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Question Number 15504 by rish@bh last updated on 11/Jun/17

Solve ⌈x^2 ⌉=(⌊x⌋)^2 +2x

$$\mathrm{Solve}\:\lceil{x}^{\mathrm{2}} \rceil=\left(\lfloor{x}\rfloor\right)^{\mathrm{2}} +\mathrm{2}{x} \\ $$

Commented by rish@bh last updated on 11/Jun/17

I got only 0 and .5 but my book says ((n+1)/2) and 0

$$\mathrm{I}\:\mathrm{got}\:\mathrm{only}\:\mathrm{0}\:\mathrm{and}\:.\mathrm{5}\:\mathrm{but}\:\mathrm{my}\:\mathrm{book} \\ $$$$\mathrm{says}\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{0} \\ $$

Commented by mrW1 last updated on 11/Jun/17

your book is wrong. you can check with n=1⇒x=1⇒1^2 =1^2 +2 !!!

$$\mathrm{your}\:\mathrm{book}\:\mathrm{is}\:\mathrm{wrong}.\:\mathrm{you}\:\mathrm{can}\:\mathrm{check} \\ $$$$\mathrm{with}\:\mathrm{n}=\mathrm{1}\Rightarrow\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{1}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}\:!!! \\ $$

Commented by rish@bh last updated on 11/Jun/17

Thanks!

$$\mathrm{Thanks}! \\ $$

Answered by mrW1 last updated on 11/Jun/17

2x=⌈x^2 ⌉−(⌊x⌋)^2 =integer ⇒x=integer or integer+0.5 let x=n+f with f=0 or 0.5 (1) f=0 x=n n^2 =n^2 +2n ⇒n=0 ⇒x=0 (2) f=0.5 x=n+0.5 x^2 =n^2 +n+0.25 ⌈x^2 ⌉=n^2 +n+1 ⌊x⌋=n 2x=2n+1 n^2 +n+1=n^2 +2n+1 n=0 ⇒x=0.5

$$\mathrm{2x}=\lceil{x}^{\mathrm{2}} \rceil−\left(\lfloor{x}\rfloor\right)^{\mathrm{2}} =\mathrm{integer} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{integer}\:\mathrm{or}\:\mathrm{integer}+\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{let}\:\mathrm{x}=\mathrm{n}+\mathrm{f}\:\mathrm{with}\:\mathrm{f}=\mathrm{0}\:\mathrm{or}\:\mathrm{0}.\mathrm{5} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{f}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{n} \\ $$$$\mathrm{n}^{\mathrm{2}} =\mathrm{n}^{\mathrm{2}} +\mathrm{2n} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}=\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{x}=\mathrm{n}+\mathrm{0}.\mathrm{5} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{0}.\mathrm{25} \\ $$$$\lceil\mathrm{x}^{\mathrm{2}} \rceil=\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1} \\ $$$$\lfloor\mathrm{x}\rfloor=\mathrm{n} \\ $$$$\mathrm{2x}=\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{n}^{\mathrm{2}} +\mathrm{n}+\mathrm{1}=\mathrm{n}^{\mathrm{2}} +\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{n}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{0}.\mathrm{5} \\ $$

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