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The-possible-value-of-x-that-satisfy-x-2-x-2-2x-x-2-2x-1-2017-x-R-and-x-2-2x-




Question Number 56719 by naka3546 last updated on 22/Mar/19
The  possible  value  of  x  that  satisfy  :          x^2  + ⌊x^2  − 2x⌋ + ⌈x^2  + 2x + 1⌉  =  2017  ,  x ∈  R^+   and   ⌊x^2 ⌋ − ⌈2x⌉  =  ...
$${The}\:\:{possible}\:\:{value}\:\:{of}\:\:{x}\:\:{that}\:\:{satisfy}\:\:: \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2017}\:\:,\:\:{x}\:\in\:\:\mathbb{R}^{+} \\ $$$${and}\:\:\:\lfloor{x}^{\mathrm{2}} \rfloor\:−\:\lceil\mathrm{2}{x}\rceil\:\:=\:\:… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
x=25.923  ⌊(25.923)^2 ⌋−⌈2×25.923⌉  =⌊672.001929⌋−⌈51.846⌉  =672−52  =620
$${x}=\mathrm{25}.\mathrm{923} \\ $$$$\lfloor\left(\mathrm{25}.\mathrm{923}\right)^{\mathrm{2}} \rfloor−\lceil\mathrm{2}×\mathrm{25}.\mathrm{923}\rceil \\ $$$$=\lfloor\mathrm{672}.\mathrm{001929}\rfloor−\lceil\mathrm{51}.\mathrm{846}\rceil \\ $$$$=\mathrm{672}−\mathrm{52} \\ $$$$=\mathrm{620} \\ $$
Commented by naka3546 last updated on 22/Mar/19
Sir, the  answer  x  not  suitable  for  the  equation  above .  Please,  recheck  the  answer .
$${Sir},\:{the}\:\:{answer}\:\:{x}\:\:{not}\:\:{suitable}\:\:{for}\:\:{the}\:\:{equation}\:\:{above}\:. \\ $$$${Please},\:\:{recheck}\:\:{the}\:\:{answer}\:. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/Mar/19
ok   let me check...
$${ok}\:\:\:{let}\:{me}\:{check}… \\ $$
Answered by mr W last updated on 23/Mar/19
x^2  + ⌊x^2  − 2x⌋ + ⌈x^2  + 2x + 1⌉  =  2017  x^2 +integer+integer=integer  ⇒x^2 =integer, say x^2 =n  n+n+⌊−2(√n)⌋+n+⌈2(√n)⌉+1=2017  since ⌊−x⌋=−⌈x⌉  ⇒⌊−2(√n)⌋+⌈2(√n)⌉=0  ⇒3n=2016  ⇒n=672  ⇒x=(√n)=(√(672))=4(√(42))≈25.923  ⌊x^2 ⌋−⌈2x⌉=672−52=620
$${x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2017} \\ $$$${x}^{\mathrm{2}} +{integer}+{integer}={integer} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={integer},\:{say}\:{x}^{\mathrm{2}} ={n} \\ $$$${n}+{n}+\lfloor−\mathrm{2}\sqrt{{n}}\rfloor+{n}+\lceil\mathrm{2}\sqrt{{n}}\rceil+\mathrm{1}=\mathrm{2017} \\ $$$${since}\:\lfloor−{x}\rfloor=−\lceil{x}\rceil \\ $$$$\Rightarrow\lfloor−\mathrm{2}\sqrt{{n}}\rfloor+\lceil\mathrm{2}\sqrt{{n}}\rceil=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{n}=\mathrm{2016} \\ $$$$\Rightarrow{n}=\mathrm{672} \\ $$$$\Rightarrow{x}=\sqrt{{n}}=\sqrt{\mathrm{672}}=\mathrm{4}\sqrt{\mathrm{42}}\approx\mathrm{25}.\mathrm{923} \\ $$$$\lfloor{x}^{\mathrm{2}} \rfloor−\lceil\mathrm{2}{x}\rceil=\mathrm{672}−\mathrm{52}=\mathrm{620} \\ $$
Commented by naka3546 last updated on 23/Mar/19
but,  if  x  is  substituted  to  equation  above .  Is  x  right  ?    x^2  + ⌊x^2  − 2x⌋ + ⌈x^2  + 2x + 1⌉  =  2015    how  is  it  happen ?
$${but},\:\:{if}\:\:{x}\:\:{is}\:\:{substituted}\:\:{to}\:\:{equation}\:\:{above}\:. \\ $$$${Is}\:\:{x}\:\:{right}\:\:?\:\: \\ $$$${x}^{\mathrm{2}} \:+\:\lfloor{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\rfloor\:+\:\lceil{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:+\:\mathrm{1}\rceil\:\:=\:\:\mathrm{2015}\:\: \\ $$$${how}\:\:{is}\:\:{it}\:\:{happen}\:? \\ $$
Commented by mr W last updated on 23/Mar/19
how did you get 2015? this is how it is:  x^2 =672  x=(√(672))≈25.92  2x≈51.8  ⌊x^2 −2x⌋=⌊672−51.8⌋=⌊620.2⌋=620  ⌈x^2 +2x+1⌉=⌈672+51.8+1⌉=⌈724.8⌉=725  x^2 +⌊x^2 −2x⌋+⌈x^2 +2x+1⌉=672+620+725=2017  so x=(√(672)) is correct.
$${how}\:{did}\:{you}\:{get}\:\mathrm{2015}?\:{this}\:{is}\:{how}\:{it}\:{is}: \\ $$$${x}^{\mathrm{2}} =\mathrm{672} \\ $$$${x}=\sqrt{\mathrm{672}}\approx\mathrm{25}.\mathrm{92} \\ $$$$\mathrm{2}{x}\approx\mathrm{51}.\mathrm{8} \\ $$$$\lfloor{x}^{\mathrm{2}} −\mathrm{2}{x}\rfloor=\lfloor\mathrm{672}−\mathrm{51}.\mathrm{8}\rfloor=\lfloor\mathrm{620}.\mathrm{2}\rfloor=\mathrm{620} \\ $$$$\lceil{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\rceil=\lceil\mathrm{672}+\mathrm{51}.\mathrm{8}+\mathrm{1}\rceil=\lceil\mathrm{724}.\mathrm{8}\rceil=\mathrm{725} \\ $$$${x}^{\mathrm{2}} +\lfloor{x}^{\mathrm{2}} −\mathrm{2}{x}\rfloor+\lceil{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\rceil=\mathrm{672}+\mathrm{620}+\mathrm{725}=\mathrm{2017} \\ $$$${so}\:{x}=\sqrt{\mathrm{672}}\:{is}\:{correct}. \\ $$

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