# Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$

How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?

I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.

I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?

Help would be appreciated! Thanks!

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\begin{align} &\ \ \ \sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}} \\ \\ &=\sqrt{(\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}\ \\ \\ &=\sqrt{3}+1+\sqrt{3}-1 \\ \\ &=\boxed{2\sqrt{3}} \end{align}

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You have the right idea in squaring and then taking the square root.

Note that $k = (\sqrt{4 + 2\sqrt{3}} + \sqrt{4 - 2\sqrt{3}})^{2} = 4 + 2\sqrt{3} + 2\sqrt{(4+2\sqrt{3})(4-2\sqrt{3})} + 4 - 2\sqrt{3}$

But this is just:

$k = 8 + 2\sqrt{16 - 12} = 12$

So

$\sqrt{k} = 2\sqrt{3}$

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Good solution, certainly much better than mine. – André Nicolas May 7 '13 at 1:31
I wouldn't say that @AndréNicolas, just less observant than yours ;) – Alex Wertheim May 7 '13 at 1:31
+1 Much better explanation than the "rabbit out of a hat" accepted answer. – wim May 7 '13 at 4:35

Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$

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Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring: \begin{align} \left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2 &=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\ &=8+2\sqrt{16-12}\\[6pt] &=12 \end{align} Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$

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Write as $\sqrt{4+2\sqrt{3}} = a+b\sqrt{3}$. Now square both sides, equate real and radical part. This gives two equations in $a$ and $b$. Now eliminate $a$, solve for $b$. Goes perfect. Same for the other term.

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Using the following formula:

$$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+c}{2}}\pm\sqrt{\frac{a-c}{2}}$$

where

$c=\sqrt{a^2-b}$

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