Question Number 134005 by mr W last updated on 26/Feb/21
$${solve}\:{x}^{\mathrm{3}} −\mathrm{2}\lfloor{x}\rfloor=\mathrm{5} \\ $$
Answered by MJS_new last updated on 26/Feb/21
![x=i[nteger part]+f[ractal part] (i+f)^3 −2i=5 f=−i+((2i+5))^(1/3) 0≤−i+((2i+5))^(1/3) ≤1 ⇒ i=1∨i=2 ⇒ f=−1+(7)^(1/3) ∨f=−2+(9)^(1/3) ⇒ x=(7)^(1/3) ∨x=(9)^(1/3)](https://www.tinkutara.com/question/Q134027.png)
$${x}={i}\left[\mathrm{nteger}\:\mathrm{part}\right]+{f}\left[\mathrm{ractal}\:\mathrm{part}\right] \\ $$$$\left({i}+{f}\right)^{\mathrm{3}} −\mathrm{2}{i}=\mathrm{5} \\ $$$${f}=−{i}+\sqrt[{\mathrm{3}}]{\mathrm{2}{i}+\mathrm{5}} \\ $$$$\mathrm{0}\leqslant−{i}+\sqrt[{\mathrm{3}}]{\mathrm{2}{i}+\mathrm{5}}\leqslant\mathrm{1} \\ $$$$\Rightarrow\:{i}=\mathrm{1}\vee{i}=\mathrm{2} \\ $$$$\Rightarrow\:{f}=−\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{7}}\vee{f}=−\mathrm{2}+\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$$$\Rightarrow\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{7}}\vee{x}=\sqrt[{\mathrm{3}}]{\mathrm{9}} \\ $$
Commented by mr W last updated on 26/Feb/21
$${great}!\:{thanks}! \\ $$
Commented by MJS_new last updated on 26/Feb/21
$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\lfloor{z}\rfloor\:\mathrm{and}\:\lceil{z}\rceil\:\mathrm{are}\:\mathrm{defined}\:\mathrm{for} \\ $$$${z}\in\mathbb{C},\:\mathrm{maybe}\:\lfloor{a}×{b}\mathrm{i}\rfloor=\lfloor{a}\rfloor+\lfloor{b}\rfloor\mathrm{i}?\:\mathrm{anyway}\:\mathrm{I} \\ $$$$\mathrm{haven}'\mathrm{t}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{for}\:{x}\notin\mathbb{R} \\ $$