 1) = (-\infty, -3] \cup [0, 1)$

Thus, the solution set for $a$ is $(-\infty, -3] \cup [0, 1)$.

Step 6: Determine the Values of $\alpha$, $\beta$, and $\gamma$

We are given that the solution set is $(-\infty, -\alpha] \cup [\beta, \gamma)$. Comparing this to our result, we have: $-\alpha = -3 \Rightarrow \alpha = 3$ $\beta = 0$ $\gamma = 1$

Step 7: Calculate $2\alpha + \beta + \gamma$

$2\alpha + \beta + \gamma = 2(3) + 0 + 1 = 6 + 0 + 1 = 7$

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when manipulating inequalities, especially when multiplying or dividing by negative numbers.
• Critical Points: When solving rational inequalities, remember to consider both the zeros of the numerator and the denominator as critical points.
• Intersection vs. Union: Ensure you are using the correct set operation (intersection or union) based on the problem's requirements. In this case, all conditions must be satisfied, so we need the intersection.

Summary

The problem required finding the values of $a$ for which the given quadratic equation has positive roots. This was achieved by applying the conditions for real and positive roots: $D \ge 0$, $-\frac{B}{A} > 0$, and $\frac{C}{A} > 0$. Solving the resulting inequalities and finding their intersection yielded the solution set for $a$. Finally, the values of $\alpha$, $\beta$, and $\gamma$ were determined, and $2\alpha + \beta + \gamma$ was calculated.

The final answer is \boxed{7}.