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Question Number 29161 by abdo imad last updated on 04/Feb/18

let give f(x)=(√(x−1+2(√(x−2)))) +(√(x−1−2(√(x−2)))) 1) simlify f(x) 2) solve inside N^2 the equation f(x)=y.

$${let}\:{give}\:{f}\left({x}\right)=\sqrt{{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}−\mathrm{2}}}\:\:+\sqrt{{x}−\mathrm{1}−\mathrm{2}\sqrt{{x}−\mathrm{2}}} \\ $$$$\left.\mathrm{1}\right)\:{simlify}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{solve}\:{inside}\:\mathbb{N}^{\mathrm{2}} \:{the}\:{equation}\:{f}\left({x}\right)={y}. \\ $$

Commented by abdo imad last updated on 06/Feb/18

we musthave x≥2 and for that f(x)=(√(x−2 +2(√(x−2))+1)) +(√(x−2−2(√(x−2))+1)) =(√(((√(x−2))+1)^2 )) +(√(((√(x−2))−1)^2 )) =∣(√(x−2))+1∣ +∣(√(x−2))−1∣=(√(x−2)) +1 +∣(√(x−2))−1∣ so if x≥3 f(x)=(√(x−2))+1 +(√(x−2))−1=2(√(x−2)) if 2≤x≤3 f(x)= (√(x−2))+1 +1−(√(x−2)) =2 2)for x integr and x≥3 f(x)=y ⇔y=2(√(x−2)) ⇒y^2 =4(x−2)⇒y even ⇒y=2k ,kintegr ⇒2k=4(x−2)⇒k=2(x−2) ⇒k=2p p from N⇒ x−2=p and y=4p ⇒x=p+2 and y=4p .

$${we}\:{musthave}\:{x}\geqslant\mathrm{2}\:{and}\:{for}\:{that}\:{f}\left({x}\right)=\sqrt{{x}−\mathrm{2}\:+\mathrm{2}\sqrt{{x}−\mathrm{2}}+\mathrm{1}}\:\:+\sqrt{{x}−\mathrm{2}−\mathrm{2}\sqrt{{x}−\mathrm{2}}+\mathrm{1}} \\ $$$$=\sqrt{\left(\sqrt{{x}−\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }\:+\sqrt{\left(\sqrt{{x}−\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\mid\sqrt{{x}−\mathrm{2}}+\mathrm{1}\mid\:+\mid\sqrt{{x}−\mathrm{2}}−\mathrm{1}\mid=\sqrt{{x}−\mathrm{2}}\:+\mathrm{1}\:+\mid\sqrt{{x}−\mathrm{2}}−\mathrm{1}\mid\:{so} \\ $$$${if}\:{x}\geqslant\mathrm{3}\:\:\:{f}\left({x}\right)=\sqrt{{x}−\mathrm{2}}+\mathrm{1}\:+\sqrt{{x}−\mathrm{2}}−\mathrm{1}=\mathrm{2}\sqrt{{x}−\mathrm{2}} \\ $$$${if}\:\mathrm{2}\leqslant{x}\leqslant\mathrm{3}\:\:{f}\left({x}\right)=\:\sqrt{{x}−\mathrm{2}}+\mathrm{1}\:+\mathrm{1}−\sqrt{{x}−\mathrm{2}}\:=\mathrm{2} \\ $$$$\left.\mathrm{2}\right){for}\:{x}\:{integr}\:{and}\:{x}\geqslant\mathrm{3}\:{f}\left({x}\right)={y}\:\Leftrightarrow{y}=\mathrm{2}\sqrt{{x}−\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{4}\left({x}−\mathrm{2}\right)\Rightarrow{y}\:{even}\:\Rightarrow{y}=\mathrm{2}{k}\:,{kintegr} \\ $$$$\Rightarrow\mathrm{2}{k}=\mathrm{4}\left({x}−\mathrm{2}\right)\Rightarrow{k}=\mathrm{2}\left({x}−\mathrm{2}\right)\:\Rightarrow{k}=\mathrm{2}{p}\:{p}\:{from}\:{N}\Rightarrow \\ $$$${x}−\mathrm{2}={p}\:{and}\:{y}=\mathrm{4}{p}\:\Rightarrow{x}={p}+\mathrm{2}\:{and}\:{y}=\mathrm{4}{p}\:. \\ $$

Answered by $@ty@m last updated on 05/Feb/18

let (√(x−2))=y⇒y^2 =x−2 (√(x−1+2(√(x−2)))) +(√(x−1−2(√(x−2)))) =(√(y^2 +2−1+2y))+(√(y^2 +2−1−2y)) =(√(y^2 +1+2y))+(√(y^2 +1−2y)) =y+1+y−1 =2y =2(√(x−2))

$${let}\:\sqrt{{x}−\mathrm{2}}={y}\Rightarrow{y}^{\mathrm{2}} ={x}−\mathrm{2} \\ $$$$\sqrt{{x}−\mathrm{1}+\mathrm{2}\sqrt{{x}−\mathrm{2}}}\:\:+\sqrt{{x}−\mathrm{1}−\mathrm{2}\sqrt{{x}−\mathrm{2}}} \\ $$$$=\sqrt{{y}^{\mathrm{2}} +\mathrm{2}−\mathrm{1}+\mathrm{2}{y}}+\sqrt{{y}^{\mathrm{2}} +\mathrm{2}−\mathrm{1}−\mathrm{2}{y}} \\ $$$$=\sqrt{{y}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{y}}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{y}} \\ $$$$={y}+\mathrm{1}+{y}−\mathrm{1} \\ $$$$=\mathrm{2}{y} \\ $$$$=\mathrm{2}\sqrt{{x}−\mathrm{2}} \\ $$

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