Mathematics · Quadratic Equations

JEE Main 2025 — 2 April, Evening Shift — Question 43

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 $(-\infty,-\alpha] \cup[\beta, \gamma)$ , then $2 \alpha+\beta+\gamma$ is equal to $\_\_\_\_$ .

Answer: 7

Numerical answer — enter this value.

$f(x)=(1-a) x^{2}+2(a-3) x+9, f(0)=9>0$

D \geq 0 \Rightarrow 4(a-3)^{2} \geq 4(1-a) \cdot 9

\begin{gathered} \Rightarrow a \in(-\infty,-3] \cup[0, \infty)…(i) \end{gathered}

\begin{gathered} x_{1}+x_{2}=\frac{-2(a-3)}{1-a}, x_{1} x_{2}=\frac{9}{1-a} …(ii) \end{gathered}

$x_{1}+x_{2}>0 \Rightarrow \frac{a-3}{a-1}>0 \Rightarrow a \in(-\infty, 1) \cup(3, \infty)$ .

\begin{gathered} x_{1} x_{2}>0 \Rightarrow 1-a>0 \Rightarrow a \in(-\infty, 1) . …(iii) \end{gathered}

$\Rightarrow$ Interaction of (i), (ii) and (iii)

\begin{aligned} & a \in(-\infty,-3] \cup[0,1) \\& \Rightarrow \alpha=3, \beta=0, \gamma=1 \Rightarrow 2 \alpha+\beta+\gamma=7 \end{aligned}

Answer key and solution verified before publishing.

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