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

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 $\sqrt{12}$.

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  • $\begingroup$ Note that $\sqrt{a+b}+\sqrt{a-b}\neq\sqrt{2a},$ which is where I think you made your mistake. $\endgroup$ – Arcturus Jul 26 '16 at 19:04
  • $\begingroup$ Observe that the minimal polynomial for roots $6±2\sqrt5$ is $x^2-12x+16$ $\endgroup$ – Zack Ni Jul 27 '16 at 11:36
  • $\begingroup$ Consider $\sqrt{6.5 - 2.5} + \sqrt{6.5 + 2.5}$. The answer is not $\sqrt{13}$. $\endgroup$ – John Joy Jul 27 '16 at 17:05
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Set $$t=\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$ $$t^2=6-2\sqrt5+2\sqrt{(6-2\sqrt5)(6+2\sqrt5)}+6+2\sqrt5$$ $$t^2=12+2\sqrt{36-20}=12+2(4)=20$$ $$t^2=20\implies t=2\sqrt5$$

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

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

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

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  • $\begingroup$ Did you mean $\sqrt{a + b + 2\sqrt{ab}}$? $\endgroup$ – N. F. Taussig Jul 26 '16 at 15:54
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Let $\alpha = \sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}$. Then $$ \alpha^2 = 12+2\sqrt{36-20} = 20 $$ hence $\color{red}{\alpha=2\sqrt{5}}$.

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