Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$

Question

$$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$

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 $\sqrt{12}$.

Answer 1

Hint. Observe that $$ (\sqrt{5}-1)^2=6-2\sqrt{5},\quad (\sqrt{5}+1)^2=6+2\sqrt{5}. $$

Answer 2

Hint: $(\sqrt{5}\pm 1)^2 = 6 \pm 2 \sqrt{5}$.

Answer 3

Set $$t=\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$ $$t^2=6-2\sqrt5+2\sqrt{(6-2\sqrt5)(6+2\sqrt5)}+6+2\sqrt5$$ $$t^2=12+2\sqrt{36-20}=12+2(4)=20$$ $$t^2=20\implies t=2\sqrt5$$

Answer 4

Hint $\,\ \sqrt a + \sqrt b\, = \sqrt{(\sqrt a + \sqrt b)^2} = \sqrt{a+b +2{\sqrt{ab}}}.\ $ Here $\ a+b=12,\ \color{}{\sqrt{ab}} = 4.$

Answer 5

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} $.

Answer 6

Let $\alpha = \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}$. Then $$ \alpha^2 = 12+2\sqrt{36-20} = 20 $$ hence $\color{red}{\alpha=2\sqrt{5}}$.