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Question Number 131880 by Algoritm last updated on 09/Feb/21

Answered by SEKRET last updated on 09/Feb/21

 x=(5/2)

$$\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by Algoritm last updated on 09/Feb/21

prove that

$$\mathrm{prove}\:\mathrm{that} \\ $$

Commented by Raxreedoroid last updated on 09/Feb/21

4((5/2))^2 −5⌊2.5⌋+8{x}=19  x=⌊x⌋+{x}⇒{x}=x−⌊x⌋  25−10+8(2.5−⌊2.5⌋)=19  15+8(0.5)=19  19=19  ∴ x=19

$$\mathrm{4}\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{5}\lfloor\mathrm{2}.\mathrm{5}\rfloor+\mathrm{8}\left\{{x}\right\}=\mathrm{19} \\ $$$${x}=\lfloor{x}\rfloor+\left\{{x}\right\}\Rightarrow\left\{{x}\right\}={x}−\lfloor{x}\rfloor \\ $$$$\mathrm{25}−\mathrm{10}+\mathrm{8}\left(\mathrm{2}.\mathrm{5}−\lfloor\mathrm{2}.\mathrm{5}\rfloor\right)=\mathrm{19} \\ $$$$\mathrm{15}+\mathrm{8}\left(\mathrm{0}.\mathrm{5}\right)=\mathrm{19} \\ $$$$\mathrm{19}=\mathrm{19} \\ $$$$\therefore\:{x}=\mathrm{19} \\ $$

Answered by mr W last updated on 09/Feb/21

just for easy writing:  let n=⌊x⌋, f={x} with 0≤f<1  x=n+f    assume x>0.   you can assume x<0 similarly, and  you′ll see that it gives no solution.    4(n+f)^2 −5n+8f=19  4(n+1)^2 −5n>4(n+f)^2 −5n=19−8f>11  4(n+1)^2 −5n>11  4n^2 +3n−7>0  n_(1,2) =((−3±11)/8)=−(7/4), 1   n<−(7/4) (rejected) or n>1    ...(I)    4n^2 −5n<4(n+f)^2 −5n=19−8f≤19  4n^2 −5n<19  4n^2 −5n−19<0  n_(1,2) =((5±(√(329)))/8)≈−1.6, 2.9  ⇒−1.6<n<2.9   ...(II)    from (I) and (II):  1<n<2.9  ⇒n=2    4(2+f)^2 −5×2+8f=19  4f^2 +24f−13=0  f=((−24±(√(24^2 +16×13)))/8)=(1/2), −((13)/2) (rejected)  ⇒solution: n=2, f=(1/2)  ⇒x=2+(1/2)=(5/2)

$${just}\:{for}\:{easy}\:{writing}: \\ $$$${let}\:{n}=\lfloor{x}\rfloor,\:{f}=\left\{{x}\right\}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${x}={n}+{f} \\ $$$$ \\ $$$${assume}\:{x}>\mathrm{0}.\: \\ $$$${you}\:{can}\:{assume}\:{x}<\mathrm{0}\:{similarly},\:{and} \\ $$$${you}'{ll}\:{see}\:{that}\:{it}\:{gives}\:{no}\:{solution}. \\ $$$$ \\ $$$$\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}+\mathrm{8}{f}=\mathrm{19} \\ $$$$\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}{n}>\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}=\mathrm{19}−\mathrm{8}{f}>\mathrm{11} \\ $$$$\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}{n}>\mathrm{11} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} +\mathrm{3}{n}−\mathrm{7}>\mathrm{0} \\ $$$${n}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{3}\pm\mathrm{11}}{\mathrm{8}}=−\frac{\mathrm{7}}{\mathrm{4}},\:\mathrm{1} \\ $$$$\:{n}<−\frac{\mathrm{7}}{\mathrm{4}}\:\left({rejected}\right)\:{or}\:{n}>\mathrm{1}\:\:\:\:...\left({I}\right) \\ $$$$ \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}<\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{5}{n}=\mathrm{19}−\mathrm{8}{f}\leqslant\mathrm{19} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}<\mathrm{19} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{5}{n}−\mathrm{19}<\mathrm{0} \\ $$$${n}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{5}\pm\sqrt{\mathrm{329}}}{\mathrm{8}}\approx−\mathrm{1}.\mathrm{6},\:\mathrm{2}.\mathrm{9} \\ $$$$\Rightarrow−\mathrm{1}.\mathrm{6}<{n}<\mathrm{2}.\mathrm{9}\:\:\:...\left({II}\right) \\ $$$$ \\ $$$${from}\:\left({I}\right)\:{and}\:\left({II}\right): \\ $$$$\mathrm{1}<{n}<\mathrm{2}.\mathrm{9} \\ $$$$\Rightarrow{n}=\mathrm{2} \\ $$$$ \\ $$$$\mathrm{4}\left(\mathrm{2}+{f}\right)^{\mathrm{2}} −\mathrm{5}×\mathrm{2}+\mathrm{8}{f}=\mathrm{19} \\ $$$$\mathrm{4}{f}^{\mathrm{2}} +\mathrm{24}{f}−\mathrm{13}=\mathrm{0} \\ $$$${f}=\frac{−\mathrm{24}\pm\sqrt{\mathrm{24}^{\mathrm{2}} +\mathrm{16}×\mathrm{13}}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{2}},\:−\frac{\mathrm{13}}{\mathrm{2}}\:\left({rejected}\right) \\ $$$$\Rightarrow{solution}:\:{n}=\mathrm{2},\:{f}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$

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