Question Number 191962 by ajfour last updated on 04/May/23
Answered by mr W last updated on 04/May/23
![[(√((a+x)^2 −(a−x)^2 ))−(√((a+b)^2 −(a−b)^2 ))]^2 =(b+x)^2 −(2a−b−x)^2 [(√(ax))−(√(ab))]^2 =2a(x−a+b) x+2(√(bx))−2a+b=0 ⇒(√x)=−(√b)+(√(2a)) ⇒x=((√(2a))−(√b))^2 =2a+b−2(√(2ab))](https://www.tinkutara.com/question/Q191971.png)
$$\left[\sqrt{\left({a}+{x}\right)^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }−\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{x}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}−{b}−{x}\right)^{\mathrm{2}} \\ $$$$\left[\sqrt{{ax}}−\sqrt{{ab}}\right]^{\mathrm{2}} =\mathrm{2}{a}\left({x}−{a}+{b}\right) \\ $$$${x}+\mathrm{2}\sqrt{{bx}}−\mathrm{2}{a}+{b}=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{x}}=−\sqrt{{b}}+\sqrt{\mathrm{2}{a}} \\ $$$$\Rightarrow{x}=\left(\sqrt{\mathrm{2}{a}}−\sqrt{{b}}\right)^{\mathrm{2}} =\mathrm{2}{a}+{b}−\mathrm{2}\sqrt{\mathrm{2}{ab}} \\ $$
Commented by mr W last updated on 04/May/23
$${interesting}! \\ $$
Commented by ajfour last updated on 05/May/23
$$\left(\mathrm{2}\sqrt{{ax}}−\mathrm{2}\sqrt{{ab}}\right)^{\mathrm{2}} =\mathrm{4}{a}\left({x}+{b}−{a}\right) \\ $$$$\left(\sqrt{{x}}−\sqrt{{b}}\right)^{\mathrm{2}} ={x}+{b}−{a} \\ $$$$\Rightarrow\:{a}=\mathrm{2}\sqrt{{bx}} \\ $$