# How do you simplify sqrt(144x^6)?

Oct 10, 2015

This can be written as $12 {x}^{3}$

#### Explanation:

$\sqrt{144 {x}^{6}} = \sqrt{144} \cdot \sqrt{{x}^{6}} = \sqrt{12 \cdot 12} \cdot \sqrt{{x}^{6}} = 12 \cdot \sqrt{{x}^{6}} = 12 {x}^{3}$

The final step: $\sqrt{{x}^{6}} = {x}^{3}$ comes from the properties of exponents, which say, that:

1. $\sqrt[n]{a} = {a}^{\frac{1}{n}}$
2. ${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c}$

Using the properties above we can write, that:

$\sqrt{{x}^{6}} = {\left({x}^{6}\right)}^{\frac{1}{2}} = {x}^{\frac{6}{2}} = {x}^{3}$

Oct 10, 2015

$\sqrt{144 {x}^{6}} = 12 \left\mid {x}^{3} \right\mid$

#### Explanation:

If $a , b \ge 0$ then $\sqrt{a b} = \sqrt{a} \sqrt{b}$

If $a \ge 0$ then $\sqrt{{a}^{2}} = a$

$\sqrt{144 {x}^{6}} = \sqrt{144} \sqrt{{x}^{6}} = \sqrt{{12}^{2}} \sqrt{{\left\mid {x}^{3} \right\mid}^{2}} = 12 \left\mid {x}^{3} \right\mid$

The modulus operation is necessary to deal with the case $x < 0$.

Going deeper:

Note that if $a < 0$ then $\sqrt{{a}^{2}} = - a$, so in general we can write $\sqrt{{x}^{2}} = \left\mid x \right\mid$.

Any number has two square roots. If $a \ge 0$ then it has a positive square root: $\sqrt{a}$ and a negative square root: $- \sqrt{a}$.

If $a < 0$ then it has square roots $i \sqrt{- a}$ and $- i \sqrt{- a}$. We call $i \sqrt{- a}$ the principal square root, defining $\sqrt{a} = i \sqrt{- a}$.

When it comes to square roots of negative numbers, the identity $\sqrt{a b} = \sqrt{a} \sqrt{b}$ breaks down. For example:

$1 = \sqrt{1} = \sqrt{- 1 \times - 1} \ne \sqrt{- 1} \times \sqrt{- 1} = - 1$