If the set of all a∈R∖{1}, for which the roots of the equation (1−a)x2+2(a−3)x+9=0 are positive is (−∞,−α]∪[β,γ), then 2α+β+γ is equal to:
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Step-by-Step Solution
Step 1: Analyze conditions for positive roots
For a quadratic equation Ax2+Bx+C=0 to have positive roots, three conditions must be met: the discriminant D≥0, the sum of the roots -B/A > 0, and the product of the roots C/A > 0. In our given equation, A=(1−a), B=2(a−3), and C=9.
Step 2: Apply Discriminant Condition
First, we apply the discriminant condition D≥0. Substituting the coefficients, we get 4(a−3)2−36(1−a)≥0. Dividing by 4 and expanding, we have (a2−6a+9)−9+9a≥0, which simplifies to a2+3a≥0. Factoring gives a(a+3)≥0. This inequality holds when a∈(−∞,−3]∪[0,∞).
Step 3: Apply Sum of Roots Condition
Next, we apply the sum of roots condition, -B/A > 0. Substituting the coefficients, we get -2(a-3)/(1-a) > 0. This inequality can be rewritten as 2(a-3)/(a-1) > 0. This inequality holds when a∈(−∞,1)∪(3,∞).
Step 4: Apply Product of Roots Condition
Finally, we apply the product of roots condition, C/A > 0. Substituting the coefficients, we get 9/(1−a)>0. This inequality holds when 1−a>0, which means a<1. So, a∈(−∞,1).
Step 5: Find the intersection of all conditions
To find the set of all a for which the roots are positive, we need to find the intersection of the solution sets from all three conditions. The intersection is (−∞,−3]. Comparing this with the given form (−∞,−α]∪[β,γ), we find that −α=−3, so α=3. The union part [β,γ) is empty, which means β and γ are not defined in this context, or we can consider the interval to be (−∞,−3] and the union part to be empty. However, the problem implies a specific form. Let's re-evaluate the intersection carefully. The intersection of (−∞,−3]∪[0,∞) and (−∞,1)∪(3,∞) is (−∞,−3]∪[0,1)∪(3,∞). Intersecting this with (−∞,1) gives (−∞,−3]∪[0,1).
Step 6: Determine α,β,γ and calculate 2α+β+γ
From the intersection (−∞,−3]∪[0,1), we compare it with the given form (−∞,−α]∪[β,γ). This implies −α=−3, so α=3. Also, β=0 and γ=1. Now we can calculate 2α+β+γ=2(3)+0+1=6+0+1=7.