# Solve the irrational equation

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

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

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.

• Already do that and grow May 30, 2017 at 3:19
• I don't understand your comment. If you have done this, please show where it got you and where you are stuck now. May 30, 2017 at 3:51
• Ross Millikan thaks for help, "get to the roots of 3 you say, but from there on the problem is not how to reduce it, on the contrary seems to grow more " May 31, 2017 at 1:19
• I didn't say anything about the roots of $3$. If you do what I suggested, the denominators in both fractions on the left become $\sqrt 3$. Did you try it? Multiplying by the conjugate should be a standard approach for problems like this. You get terms like $x^{3/2}$ that need to be dealt with by squaring. I think you get a cubic that has a single real rational root. If you show what you have tried and where you are stuck, we can help much more. -1 May 31, 2017 at 1:58
• I will try there May 31, 2017 at 2:12

hint

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

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

• is not the same ....The denominators are different May 30, 2017 at 3:18
• @Susana After multiplying by conjugates, you will get $\sqrt {3}$ in both denomnators. May 30, 2017 at 3:22
• Salahamam_ Fatima 21, thanks for help, get to the roots of 3 you say, but from there on the problem is not how to reduce it, on the contrary seems to grow more May 31, 2017 at 1:17

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!

• Michael Rozenberg thanks for help, I really liked your approach and I understand it thanks for your time, even so you can get to the result in the conventional way - these problems always have some artifice - some idea, I have to solve it that way. May 31, 2017 at 1:24