How is $\sqrt{6}-\sqrt{2}$ equal to $2\sqrt{2-\sqrt{3}}$?

[Drain Brain]

how is $\sqrt{6}-\sqrt{2}$ equal to $2\sqrt{2-\sqrt{3}}$

please explain. Thanks!

 

[magneto1]

It does not. $\sqrt 6 - \sqrt 2 \approx 1.03528$, and $2 \sqrt 2 - \sqrt 3 \approx 1.09638$.

 

[Nono713]

It does not. $\sqrt 6 - \sqrt 2 \approx 1.03528$, and $2 \sqrt 2 - \sqrt 3 \approx 1.09638$.

Notice OP wrote $2 \sqrt{2 - \sqrt{3}}$, not $2 \sqrt{2} - \sqrt{3}$.

 

[magneto1]

Oops. The font rendering in Safari is messing up.

Then, $\sqrt 6 - \sqrt 2 = \sqrt 2 (\sqrt 3 - 1)$. Since the number are positive, use $a = \sqrt{a^2}$. $ \sqrt 2 (\sqrt 3 - 1) = \sqrt{2 (\sqrt{3}-1)^2}$, and deduce from there $2\sqrt{2-\sqrt 3}$.

 

[CellCoree]

The equality between $\sqrt{6}-\sqrt{2}$ and $2\sqrt{2-\sqrt{3}}$ can be proven using basic algebraic manipulations. First, we can expand the expression $2\sqrt{2-\sqrt{3}}$ by multiplying the 2 inside the square root:

$2\sqrt{2-\sqrt{3}} = 2\sqrt{2-1\cdot\sqrt{3}} = 2\sqrt{2-{\sqrt{3}}^2} = 2\sqrt{(2-\sqrt{3})(2+\sqrt{3})}$

Next, we can use the identity $(a-b)(a+b) = a^2-b^2$ to simplify the expression further:

$2\sqrt{(2-\sqrt{3})(2+\sqrt{3})} = 2\sqrt{2^2-{\sqrt{3}}^2} = 2\sqrt{4-3} = 2\sqrt{1} = 2$

Therefore, we have shown that $2\sqrt{2-\sqrt{3}} = 2$, which is equivalent to $\sqrt{6}-\sqrt{2}$ since they are both equal to 2. This proves that $\sqrt{6}-\sqrt{2}$ is indeed equal to $2\sqrt{2-\sqrt{3}}$.