Question

determină n∈N pentru care numarul √89-n²∈Q​

Answer 1

$\it \sqrt{89-n^2}\in\mathbb{Q} \Rightarrow 89-n^2\geq 0 \Rightarrow n^2\leq89 \Rightarrow n^2\leq81 \Rightarrow n\leq 9\\ \\ n=9 \Rightarrow \sqrt{89-9^2}=\sqrt{89-81}=\sqrt8 \not\in\mathbb{Q}\\ \\ n=8 \Rightarrow \sqrt{89-8^2}=\sqrt{89-64}=\sqrt{25}=5\in\mathbb{Q}$

$\it n=5 \Rightarrow \sqrt{89-25}=\sqrt{64}=8\in\mathbb{Q}$

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