1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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 $y$ $\endgroup$
    – MafPrivate
    Aug 26, 2019 at 16:35
  • $\begingroup$ why smaller than that value? $\sqrt(6) - \sqrt(2)$ is a valid solution and here $x=6$ that is greather than $8-4\sqrt(3)$ $\endgroup$
    – ESCM
    Aug 26, 2019 at 17:11
  • $\begingroup$ the unique restriction is that $x,y >0$, how you get that inequality for x? $\endgroup$
    – ESCM
    Aug 26, 2019 at 17:12
  • $\begingroup$ 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. $\endgroup$
    – MafPrivate
    Aug 27, 2019 at 2:10

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Why and what is the proof of that? $\endgroup$
    – ESCM
    Aug 26, 2019 at 16:28
  • $\begingroup$ Because if you square this you have $$x+y-2\sqrt{xy}.$$ This resembles the form whose root you want. $\endgroup$
    – Allawonder
    Aug 26, 2019 at 16:31
  • $\begingroup$ @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. $\endgroup$
    – Allawonder
    Aug 26, 2019 at 16:34
  • $\begingroup$ @EduardoS. In short, the signs are important! $\endgroup$
    – Allawonder
    Aug 26, 2019 at 16:40
0
$\begingroup$

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 $\sqrt{y}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .