How can $\sqrt{4+2\sqrt{3}} = 1+\sqrt{3}$?

Question

I'm a high school student and I'm currently studying a lot of Numerical Sets and Radicals then I came across with a problem that wants me to explain why $\sqrt{4+2\sqrt{3}} = 1+\sqrt{3}$, but I don't know how to do, this was my equation: $\sqrt{4} + \sqrt{2}\times\sqrt[4]{3}$ = $2 + \sqrt[4]{2^2\times3}$ = $2 + \sqrt[4]{2^2\times2+1}$ = $2 + \sqrt[4]{2^3 + 1}$. Where is the problem?

Answer 1

To eliminate a square root, you need a square.

In $$\sqrt{4+2\sqrt3},$$ $3$ is obviously not a square and we can't simplify it. But is $4+2\sqrt3$ a square ?

Here we need to make the educated guess that this expression could be the square of $$a+b\sqrt3,$$ i.e.

$$a^2+2\sqrt3ab+3b^2.$$

This is a good candidate as we get an integer term and another which is a multiple of $\sqrt3$.

By inspection, we immediately see that $a=b=1$ works, hence

$$\sqrt{4+2\sqrt3}=1+\sqrt3.$$

Answer 2

Your first step is incorrect: $$\sqrt{4+2\sqrt{3}}\neq \sqrt{4}+\sqrt{2}\times \sqrt[4]{3}$$ In general: $$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$$ Where $a,b \in \mathbb{R}^+$.

For example, $a=4, b=9$ does not satisfy.

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Here is a hint for a correct approach:

Notice that one may write: $$4+2\sqrt{3}=1+2\sqrt{3}+3=1^2+2\sqrt{3}+(\sqrt{3})^2$$ Now use the fact that: $$(a+b)^2=a^2+2ab+b^2$$