Question

For how many integers $n$ is it true that $\sqrt{n} \le \sqrt{4n - 6} < \sqrt{2n + 5}$?

Answer 1

$\bf \sqrt{n}\leqslant \sqrt{4n-6}<\sqrt{2n+5}\implies \begin{cases} \sqrt{n}\leqslant \sqrt{4n-6}\\[1em] \sqrt{n}< \sqrt{2n+5} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt{n}\leqslant \sqrt{4n-6}\implies \stackrel{\textit{squaring both sides}}{n\leqslant 4n-6}\implies 0\leqslant 4n-n-6\implies 0\leqslant 3n-6 \\\\\\ 6\leqslant 3n\implies \cfrac{6}{3}\leqslant n\implies \boxed{2\leqslant n} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}$

namely, -5, -4, -3, -2, -1, 0, 1.  Excluding "2" because n < 2.

Answer 2

Answer:

4

Step-by-step explanation:

2, 3, 4, 5