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Question Number 152764 by mnjuly1970 last updated on 01/Sep/21

Commented by prakash jain last updated on 01/Sep/21

$$1)x=n+f$$$$n⩾0$$$$n=(n+f)(n+f)={(n+f)}^2$$$$f^2+2nf+n(n-1)=0$$$$f=\frac{-2n\pm \sqrt{4n^2+4n-4n^2}}{2}$$$$f=-n\pm \sqrt{n}$$$$n=1,f=0$$$$n=0,f=0$$$$n<0$$$$n=-{(n+f)}^2$$$$f^2+2nf+n(n+1)=0$$$$f=-n\pm \sqrt{-n}$$$$f<1\Rightarrow nosolutionforn<0\left(wrong\right)$$$$\mathrm{A}\mathrm{mistake}\mathrm{pointed}\mathrm{by}\mathrm{mr}\mathrm{W}.$$$$\mathrm{for}n<0$$$$n=-1\Rightarrow f=0$$$$x=-1$$$$n=-2\Rightarrow f=2-\sqrt{2}$$$$x=n+f=-\sqrt{2}$$$$-----------$$$$x=0\mathrm{or}x=1$$$$x=-1,x=-\sqrt{2}$$

Commented by mr W last updated on 01/Sep/21

$$ithinkx=-\sqrt{2}isalsoasolution.$$$$\lfloor -\sqrt{2}\rfloor =-2$$$$-\sqrt{2}\times \mid -\sqrt{2}\mid =-2✓$$

Answered by prakash jain last updated on 01/Sep/21

$$\left(2\right)\mathrm{is}\mathrm{same}\mathrm{as}1\mathrm{put}y=x-1$$

Answered by MJS_new last updated on 02/Sep/21

$$\left(3\right)$$$$\left(a\right)\mathrm{obviously}x=-2\vee x=0\vee x=1$$$$\left(b\right)$$$$\mid \lfloor x\rfloor -1\mid <\mid x^2-1\mid \mathrm{\forall }(x<-3)\vee (x>1)$$$$\Rightarrow \mathrm{there}\mathrm{could}\mathrm{be}\mathrm{a}\mathrm{solution}-3<x<-2$$$$\Rightarrow \mid \lfloor x\rfloor -1\mid =4\wedge \mid x^2-1\mid =x^2-1$$$$\Rightarrow \mathrm{we}\mathrm{have}{x}^2-1=4\wedge -3<x<-2\Rightarrow x=-\sqrt{5}$$

Answered by mnjuly1970 last updated on 02/Sep/21

$$\lfloor x\rfloor =k$$$$x=\sqrt{k}ifx⩾0$$$$x=-\sqrt{-k}x<0$$$$k⩽\sqrt{k}<k+1\Rightarrow k=0,1$$$$\because x=0,1.......\left(★\right)$$$$k⩽-\sqrt{-k}<k+1$$$$-k-1<\sqrt{-k}⩽-k\Rightarrow k=0,-1,-2$$$$\because x=0,-1,-\sqrt{2}....\left(★\right)$$$$$$$$\left(★\right)\cup \left(★★\right)\Rightarrow x\in \{-\sqrt{2},-1,0,1\}$$

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