- #1
diredragon
- 323
- 15
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?
Last edited by a moderator: