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## Homework Statement

##\frac{x+\sqrt3}{\sqrt{x}+\sqrt{x+\sqrt3}} + \frac{x-\sqrt3}{\sqrt{x}-\sqrt{x-\sqrt3}} = \sqrt{x}##

All real solutions to this equation are found in the set:

##a) [\sqrt3, 2\sqrt3), b) (2\sqrt3, 3\sqrt3), c) (3\sqrt3, 6), d) [6, 8)##

## Homework Equations

3. The Attempt at a Solution

##\frac{x+\sqrt3}{\sqrt{x}+\sqrt{x+\sqrt3}} + \frac{x-\sqrt3}{\sqrt{x}-\sqrt{x-\sqrt3}} = \sqrt{x}##

##\frac{(x+\sqrt3)^2}{(\sqrt{x}+\sqrt{x+\sqrt3})^2} + \frac{(x-\sqrt3)^2}{(\sqrt{x}-\sqrt{x-\sqrt3})^2} + 2(\frac{x+\sqrt3}{\sqrt{x}+\sqrt{x+\sqrt3}}\frac{x-\sqrt3}{\sqrt{x}-\sqrt{x-\sqrt3}}) = x##

##\frac{(x+\sqrt3)^2(\sqrt{x}-\sqrt{x-\sqrt3})^2+(x-\sqrt3)^2(\sqrt{x}+\sqrt{x+\sqrt3})^2 + 2(x^2-3)((\sqrt{x}+\sqrt{x+\sqrt3})(\sqrt{x}-\sqrt{x-\sqrt3})}{(\sqrt{x}+\sqrt{x+\sqrt3})^2(\sqrt{x}-\sqrt{x-\sqrt3})^2}=x##

##\frac{(x+\sqrt{3})^2(\sqrt{x}-\sqrt{x-\sqrt{3}})^2+(x-\sqrt{3})^2(\sqrt{x}+\sqrt{x+\sqrt{3}})^2 + 2(x^2-3)((\sqrt{x}+\sqrt{x+\sqrt{3}})(\sqrt{x}-\sqrt{x-\sqrt{3}}))-(\sqrt{x}+\sqrt{x+\sqrt{3}})^2(\sqrt{x}-\sqrt{x-\sqrt{3}})^2x(\sqrt{x}+\sqrt{x+\sqrt{3}})^2(\sqrt{x}-\sqrt{x-\sqrt{3}})^2}=0##

##(x+\sqrt{3})^2(\sqrt{x}-\sqrt{x-\sqrt{3}})^2+(x-\sqrt{3})^2(\sqrt{x}+\sqrt{x+\sqrt{3}})^2 + 2(x^2-3)((\sqrt{x}+\sqrt{x+\sqrt{3}})(\sqrt{x}-\sqrt{x-\sqrt{3}}))-(\sqrt{x}+\sqrt{x+\sqrt{3}})^2(\sqrt{x}-\sqrt{x-\sqrt{3}})^2x=0##

there should be an easier way to do this rather than all i wrote + more and more. If you know, could you hint, rather than giving the answer?

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