# How do you simplify sqrt2(sqrt8-sqrt4)?

Jul 2, 2015

$\sqrt{2} \left(\sqrt{8} - \sqrt{4}\right) = \sqrt{2} \left(2 \sqrt{2} - 2\right) = 4 - 2 \sqrt{2}$

#### Explanation:

A couple of properties of a square root play an important role in these transformations.

First of all, square root in this case is understood in its arithmetic meaning, that is a positive value. So, $\sqrt{4}$ is equal to $2$, not $\pm 2$.

So, we can replace: $\sqrt{4} = 2$.

Secondly, keep in mind that $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$, where $a$ and $b$ are positive numbers. This can be easily proven based on the definition of $\sqrt{X}$ being a number that, if raised to the power of $2$, would produce $X$.
Indeed, let's raise $\sqrt{a} \cdot \sqrt{b}$ to the power of 2:
${\left[\sqrt{a} \cdot \sqrt{b}\right]}^{2} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{a} \cdot \sqrt{b} =$

$= \left(\sqrt{a} \cdot \sqrt{a}\right) \cdot \left(\sqrt{b} \cdot \sqrt{b}\right) = a \cdot b$,

which proves that $\sqrt{a} \cdot \sqrt{b}$ is $\sqrt{a \cdot b}$

That's why we can replace
$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2 \sqrt{2}$

Finally,
$\sqrt{2} \cdot 2 \cdot \sqrt{2} = 2 \cdot \sqrt{2} \cdot \sqrt{2} = 2 \cdot 2 = 4$