# Simplifying expression $\sqrt{6+2\sqrt{5}} - \sqrt{6-2\sqrt{5}}.$

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

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 $a=2$.

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 $$6+2\sqrt{5} = a+b+2\sqrt{ab}\implies a+b=6, \ \ ab=5\implies (a,b)=(1,5) \ \text{or} \ (5,1)$$

• How did you denest them like that? Please explain, I'm not Ramanujan :) Apr 30, 2017 at 10:03
• How do you extract those square roots? Inquiring minds want to know. Apr 30, 2017 at 10:04
• @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. Apr 30, 2017 at 10:04
• I added a note :) :) Apr 30, 2017 at 10:08

Here is how you de-nest double layer square roots. All variables below are assumed rational.

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

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

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

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:

$$$$\tag{4} a =x + y.$$$$ Equations (3) and (4) constitute a linear system for $$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.