JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 5)
Consider the equation $x^2+4 x-n=0$, where $n \in[20,100]$ is a natural number. Then the number of all distinct values of $n$, for which the given equation has integral roots, is equal to
6
5
8
7
Explanation
$$\begin{aligned} & x^2+4 x-n=0 \text { has integer roots } \\ & \Rightarrow x=\frac{-4 \pm \sqrt{16+4 n}}{2}=-2 \pm \sqrt{4+n} \end{aligned}$$
For $x$ to be integer $4+n$ must be perfect squares
$$\begin{aligned} & n \in[20,100] \\ & n+4 \in[24,104]=S \end{aligned}$$
$\left\{25,36, \ldots 10^2\right\} \in S \Rightarrow 5^2, 6^2, \ldots 10^2 \Rightarrow 6$ values of $n$
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