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!
\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}
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
$\sqrt{k} = 2\sqrt{3}$
Hint: Find the square of $1+\sqrt{3}$. ${}{}{}{}{}{}{}{}{}$
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, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
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.
You want to calculate $a+b$ where $a^2=4+2\sqrt3$ and $b^2=4-2\sqrt3$.
You could notice that $a^2$ and $b^2$ are roots of the quadratic equation $$x^2-8x+4=0.$$ (To find this equation you just need to find appropriate coefficients in quadratic formula.)
Now from Viete's formula you get \begin{align*} a^2+b^2&=8\\ a^2b^2&=4 \end{align*} which (together with the fact that $a,b>0$) leaves you with $ab=2$ and $$(a+b)^2=a^2+b^2+2ab=8+2\cdot3=12.$$ You get that $$a+b=\sqrt{12}=2\sqrt3.$$
Using the following formula:
$$\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+c}{2}}\pm\sqrt{\frac{a-c}{2}}$$
where
$c=\sqrt{a^2-b}$
