Could you give me some help on finding the roots (if any) of the following equation: $$ \frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}} $$ I tried to apply some classic approaches, but I had no luck... Could you lend me a hand? Thanks in advance!
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$$\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$$ $$\frac{\sqrt{1+\frac xa}}{\sqrt{1}+\sqrt{1+\frac xa}}=\frac{\sqrt{1-\frac xa}}{\sqrt{1}-\sqrt{1-\frac xa}}$$ $$\frac xa \mapsto y$$ $$\frac{\sqrt{1+y}}{1+\sqrt{1+y}}=\frac{\sqrt{1-y}}{1-\sqrt{1-y}}$$ $$\sqrt{1+y}-\sqrt{1-y^2}=\sqrt{1-y}+\sqrt{1-y^2}$$ $$\sqrt{1+y}-\sqrt{1-y}=2\sqrt{1-y^2}$$ $$1+y+1-y-2\sqrt{1-y^2}=4(1-y^2)$$ $$4y^2-2=2\sqrt{1-y^2}$$ $$2y^2-1=\sqrt{1-y^2}$$ $$4y^4-4y^2+1=1-y^2$$ $$4y^4-3y^2=0$$ $$y^2(4y^2-3)=0$$ $$y=0,\quad y=\pm\frac{\sqrt3}2$$ $$x=0,\quad x=\pm\frac{\sqrt3}2a$$
Kay K.
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2Did you assume that $a=1$? How did you introduce $y$ into the equation? – J.Gudal Dec 10 '15 at 01:22
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2I divided both sides by $\sqrt a$, and then substitued $\frac xa$ by $y$. I'll add it. – Kay K. Dec 10 '15 at 01:23
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Thank you very much @KayK., that was a nice trick! – nullgeppetto Dec 10 '15 at 01:26
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@nullgeppetto Thanks. – Kay K. Dec 10 '15 at 01:27
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@Kay K. That's very clever. Nice work. Just to clarify, $a$ is non zero since substituting $a=0$ leads to $1=-1$ – J.Gudal Dec 10 '15 at 01:27
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@KayK. I dont't mean to spam you, but since you helped me here, could you have a look at an induction proof problem I just posted here? I'm preparing for some exams and I am becoming a great spammer... Apologies, truly... – nullgeppetto Dec 10 '15 at 02:18
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@J.Gudal I think it will add clarity. Thanks. – Kay K. Dec 10 '15 at 02:25
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@nullgeppetto No problem, but it seems that somebody already answered it. – Kay K. Dec 10 '15 at 02:25
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@KayK. Thanks anyway! – nullgeppetto Dec 10 '15 at 02:27