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# solve-x-3-2-x-5-

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