# How to equalize correctly?

If i have this number:

$$2 \sqrt{2-\sqrt{3}}$$ and i want to find some $$x,y$$ nonzero real numbers such that $$2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$$

And for that, i do this:

$$(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$$

$$4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$$

$$(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$$

Then:

$$i) 8 = x+y$$,

$$ii)-4\sqrt{3} = 2\sqrt{xy} => -2\sqrt{3}=\sqrt{xy}$$

$$ii) = (-2\sqrt{3})^2= (\sqrt{xy})^2 => 4\cdot 3=xy , x = 12/y$$

And solving the equation $$y^2-8y+12=0$$ gives $$y_{1,2} = \{6,2\}$$

But $$2\sqrt{2-\sqrt{3}} \neq \sqrt{6} + \sqrt{2}$$

I know that the correct value must be $$\sqrt{6} - \sqrt{2}$$ but my result is different. What is wrong with my development?

• Actually, there are infinitely many solutions because $x,y$ are real numbers, any $x$ that is smaller than $8-4\sqrt{3}$ can get a new $y$ Aug 26, 2019 at 16:35
• why smaller than that value? $\sqrt(6) - \sqrt(2)$ is a valid solution and here $x=6$ that is greather than $8-4\sqrt(3)$
– ESCM
Aug 26, 2019 at 17:11
• the unique restriction is that $x,y >0$, how you get that inequality for x?
– ESCM
Aug 26, 2019 at 17:12
• It is because you want $\sqrt{x}+\sqrt{y}=2\sqrt{2-\sqrt{3}}$ at first. Actually, $\sqrt{6}-\sqrt{2}$ is the simplest way to simplify it, but there are infinitely many ways to express it. Aug 27, 2019 at 2:10

## 2 Answers

You should assume a binomial of the form whose root you desire. In this case, you should have supposed the root is $$\sqrt x -\sqrt y$$ instead.

To be specific, the problem in the above calculation is in your step ii, where you set $$-\sqrt{12}=\sqrt{xy}.$$ But this is impossible if you're dealing only with real numbers. It seems you need to note that the symbol $$\sqrt{}$$ denotes a function which, by definition, assumes nonnegative values. Thus, you can see that your equation is false, for it says a negative number is equal to a nonnegative one. That's a contradiction.

• Why and what is the proof of that?
– ESCM
Aug 26, 2019 at 16:28
• Because if you square this you have $$x+y-2\sqrt{xy}.$$ This resembles the form whose root you want. Aug 26, 2019 at 16:31
• @EduardoS. Let's focus on $$2-\sqrt 3.$$ Then if you suppose the root is of the form $\sqrt x+\sqrt y,$ you would need to have $-\sqrt 3=\sqrt{4xy},$ which is not possible in the real field. Aug 26, 2019 at 16:34
• @EduardoS. In short, the signs are important! Aug 26, 2019 at 16:40

I can write $$2\sqrt{2-\sqrt{3}} = \sqrt{y}+\sqrt{x}$$ as: $$2\sqrt{(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})$$ or $$2\sqrt{(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})$$. This can be easily obtained by the system: $$\left\{\begin{matrix} a^2+b^2=2 \\2ab=-\sqrt{3} \end{matrix}\right.$$ So: $$\sqrt{x}=-\frac{\sqrt2}{2}$$, but this is impossibile because the square root can't be negative: same arguments for $$\sqrt{y}$$.