An algebra problem by s p

$\begin{aligned}\sqrt{n}+\sqrt{n+1}<11 & \implies 2n+1+2\sqrt{n(n+1)}<121 ~~~~ \color{#3D99F6}{\text{(squaring both sides)}} \\& \implies n+\sqrt{n(n+1)}<60\\& \implies \sqrt{n(n+1)}<60-n\\& \implies n^2+n<3600+n^2-120n~~~~ \color{#3D99F6}{\text{(squaring both sides)}}\\& \implies 121n<3600 \\& \implies n<29.752\cdots \implies n\leq 29 ~~~~ \color{#20A900}{\text{Since,}~ n\in\mathbb{N}}\end{aligned}$

Thus, $\boxed{29}~ \text{positive integers are possible.}$

Relevant wiki: Mean Value Theorem

This is not intended to be a straight-forward or simple solution. It's just intended to show a different way to solve it.

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Since $f(x) = \sqrt{x} + \sqrt{x+1}$ is an increasing function of $x \geq 0$ , the solution to $\sqrt{x}+\sqrt{x+1} < 11$ will have the form $x < M = f^{-1}(11)$ . To estimate $M,$ first note that $\sqrt{25}+\sqrt{25+1} < \sqrt{25} + \sqrt{36} = 11 < \sqrt{35}+\sqrt{35+1}$ shows that $25 < M < 35$ . To further refine this estimate, we can use the mean value theorem to find some $c$ such that $25 < M < c < M+1 < 36$ and $11 - 2\sqrt{M} = \sqrt{M} + \sqrt{M+1} - 2\sqrt{M} = \sqrt{M+1} - \sqrt{M} = \frac{1}{2\sqrt{c}} \in \left(\frac{1}{2\sqrt{36}}, \frac{1}{2\sqrt{25}}\right) = \left(\frac{1}{12},\frac{1}{10}\right)$ Then solving $\frac{1}{12} < 11-2\sqrt{M} < \frac{1}{10} \iff \frac{109}{20} < \sqrt{M} < \frac{131}{24} \iff \left(\frac{109}{20}\right)^2 < M < \left(\frac{131}{24}\right)^2$

From this, all we need to note is that $29 < \left(\frac{109}{20}\right)^2 < M < \left(\frac{131}{24}\right)^2 < 30,$ as this shows if $n$ is a positive integer, then $\sqrt{n} + \sqrt{n+1} < 11 \iff n < M \iff n \leq 29$ so there are $\boxed{29}$ such positive integer solutions.