JEE Main 2025 — Quadratic Equation Question with Solution
JEE Main 2025 (2 Apr Shift 2)
Question
If the set of all , for which the roots of the equation are positive is , then is equal to _______ .
Enter your answer
Show full solutionCorrect answer: 7
Correct answer
7
Step-by-step explanation
Both the roots are positive
$\begin{aligned}
& D \geq 0 \\ & 4(a-3)^2-4 \times 9(1-a) \geq 0 \\ & a^2-6 a+9-9+9 a \geq 0 \\ & a^2+3 a \geq 0 \\ & a(a+3) \geq 0
\end{aligned}\mathrm{a} \in(-\infty,-3] \cup[0, \infty)\begin{aligned} & -\frac{\mathrm{b}}{2 \mathrm{a}} \gt 0 \\ & \frac{2(\mathrm{a}-3)}{2(\mathrm{a}-1)} \gt 0\end{aligned}a \in(-\infty, 1) \cup(3, \infty)f(0)=9 \gt 0\cap\begin{aligned}
& a \in(-\infty,-3] \cup[0,1) \\ & 2 \alpha+\beta+\gamma-6+0+1=7
\end{aligned}$
$\begin{aligned}
& D \geq 0 \\ & 4(a-3)^2-4 \times 9(1-a) \geq 0 \\ & a^2-6 a+9-9+9 a \geq 0 \\ & a^2+3 a \geq 0 \\ & a(a+3) \geq 0
\end{aligned}\mathrm{a} \in(-\infty,-3] \cup[0, \infty)\begin{aligned} & -\frac{\mathrm{b}}{2 \mathrm{a}} \gt 0 \\ & \frac{2(\mathrm{a}-3)}{2(\mathrm{a}-1)} \gt 0\end{aligned}a \in(-\infty, 1) \cup(3, \infty)f(0)=9 \gt 0\cap\begin{aligned}
& a \in(-\infty,-3] \cup[0,1) \\ & 2 \alpha+\beta+\gamma-6+0+1=7
\end{aligned}$
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Quadratic Equation chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2025, covering the Quadratic Equation chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.