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 $2\sqrt{2-\sqrt 3}$.