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I'm stuck. Squaring it will change it's value. Is there any general method of simplifying expressions of the form $$\sqrt{a+b}-\sqrt{a-b} = c \ \ ?$$

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It's clearly positive. Its square is $$a^2=6+2\sqrt 5+6-2\sqrt 5-2\sqrt{(6+2\sqrt5)(6-2\sqrt5)} =12-2\sqrt{16}=4$$ so $a=2$.

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If I were you I would tackle the individual square roots to find$$\sqrt{6+2\sqrt{5}} = \sqrt{5}+1$$ $$\sqrt{6-2\sqrt{5}} = \sqrt{5}-1$$ and see that the difference is $2$.

Note: It is not always possible to denest a square root like this, but if our goal is to try to find integers $a$, $b$ such that $\sqrt{6+2\sqrt{5}}=\sqrt{a}+\sqrt{b}$, we can try to solve $a$, $b$ by $$6+2\sqrt{5} = a+b+2\sqrt{ab}\implies a+b=6, \ \ ab=5\implies (a,b)=(1,5) \ \text{or} \ (5,1)$$

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    $\begingroup$ How did you denest them like that? Please explain, I'm not Ramanujan :) $\endgroup$
    – Parseval
    Apr 30, 2017 at 10:03
  • $\begingroup$ How do you extract those square roots? Inquiring minds want to know. $\endgroup$ Apr 30, 2017 at 10:04
  • $\begingroup$ @Parseval Once you see that $6=5+1$, you may recognize $x^2+2x+1$ with $x=\sqrt{5}$. However, the other answer is less tricky. $\endgroup$ Apr 30, 2017 at 10:04
  • $\begingroup$ I added a note :) :) $\endgroup$
    – Lazy Lee
    Apr 30, 2017 at 10:08
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Here is how you de-nest double layer square roots. All variables below are assumed rational.

Start with

\begin{equation} \tag{1} \sqrt{a+ \sqrt{b}} = \sqrt {x} + \sqrt{y}, \end{equation} which has the conjugate relation \begin{equation} \tag{2} \sqrt{a - \sqrt{b}} = \sqrt{x} -\sqrt{y}, \end{equation} where $x\ge y $ wlog.

Multiply these two relations together, and use the difference of squares factorization:

\begin{equation} \tag{3} \sqrt {a^2-b} \equiv r = x-y. \end{equation}

The quantity $a^2-b$ must be a perfect square if the de-nesting is to succeed. If so, call its square root $r $.

Now using the uniqueness of quadratic sure, square (1) and equate rational parts, getting:

\begin{equation} \tag{4} a =x + y. \end{equation} Equations (3) and (4) constitute a linear system for $x$ and $y$, which is easily solved, thus \begin{equation*} x = \frac{a + r}{2} \qquad \text{and} \qquad y = \frac{a - r}{2}. \end{equation*} Plug these results into the given problem ($a=6, b=20$) and infer that the de-nested roots are:

$$\sqrt {6+2\sqrt {5}} =\sqrt {5}+1 \qquad \text{and} \qquad \sqrt {6-2\sqrt {5}} =\sqrt {5}-1 $$ with the difference of $2$.

Note:If in this development we allow negative values of $b $ and drop the rationality requirement on $r$, this is also a method to extract square roots of complex numbers.

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