JEE Main 2025 — Quadratic Equation Question with Solution
JEE Main 2025 (4 Apr Shift 1)
Question
Consider the equation , where is a natural number. Then the number of all distinct values of , for which the given equation has integral roots, is equal to
Choose an option
Show full solutionCorrect option: C
Correct answer
C6
Step-by-step explanation
$\begin{aligned}
& \mathrm{x}^2+4 \mathrm{x}+4=\mathrm{n}+4 \\ & (\mathrm{x}+2)^2=\mathrm{n}+4 \\ & \mathrm{x}=-2 \pm \sqrt{\mathrm{n}+4} \\ & \because 20 \leq \mathrm{n} \leq 100 \\ & \sqrt{24} \leq \sqrt{\mathrm{n}+4} \leq \sqrt{104} \\ & \Rightarrow \sqrt{\mathrm{n}+4} \in\{5,6,7,8,9,10\}
\end{aligned}\therefore ' 6n$ ' are possible
& \mathrm{x}^2+4 \mathrm{x}+4=\mathrm{n}+4 \\ & (\mathrm{x}+2)^2=\mathrm{n}+4 \\ & \mathrm{x}=-2 \pm \sqrt{\mathrm{n}+4} \\ & \because 20 \leq \mathrm{n} \leq 100 \\ & \sqrt{24} \leq \sqrt{\mathrm{n}+4} \leq \sqrt{104} \\ & \Rightarrow \sqrt{\mathrm{n}+4} \in\{5,6,7,8,9,10\}
\end{aligned}\therefore ' 6n$ ' are possible
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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.