# Find the set that contains the real solution to an equation

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1. Dec 31, 2015

### diredragon

1. The problem statement, all variables and given/known data
$\frac{x+\sqrt3}{\sqrt{x}+\sqrt{x+\sqrt3}} + \frac{x-\sqrt3}{\sqrt{x}-\sqrt{x-\sqrt3}} = \sqrt{x}$
All real solutions to this equation are found in the set:
$a) [\sqrt3, 2\sqrt3), b) (2\sqrt3, 3\sqrt3), c) (3\sqrt3, 6), d) [6, 8)$
2. Relevant 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: Dec 31, 2015
2. Dec 31, 2015

### diredragon

ok i have no idea how all this got messed up

3. Dec 31, 2015

### Samy_A

First separately rationalizing the denominators in the two terms of the LHS, and only then squaring both sides, leeds to a manageable equation (that has to be squared again after simplifying).
There may be a smarter way to do this, though. For now it escapes me.

Last edited: Dec 31, 2015
4. Dec 31, 2015

### Staff: Mentor

@diredragon, I fixed your LaTeX. Some obvious errors:
1. Don't use \Bigg or a variant -- That's not anything that our system recognizes, I don't believe. This could have been the source of all of those BBCode bold tags, but I'm not sure. In any case, you can't mix LaTeX and BBCode.
2. Don't write \sqrt3, as you had in many places -- this has to be \sqrt{3}.

Take a look at your first post to verify that my changes are what you intended.

Last edited: Dec 31, 2015
5. Dec 31, 2015

### epenguin

And I have no idea what the question is.
OK, it's not stated but with a bit of goodwill...
It might have been solve that equation. In which case just a normal adding of fractions procedure already will give you some rationalization.
Or probably now I think about it the question could be to show that the pair of values all a, or the pair b, etc. are solutions. It would seem fairly easy to check any of these half dozen values. However I think it would probably be less work if you rationalized at least partly of the start anyway. Although you ought to be able to show that one of these pairs is a pair of solutions, I don't see how you can show that it is the only real solution pair unless you could get out of the equation the corresponding quadratic factor, and even then,... Well we will have to see what it looks like and if there is another factor maybe it's easy to show that has no real roots. But anyway get started. At the moment we don't even know what you've done.

Last edited: Jan 1, 2016
6. Jan 1, 2016

### SammyS

Staff Emeritus
The question only asks for the interval in which any solution (real) may be found.

I would start by finding the domain of each term.

7. Jan 1, 2016

### epenguin

My unmathematicalness - those brackets are intervals - I thought they must be roots! I think the first steps will be the same though, and you will probably get something relatively simple.

8. Jan 1, 2016

### PeroK

Here's an idea: let $x = \alpha y$ where $\alpha = \sqrt{3}$. Things simplify sufficiently without squaring. And you get a general result for $\alpha$ into the bargain.

9. Jan 1, 2016

### epenguin

Now you have adjusted the formatting I can see your working. Yes that is unnecessarily complicated (Plus when you square things you have to be careful sometimes not to introduce false 'solutions'). A simpler way is what I previously suggested - I think I can see it works out to something relatively simple.

10. Jan 1, 2016

### Staff: Mentor

The problem is not stated clearly, IMO.
A clearer question would be "All real solutions ... are found in the interval: "
Of course, an interval is a set, but their choices use interval notation.

11. Jan 2, 2016

### diredragon

After rationalising and squaring i get:
$3(x - \sqrt{3})^2(\sqrt{x} + \sqrt{x + \sqrt{3}})^2 + 3(x + \sqrt{3})^2(\sqrt{x} -\sqrt{x - \sqrt{3}})^2 - 6(x^2 - 3)(\sqrt{x} + \sqrt{x + \sqrt{3}})(\sqrt{x} - \sqrt{x -\sqrt{3}}) - 9x = 0$
I made out some assumption here:
$x \ge \sqrt{3}$ otherwise we dont get a real solutions because of $\sqrt{x - \sqrt{3}}$
But i cant get a uniform solution here. If this is the form of the equation after the simplification, and if indeed $x \ge \sqrt{3}$ than the $-9x$ never equals $0$. Im stuck

12. Jan 2, 2016

### PeroK

I recommend setting $x = \sqrt{3}y$ as in post #8. You should be able to get rid of all the $\sqrt{3}$ factors. Then it's a bit easier to work with the equation in $y$.

I must admit, I couldn't see the point in squaring things. The solution comes out without squaring.

13. Jan 2, 2016

### SammyS

Staff Emeritus
What did you get after simplifying, but before squaring?

14. Jan 2, 2016

### epenguin

I take back what I said before, there was something I didn't notice, and I don't think that combining the fractions as first step gives anything very simple.

You get some simplification bye rationalizing the fractions separately, but the result is still pretty ugly and intractable. I don't see any symmetry or any other feature of the expression that enables you to do it in any other way than by elaborating it as a polynomial, but this still looks complicated and nasty.

I invite the 0P to go back and check that he has transcribed the formula of problem exactly, and also I feel it is time we were sure what the question is! So please transcribe that exactly also.

15. Jan 2, 2016

### PeroK

You should try $x = \sqrt{3}y$ and it simplifies to something more manageable!

16. Jan 2, 2016

### epenguin

Tractable? I think I tracted it. But still requiring several squarings?

17. Jan 2, 2016

### PeroK

No squaring! Just a tiny bit of algebraic sleight of hand.

18. Jan 2, 2016

### SammyS

Staff Emeritus
Remember: The problem statement merely asks for finding the interval in which the solution(s) is found.

19. Jan 3, 2016

### diredragon

SammyS is right. I have written down the problem as i have read it and i agree that it asks for the interval in which the solution is found.
Im gonna try the $x = \sqrt{3}y$ next

20. Jan 3, 2016

### epenguin

Actually I'm not convinced that you need to simplify this expression at all to solve the problems. After all the arguments that you used to convince yourself that there are no real roots below √3 didn't really need this I think.

The algorithm guaranteed to tell you the number of real roots in an interval is that of Sturm or something like it. But that is pretty heavy calculation here. I think rather the question is inviting you to use your mathematical nous and a bit of luck to just solve it quicker way.

I'll give you a starter hint. It's quite easy to test if there is an odd number of routes in any interval.

That may then suggest things to try; at that point at least try to come up with a plan.