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Find the greatest integer $N$ such that $N<\dfrac{1}{\sqrt{33+\sqrt{128}}+\sqrt{2}-8}$.

The way I did it is this: first, I rewrote the biggest square root as $\sqrt{1+2*16+8\sqrt{2}}$. Then I made the substitution $x=\sqrt{2}$, so this became $\sqrt{16x^2+8x+1}$, which factors to $\sqrt{(4x+1)^2}$ which is equal to $4x+1$. Then I substituted back: $\dfrac{1}{4\sqrt{2}+1+\sqrt {2}-8}$, or $\dfrac 1{5 \sqrt{2}-7}$. This is where I have a problem. I am able to find the answer because I know that $1.4<\sqrt 2 <1.5$. However, is there any way to find this answer without knowing this? Thanks!

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  • $\begingroup$ It should read "...which is equal to $|4x+1|$..." $\endgroup$ – Pedro Tamaroff May 30 '13 at 20:22
  • $\begingroup$ I guess the positive root was already assumed $\endgroup$ – Ovi May 30 '13 at 20:29
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Picking up from where you left off, we have: $$ \begin{align*} N &= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \right\rfloor\\ &= \left\lfloor \dfrac{1}{5\sqrt{2}-7} \cdot \dfrac{5\sqrt{2}+7}{5\sqrt{2}+7} \right\rfloor\\ &= \left\lfloor \dfrac{5\sqrt{2}+7}{50-49} \right\rfloor\\ &= \lfloor 5\sqrt{2}+7 \rfloor\\ &= \lfloor \sqrt{50}+7 \rfloor\\ &= \lfloor \sqrt{50} \rfloor+7 \\ &= \sqrt{49}+7 \\ &= 14 \\ \end{align*}$$

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Well you can show that $1.4<\sqrt 2<1.5$, just use the following inequalities $49<50$ and $8<9$. (Hint: take square root in both of them)

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