Consider the equation ${x}^{2}+4x-n=0$, where $\mathrm{n}\in [20,100]$ is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to [JEE Main 2025, 4 Apr (Shift 1)]

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Correct answer: c) $6$. x² + 4x + 4 = n + 4 (x + 2)² = n + 4 x = -2 ± √n + 4 ∵ 20 ≤ n ≤ 100 √24 ≤ √n + 4 ≤ √104 ⇒ √n + 4 ∈ {5,6,7,8,9,10} ∴ '6' integral values of 'n' are possible

Consider the equation \({x}^{2}+4x-n=0\), where \(\mathrm{n}\in [20,100]\) is a natural number. Then the number of all d…

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