Solve the irrational equation

Question

I have tried to solve this equation but it grows, grows and grows and I do not come to anything clear, some idea or trick to solve it Thank you

$$\frac{x+\sqrt{3}}{\sqrt{x}+\sqrt{x+\sqrt{3}}}+\frac{x-\sqrt{3}}{\sqrt{x}-\sqrt{x-\sqrt{3}}} =\sqrt{x}$$

Answer 1

hint

$$(\sqrt {a}+\sqrt {a+b})(\sqrt {a}-\sqrt {a+b})=-b $$

$$(\sqrt {a}-\sqrt {a-c})(\sqrt {a}+\sqrt {a-c})=c $$

Answer 2

Multiply each fraction on the left with the conjugate of its denominator, so the first by $\frac {\sqrt x - \sqrt{x+\sqrt 3}}{\sqrt x - \sqrt{x+\sqrt 3}}$ There will be lots of simplification.

Answer 3

The domain gives $x\geq\sqrt3$.

We need to solve $$(x+\sqrt3)\left(\sqrt{x}-\sqrt{x-\sqrt3}\right)+(x-\sqrt3)\left(\sqrt{x}+\sqrt{x+\sqrt3}\right)=$$ $$=\sqrt{x}\left(\sqrt{x}-\sqrt{x-\sqrt3}\right)\left(\sqrt{x}+\sqrt{x+\sqrt3}\right)$$ or $$\sqrt{x^3}+\sqrt{x^2-3}\sqrt{x}=\sqrt{3}\left(\sqrt{x+\sqrt3}+\sqrt{x-\sqrt3}\right)$$ or $$\sqrt{x}\left(x+\sqrt{x^2-3}\right)=\sqrt3\cdot\sqrt{2x+2\sqrt{x^2-3}}$$ or $$\sqrt{x}\sqrt{x+\sqrt{x^2-3}}=\sqrt6.$$ Let $f(x)=\sqrt{x^2+x\sqrt{x^2-3}}.$

Hence, $f$ is an increasing function on $[\sqrt3,+\infty)$.

Thus, our equation has maximum one root.

But $2$ is a root, which says that it's an unique root.

Done!