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Question Number 173572 by mnjuly1970 last updated on 13/Jul/22

     Solve   equation          2⌊x⌋ = (x^( 2) /2)         (x ∈ R )

$$\: \\ $$$$\:\:\mathrm{S}{olve}\:\:\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{2}\lfloor{x}\rfloor\:=\:\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:\:\:\:\:\:\:\:\:\left({x}\:\in\:\mathbb{R}\:\right) \\ $$$$ \\ $$

Answered by MJS_new last updated on 13/Jul/22

x=i+f, i∈Z∧0≤f<1  4i=(i+f)^2   f=−i±2(√i)  ⇒ i≥0  0≤f<1 ⇒ f=−i+2(√i)∧i≤4∧i≠1  ⇒ i∈{0, 2, 3, 4}  ⇒ x∈{0, 2(√2), 2(√3), 4}

$${x}={i}+{f},\:{i}\in\mathbb{Z}\wedge\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{4}{i}=\left({i}+{f}\right)^{\mathrm{2}} \\ $$$${f}=−{i}\pm\mathrm{2}\sqrt{{i}} \\ $$$$\Rightarrow\:{i}\geqslant\mathrm{0} \\ $$$$\mathrm{0}\leqslant{f}<\mathrm{1}\:\Rightarrow\:{f}=−{i}+\mathrm{2}\sqrt{{i}}\wedge{i}\leqslant\mathrm{4}\wedge{i}\neq\mathrm{1} \\ $$$$\Rightarrow\:{i}\in\left\{\mathrm{0},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right\} \\ $$$$\Rightarrow\:{x}\in\left\{\mathrm{0},\:\mathrm{2}\sqrt{\mathrm{2}},\:\mathrm{2}\sqrt{\mathrm{3}},\:\mathrm{4}\right\} \\ $$

Commented by mnjuly1970 last updated on 14/Jul/22

thank you so much..

$${thank}\:{you}\:{so}\:{much}.. \\ $$

Answered by mr W last updated on 13/Jul/22

n=⌊x⌋≥0 ⇒x≥0  4n=(n+f)^2 ≥n^2  ⇒n≤4  4n=(n+f)^2 <(n+1)^2  ⇒n≠1  ⇒n=0,2,3,4  ⇒x=2(√n)=0; 2(√2); 2(√3); 4

$${n}=\lfloor{x}\rfloor\geqslant\mathrm{0}\:\Rightarrow{x}\geqslant\mathrm{0} \\ $$$$\mathrm{4}{n}=\left({n}+{f}\right)^{\mathrm{2}} \geqslant{n}^{\mathrm{2}} \:\Rightarrow{n}\leqslant\mathrm{4} \\ $$$$\mathrm{4}{n}=\left({n}+{f}\right)^{\mathrm{2}} <\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow{n}\neq\mathrm{1} \\ $$$$\Rightarrow{n}=\mathrm{0},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{2}\sqrt{{n}}=\mathrm{0};\:\mathrm{2}\sqrt{\mathrm{2}};\:\mathrm{2}\sqrt{\mathrm{3}};\:\mathrm{4} \\ $$

Commented by mnjuly1970 last updated on 14/Jul/22

 grateful sir

$$\:{grateful}\:{sir} \\ $$

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