Question Number 222076 by Tawa11 last updated on 16/Jun/25
Answered by Ghisom last updated on 17/Jun/25
![different methods, I like this one: x=I+f; I∈Z∧f∈R∧0≤f<1 4(I+f)^2 −40I+51=0 solving for f: f=−I+((√(40I−51))/2) [ _(f<0 which is not possible)^(the “−”solution leads to) ] we know 0≤f<1 ⇒ 0≤−I+((√(40I−51))/2)<1 −I+((√(40I−51))/2)=0 ⇒ I=(3/2)∨I=((17)/2) −I+((√(40I−51))/2)=1 ⇒ I=(5/2)∨I=((11)/2) ⇒ (3/2)≤I<(5/2)∨((11)/2)<I≤((17)/2) but I∈Z ⇒ 2≤I≤2∨6≤I≤8 ⇒ I∈{2, 6, 7, 8} x=I+f=((√(40I−51))/2) x∈{((√(29))/2), ((3(√(21)))/2), ((√(229))/2), ((√(269))/2)}](https://www.tinkutara.com/question/Q222082.png)
$$\mathrm{different}\:\mathrm{methods},\:\mathrm{I}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}: \\ $$$${x}={I}+{f};\:{I}\in\mathbb{Z}\wedge{f}\in\mathbb{R}\wedge\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{4}\left({I}+{f}\right)^{\mathrm{2}} −\mathrm{40}{I}+\mathrm{51}=\mathrm{0} \\ $$$$\mathrm{solving}\:\mathrm{for}\:{f}: \\ $$$${f}=−{I}+\frac{\sqrt{\mathrm{40}{I}−\mathrm{51}}}{\mathrm{2}}\:\:\:\:\:\left[\:_{{f}<\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}} ^{\mathrm{the}\:“−''\mathrm{solution}\:\mathrm{leads}\:\mathrm{to}} \right] \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{0}\leqslant{f}<\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{0}\leqslant−{I}+\frac{\sqrt{\mathrm{40}{I}−\mathrm{51}}}{\mathrm{2}}<\mathrm{1} \\ $$$$\:\:\:\:\:−{I}+\frac{\sqrt{\mathrm{40}{I}−\mathrm{51}}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow\:{I}=\frac{\mathrm{3}}{\mathrm{2}}\vee{I}=\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$\:\:\:\:\:−{I}+\frac{\sqrt{\mathrm{40}{I}−\mathrm{51}}}{\mathrm{2}}=\mathrm{1}\:\Rightarrow\:{I}=\frac{\mathrm{5}}{\mathrm{2}}\vee{I}=\frac{\mathrm{11}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\Rightarrow \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{I}<\frac{\mathrm{5}}{\mathrm{2}}\vee\frac{\mathrm{11}}{\mathrm{2}}<{I}\leqslant\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$\mathrm{but}\:{I}\in\mathbb{Z}\:\Rightarrow \\ $$$$\mathrm{2}\leqslant{I}\leqslant\mathrm{2}\vee\mathrm{6}\leqslant{I}\leqslant\mathrm{8} \\ $$$$\Rightarrow \\ $$$${I}\in\left\{\mathrm{2},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8}\right\} \\ $$$${x}={I}+{f}=\frac{\sqrt{\mathrm{40}{I}−\mathrm{51}}}{\mathrm{2}} \\ $$$${x}\in\left\{\frac{\sqrt{\mathrm{29}}}{\mathrm{2}},\:\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{229}}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{269}}}{\mathrm{2}}\right\} \\ $$
Commented by MathematicalUser2357 last updated on 17/Jun/25
$${how}\:{did}\:{you}\:{add}\:{borderless}\:{table}\begin{matrix}{{like}}\\{{this}}\end{matrix} \\ $$
Commented by Tawa11 last updated on 17/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 17/Jun/25
$${say}\:\lfloor{x}\rfloor={n} \\ $$$${x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} −\mathrm{40}{n}+\mathrm{51}=\mathrm{0} \\ $$$$\mathrm{40}{n}−\mathrm{51}=\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} \geqslant\mathrm{4}{n}^{\mathrm{2}} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{40}{n}+\mathrm{51}\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{2}}\leqslant{n}\leqslant\frac{\mathrm{17}}{\mathrm{2}}\:\Rightarrow\mathrm{2}\leqslant{n}\leqslant\mathrm{8}\:\:\:…\left({i}\right) \\ $$$$\mathrm{40}{n}−\mathrm{51}=\mathrm{4}\left({n}+{f}\right)^{\mathrm{2}} <\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{n}^{\mathrm{2}} −\mathrm{32}{n}+\mathrm{55}>\mathrm{0} \\ $$$$\Rightarrow{n}<\frac{\mathrm{5}}{\mathrm{2}}\:\vee\:{n}>\frac{\mathrm{11}}{\mathrm{2}}\:\Rightarrow{n}\leqslant\mathrm{2}\:\vee\:{n}\geqslant\mathrm{6}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow{n}=\mathrm{2},\:\mathrm{6},\:\mathrm{7},\:\mathrm{8} \\ $$$${x}=\sqrt{\frac{\mathrm{40}{n}−\mathrm{51}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{29}}}{\mathrm{2}},\:\frac{\mathrm{3}\sqrt{\mathrm{21}}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{229}}}{\mathrm{2}},\:\frac{\sqrt{\mathrm{269}}}{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 17/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$