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

Both the roots are positive.
For the quadratic equation \( (1 - a)x^2 + 2(a - 3)x + 9 = 0 \), we use the discriminant condition: \[ D \geq 0 \] This condition is satisfied for: \[ 4(a - 3)^2 - 4 \cdot (a - 3) \cdot 9 \geq 0 \] \[ a^2 - 6a + 9 + 9a + 9 \geq 0 \] \[ a^2 + 3a \geq 0 \] This gives: \[ a(a + 3) \geq 0 \quad \text{(Equation (i))} \] Now solving for \( a \): \[ a \in (-\infty, -3] \cup [0, \infty) \] Next, we apply the condition for the sum and product of the roots: \[ -\frac{b}{2a}>0 \] This gives: \[ \frac{2(a - 3)}{2(a - 1)}>0 \] Which implies: \[ a \in (-\infty, 1) \quad \text{(Equation (ii))} \] Therefore, combining these two conditions: \[ a \in (-\infty, -3] \cup [0, 1) \] Substituting into the given equation: \[ 2\alpha + \beta + \gamma = 7 \] Thus, the final value is: \[ 2\alpha + \beta + \gamma = 7 \]