If the set of all , for which the roots of the equation are positive is , then is equal to __________. [2025]

Question

If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

Smart Achievers · Accepted Answer

(7)

Let $x_{1}$ and $x_{2}$ be the roots of $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ .

Since, $x_{1}, x_{2} > 0$

$∴ x_{1} + x_{2} > 0 and x_{1} x_{2} > 0$

$\Rightarrow \frac{- 2 (a - 3)}{1 - a} > 0 \Rightarrow \frac{a - 3}{1 - a} < 0$

$\Rightarrow a \in (- \infty, 1) \cup (3, \infty)$

Also, $x_{1} x_{2} > 0 \Rightarrow \frac{9}{1 - a} > 0 \Rightarrow 1 - a > 0 \Rightarrow a < 1$

On combining both conditions, we get $a \in (- \infty, 1)$

Now, for real roots, discriminant must be non negative.

i.e., $(2 {(a - 3))}^{2} - 4 (1 - a) 9 \geq 0$

$\Rightarrow 4 (a^{2} - 6 a + 9) - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} - 24 a + 36 - 36 + 36 a \geq 0$

$\Rightarrow 4 a^{2} + 12 a \geq 0 \Rightarrow a (a + 3) \geq 0$

$\Rightarrow a \in (- \infty, - 3] \cup [0, \infty)$

Combining all the conditions, we get $a \in (- \infty, - 3] \cup [0, 1)$

$∴ α = 3, β = 0 and γ = 1$

$∴ 2 α + β + γ = 2 \times 3 + 0 + 1 = 7$

Solution

Q.
If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]

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Solution

Q. 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 __________. [2025]

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Q.
If the set of all $a \in R - {1}$ , for which the roots of the equation $(1 - a) x^{2} + 2 (a - 3) x + 9 = 0$ are positive is $(- \infty, - α] \cup [β, γ)$ , then $2 α + β + γ$ is equal to __________. [2025]