4(1-a) \cdot 9 \end{aligned}$$

$\Rightarrow a \in(-\infty,-3] \cup[0, \infty)\quad\text{..... (i)}$

JEE Main 2025 (Online) 2nd April Evening Shift Mathematics - Quadratic Equation and Inequalities Question 21 English Explanation

$$\begin{aligned} & x_1+x_2=\frac{-2(a-3)}{1-a}, x_1 x_2=\frac{9}{1-a} \\ & x_1+x_2>0 \Rightarrow \frac{a-3}{a-1}>0 \Rightarrow a \in(-\infty, 1) \cup(3, \infty) \ldots \text { (ii) } \end{aligned}$$

$x_1 x_2>0 \Rightarrow 1-a>0 \Rightarrow a \in(-\infty, 1) \ldots($ iii $)$

$$\begin{aligned} &\Rightarrow \text { Interaction of (i), (ii) and (iii) }\\ &\begin{aligned} & a \in(-\infty,-3] \cup[0,1) \\ \Rightarrow \quad & \alpha=3, \beta=0, \gamma=1 \Rightarrow 2 \alpha+\beta+\gamma=7 \end{aligned} \end{aligned}$$

19:T530,

$$\begin{aligned} & f(x)=(1-a) x^2+2(a-3) x+9, f(0)=9>0 \\ & D \geq 0 \Rightarrow 4(a-3)^2 \geq 4(1-a) \cdot 9 \end{aligned}$$

$\Rightarrow a \in(-\infty,-3] \cup[0, \infty)\quad\text{..... (i)}$

JEE Main 2025 (Online) 2nd April Evening Shift Mathematics - Quadratic Equation and Inequalities Question 21 Hindi Explanation

$$\begin{aligned} & x_1+x_2=\frac{-2(a-3)}{1-a}, x_1 x_2=\frac{9}{1-a} \\ & x_1+x_2>0 \Rightarrow \frac{a-3}{a-1}>0

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