0
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Already do that and grow $\endgroup$ – zeros May 30 '17 at 3:19
  • $\begingroup$ I don't understand your comment. If you have done this, please show where it got you and where you are stuck now. $\endgroup$ – Ross Millikan May 30 '17 at 3:51
  • $\begingroup$ 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 " $\endgroup$ – zeros May 31 '17 at 1:19
  • $\begingroup$ 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 $\endgroup$ – Ross Millikan May 31 '17 at 1:58
  • $\begingroup$ I will try there $\endgroup$ – zeros May 31 '17 at 2:12
1
$\begingroup$

hint

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

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

$\endgroup$
  • $\begingroup$ is not the same ....The denominators are different $\endgroup$ – zeros May 30 '17 at 3:18
  • 1
    $\begingroup$ @Susana After multiplying by conjugates, you will get $\sqrt {3} $ in both denomnators. $\endgroup$ – hamam_Abdallah May 30 '17 at 3:22
  • $\begingroup$ 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 $\endgroup$ – zeros May 31 '17 at 1:17
0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – zeros May 31 '17 at 1:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.