Question Number 131880 by Algoritm last updated on 09/Feb/21
Answered by SEKRET last updated on 09/Feb/21
$$\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$
Commented by Algoritm last updated on 09/Feb/21
$$\mathrm{prove}\:\mathrm{that} \\ $$
Commented by Raxreedoroid last updated on 09/Feb/21
$$\mathrm{4}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5}\lfloor\mathrm{2}.\mathrm{5}\rfloor+\mathrm{8}\left\{{x}\right\}=\mathrm{19} \\ $$$${x}=\lfloor{x}\rfloor+\left\{{x}\right\}\Rightarrow\left\{{x}\right\}={x}−\lfloor{x}\rfloor \\ $$$$\mathrm{25}−\mathrm{10}+\mathrm{8}\left(\mathrm{2}.\mathrm{5}−\lfloor\mathrm{2}.\mathrm{5}\rfloor\right)=\mathrm{19} \\ $$$$\mathrm{15}+\mathrm{8}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{19} \\ $$$$\mathrm{19}=\mathrm{19} \\ $$$$\therefore\:{x}=\mathrm{19} \\ $$
Answered by mr W last updated on 09/Feb/21
$${just}\:{for}\:{easy}\:{writing}: \\ $$$${let}\:{n}=\lfloor{x}\rfloor,\:{f}=\left\{{x}\right\}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${x}={n}+{f} \\ $$$$ \\ $$$${assume}\:{x}>\mathrm{0}.\: \\ $$$${you}\:{can}\:{assume}\:{x}<\mathrm{0}\:{similarly},\:{and} \\ $$$${you}'{ll}\:{see}\:{that}\:{it}\:{gives}\:{no}\:{solution}. \\ $$$$ \\ $$$$\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{8}{f}=\mathrm{19} \\ $$$$\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}{n}>\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}=\mathrm{19}−\mathrm{8}{f}>\mathrm{11} \\ $$$$\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}{n}>\mathrm{11} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{7}>\mathrm{0} \\ $$$${n}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{3}\pm\mathrm{11}}{\mathrm{8}}=−\frac{\mathrm{7}}{\mathrm{4}},\:\mathrm{1} \\ $$$$\:{n}<−\frac{\mathrm{7}}{\mathrm{4}}\:\left({rejected}\right)\:{or}\:{n}>\mathrm{1}\:\:\:\:…\left({I}\right) \\ $$$$ \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}<\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}=\mathrm{19}−\mathrm{8}{f}\leqslant\mathrm{19} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}<\mathrm{19} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}−\mathrm{19}<\mathrm{0} \\ $$$${n}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{329}}}{\mathrm{8}}\approx−\mathrm{1}.\mathrm{6},\:\mathrm{2}.\mathrm{9} \\ $$$$\Rightarrow−\mathrm{1}.\mathrm{6}<{n}<\mathrm{2}.\mathrm{9}\:\:\:…\left({II}\right) \\ $$$$ \\ $$$${from}\:\left({I}\right)\:{and}\:\left({II}\right): \\ $$$$\mathrm{1}<{n}<\mathrm{2}.\mathrm{9} \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{4}\left(\mathrm{2}+{f}\right)^{\mathrm{2}} −\mathrm{5}×\mathrm{2}+\mathrm{8}{f}=\mathrm{19} \\ $$$$\mathrm{4}{f}^{\mathrm{2}} +\mathrm{24}{f}−\mathrm{13}=\mathrm{0} \\ $$$${f}=\frac{−\mathrm{24}\pm\sqrt{\mathrm{24}^{\mathrm{2}} +\mathrm{16}×\mathrm{13}}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}},\:−\frac{\mathrm{13}}{\mathrm{2}}\:\left({rejected}\right) \\ $$$$\Rightarrow{solution}:\:{n}=\mathrm{2},\:{f}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$