<s>[INST] Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.
 [/INST] More generally,
if
$t 
=\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}}
$,
$\begin{array}\\
t^2
&=(\sqrt{a+\sqrt{b}}+\sqrt{a-\sqrt{b}})^2\\
&=a+\sqrt{b}+2(\sqrt{a+\sqrt{b}}\sqrt{a-\sqrt{b}})+a-\sqrt{b}\\
&=2a+2\sqrt{(a+\sqrt{b})(a-\sqrt{b})}\\
&=2a+2\sqrt{a^2-b}\\
\text{so}\\
t
&=\sqrt{2a+2\sqrt{a^2-b}}\\
\end{array}
$
In this case,
$a=6$
and
$b=20$
so
$t
=\sqrt{2\cdot 6+2\sqrt{36-20}}
=\sqrt{12+8}
=\sqrt{20}
=2\sqrt{5}
$.
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