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x-i-f-i-N-f-0-1-x-2-i-f-2-i-2-2if-f-2-x-R-i-2f-2-x-2-2i-f-2-2if-0-x-2-i-2-f-2-




Question Number 1034 by 123456 last updated on 21/May/15
x=i+f,i∈N,f∈[0,1)  x^2 =(i+f)^2 =i^2 +2if+f^2   x∈R_+   i+2f=2⇒x^2 =2i+f^2   2if=0⇒x^2 =i^2 +f^2
$${x}={i}+{f},{i}\in\mathbb{N},{f}\in\left[\mathrm{0},\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} =\left({i}+{f}\right)^{\mathrm{2}} ={i}^{\mathrm{2}} +\mathrm{2}{if}+{f}^{\mathrm{2}} \\ $$$${x}\in\mathbb{R}_{+} \\ $$$${i}+\mathrm{2}{f}=\mathrm{2}\Rightarrow{x}^{\mathrm{2}} =\mathrm{2}{i}+{f}^{\mathrm{2}} \\ $$$$\mathrm{2}{if}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} ={i}^{\mathrm{2}} +{f}^{\mathrm{2}} \\ $$
Answered by prakash jain last updated on 21/May/15
2if=0⇒i=0∨f=0  ∵f<1, i+2f=2  ⇒i>0  f=0,i=2  x^2 =2^2 ⇒x=±2
$$\mathrm{2}{if}=\mathrm{0}\Rightarrow{i}=\mathrm{0}\vee{f}=\mathrm{0} \\ $$$$\because{f}<\mathrm{1},\:{i}+\mathrm{2}{f}=\mathrm{2}\:\:\Rightarrow{i}>\mathrm{0} \\ $$$${f}=\mathrm{0},{i}=\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}} \Rightarrow{x}=\pm\mathrm{2} \\ $$

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