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For this question,

Show the following irrational-looking expressions are actually rational numbers.

(a) $\sqrt{4+2\sqrt{3}}-\sqrt{3}$, and

(b) ...

I solved it as follows:

$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\ x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\ (x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\ \end{align}\\ x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\ (x-1)(x+(1+2\sqrt{3}))=0.$$

My question is that, there are two numbers satisfying $x = \sqrt{4+2\sqrt{3}}-\sqrt{3}$, but one of them is irrational. Then, how can we say $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational as a whole?

eca2ed291a2f572f66f4a5fcf57511
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    Because $4+2\sqrt3=(\sqrt3)^2+2\sqrt3+1$ hence $\sqrt{4+2\sqrt3}=\sqrt3+1$. – Did Oct 04 '17 at 08:07
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    Additionally, $x=-1-2\sqrt3$ does is not a solution to your original equality –  Oct 04 '17 at 08:09
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    There aren't two numbers satisfying $x = \sqrt{4+2\sqrt{3}}-\sqrt{3}$, there is only one. You introduce a second solution when you square in the third line. – Arthur Oct 04 '17 at 08:10
  • @Did Will it be wrong to say: $4+2\sqrt3=(-\sqrt3)^2+2\cdot(-\sqrt3).(-1)+(-1)^2 \implies \sqrt(4+2\sqrt3)=-\sqrt3-1$? – SchrodingersCat Oct 04 '17 at 08:10
  • @Arthur I'm not sure what do you mean by a second solution... – eca2ed291a2f572f66f4a5fcf57511 Oct 04 '17 at 08:11
  • @SchrodingersCat, shouldn't the root be a positive number? –  Oct 04 '17 at 08:12
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    @SchrodingersCat Yes, that is wrong to say, but not for any deep arithmetical reasons. What you write the is wrong because we define square roots to be positive. – Arthur Oct 04 '17 at 08:12
  • @Arthur That's the convention. I wanted to point it out. – SchrodingersCat Oct 04 '17 at 08:13
  • @Il-seobBae, there are two solutions to ${(x+\sqrt3)}^2={\sqrt{4+2\sqrt3}}^2$ –  Oct 04 '17 at 08:13
  • @SchrodingersCat What Arthur said. Actually, I find your comment rather frightening... Was it meant to be a joke? – Did Oct 04 '17 at 08:13
  • @SchrodingersCat Sorry but what it is exactly that you wanted to point out at me? – Did Oct 04 '17 at 08:14
  • @Did Nothing to you. To the OP. That we only consider the positive square root. – SchrodingersCat Oct 04 '17 at 08:15
  • @SchrodingersCat So, you wish to address a comment at the OP, hence you start the comment by my username? – Did Oct 04 '17 at 08:15
  • And by a second solution I mean this (illustrated by a slightly simplified and inherently circular example): If we start with $x=\sqrt4$, and want to solve it, then we can square it to get $x^2=4$, then use the quadratic formula to obtain $x=\pm 2$. But the original equation only had one solution, namely $2$. What happened was that the moment I squared I became unable to tell the difference between whether the original equation was $x=\sqrt4$, or $-x=\sqrt4$. The new equation $x^2=4$ accommodates them both. – Arthur Oct 04 '17 at 08:16
  • @Did I took your username because, you posted something similar. I just wanted to point out to OP, taking into consideration your post, that what you did is right ( $\sqrt{}$ is defined as $+\sqrt{}$), but if he considered $\sqrt{}$ as $-\sqrt{}$, then he would get the wrong result. That is what I showed. – SchrodingersCat Oct 04 '17 at 08:23
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    @SchrodingersCat OK. Please try to be less confusing next time. – Did Oct 04 '17 at 08:31

3 Answers3

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$\sqrt{4+2\sqrt{3}}=\sqrt{1+2\sqrt{3}+3}=\sqrt{(1+\sqrt{3})^2}=1+\sqrt{3}$

Duchamp Gérard H. E.
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A classical trap: when you square the members of an equation, you introduce alien solutions.

Going from $$x+a=b$$ to $$x^2+2ax+a^2=b,$$

you introduce

$$x+a=-b$$

that has noting to do with the original problem.

In the given question, the irrational parts of $a$ and $b$ cancel each other in $b-a$, but not in $-b-a$.

1

With respect to your written
note that $$\bf{x = \sqrt{4+2\sqrt{3}}-\sqrt{3}=\sqrt{4+\sqrt{12}}-\sqrt{3} > 0}$$ $$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\ x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\ (x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\ \end{align}\\ x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\ (x-1)(x+(1+2\sqrt{3}))=0.$$ now you have two roots ,but $\bf{x >0}$ so $$x-1=0 \to x=1 \checkmark \\x=-1-2\sqrt 3 <0 \times$$

Khosrotash
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