# How do you simplify sqrt80+sqrt20?

Sep 24, 2015

$\sqrt{80} + \sqrt{20} = 6 \sqrt{5}$

#### Explanation:

Fully factorising the terms in the square roots to their prime factors (to assist clarity), it might be noted

$\sqrt{80} = \sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 5}$

and

$\sqrt{20} = \sqrt{2 \cdot 2 \cdot 5}$

Collecting terms that might be expressed as squares, it might be noted

$\sqrt{80} = \sqrt{{4}^{2} \cdot 5}$

and

$\sqrt{20} = \sqrt{{2}^{2} \cdot 5}$

Taking the squared terms outside the radical (taking the square roots whilst doing this, and presuming that the use of the radical sign implies that only the positive square root is intended)

$\sqrt{80} = 4 \sqrt{5}$

and

$\sqrt{20} = 2 \sqrt{5}$

Both terms now comprise the same radical ($\sqrt{5}$) but with different coefficients. As the radical is the same, they may be added

$\sqrt{80} + \sqrt{20} = 4 \sqrt{5} + 2 \sqrt{5}$

$= \left(4 + 2\right) \sqrt{5}$

$= 6 \sqrt{5}$, as required.