Question

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 ...........

Hint:

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.

Answer 1

Correct Answer: 7

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: \[2\alpha + \beta + \gamma = 7\]