Question Number 133469 by Eric002 last updated on 22/Feb/21
$${find}\:{x}\:{in}\:{terms}\:{of}\:{a} \\ $$$$\frac{\mathrm{1}+{x}−\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }}{\mathrm{1}+{x}+\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }}={a}^{\mathrm{3}} \frac{\sqrt{\mathrm{2}+{x}}+\sqrt{{x}}}{\:\sqrt{\mathrm{2}+{x}}−\sqrt{{x}}} \\ $$
Answered by EDWIN88 last updated on 22/Feb/21
![(([(1+x)−(√(2x+x^2 )) ]^2 )/((1+x)^2 −(2x+x^2 )))= (((1+x)^2 +(2x+x^2 )−2(x+1)(√(2x+x^2 )))/1) = 2x^2 +4x+1−2(x+1)(√(2x+x^2 )) (([ (√(2+x))+(√x) ]^2 )/((2+x)−x)) = ((2x+2+2(√(2x+x^2 )))/2) = x+1+(√(2x+x^2 )) a^3 = ((2x^2 +4x+1−2(x+1)(√(2x+x^2 )))/(x+1+(√(2x+x^2 )))) a^3 = (([(x+1)−(√(2x+x^2 )) ]^2 )/((x+1)+(√(2x+x^2 )))) = [(x+1)−(√(2x+x^2 )) ]^3 ⇔x+1−(√(2x+x^2 )) = a ⇒(x+1)^2 = [ a+(√(2x+x^2 )) ]^2 ⇒1 = a^2 +2a(√(2x+x^2 )) ⇒(1−a^2 )^2 = 4a^2 (2x+x^2 ) ⇒(((1−a^2 )/(2a)))^2 +1=(x+1)^2 ⇒x = (√((a^4 +2a^2 +1)/(4a^2 ))) −1 ⇒x = ((a^2 +1)/(2a))−1=((a^2 −2a+1)/(2a))](Q133480.png)
$$\frac{\left[\left(\mathrm{1}+\mathrm{x}\right)−\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} }\:\right]^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} −\left(\mathrm{2x}+\mathrm{x}^{\mathrm{2}} \right)}=\:\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} +\left(\mathrm{2x}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}} \\ $$$$=\:\mathrm{2x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{1}−\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\frac{\left[\:\sqrt{\mathrm{2}+\mathrm{x}}+\sqrt{\mathrm{x}}\:\right]^{\mathrm{2}} }{\left(\mathrm{2}+\mathrm{x}\right)−\mathrm{x}}\:=\:\frac{\mathrm{2x}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} }}{\mathrm{2}}\:=\:\mathrm{x}+\mathrm{1}+\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} } \\ $$$$ \\ $$$${a}^{\mathrm{3}} \:=\:\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}−\mathrm{2}\left({x}+\mathrm{1}\right)\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }}{{x}+\mathrm{1}+\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }} \\ $$$${a}^{\mathrm{3}} =\:\frac{\left[\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }\:\right]^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)+\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }}\:=\:\left[\left({x}+\mathrm{1}\right)−\sqrt{\mathrm{2}{x}+{x}^{\mathrm{2}} }\:\right]^{\mathrm{3}} \\ $$$$\Leftrightarrow\mathrm{x}+\mathrm{1}−\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{a}\: \\ $$$$\Rightarrow\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\left[\:\mathrm{a}+\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} }\:\right]^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}\:=\:\mathrm{a}^{\mathrm{2}} +\mathrm{2a}\sqrt{\mathrm{2x}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{a}^{\mathrm{2}} \right)^{\mathrm{2}} =\:\mathrm{4a}^{\mathrm{2}} \left(\mathrm{2x}+\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left(\frac{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }{\mathrm{2a}}\right)^{\mathrm{2}} +\mathrm{1}=\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{x}\:=\:\sqrt{\frac{\mathrm{a}^{\mathrm{4}} +\mathrm{2a}^{\mathrm{2}} +\mathrm{1}}{\mathrm{4a}^{\mathrm{2}} }}\:−\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2a}}−\mathrm{1}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{2a}+\mathrm{1}}{\mathrm{2a}} \\ $$
Commented by Eric002 last updated on 22/Feb/21
$${well}\:{done} \\ $$