# How do you simplify \sqrt { 225x ^ { 18} y ^ { 22} }?

Dec 11, 2016

$15 {x}^{9} {y}^{11}$

#### Explanation:

Recall : $\sqrt{x} = {x}^{\frac{1}{2}} , \mathmr{and} \sqrt{{x}^{2}} = {\left({x}^{2}\right)}^{\frac{1}{2}} = {x}^{2 \cdot \frac{1}{2}} = x$,

$\sqrt{225 {x}^{18} {y}^{22}} = \sqrt{15 \cdot 15 \cdot {x}^{18} \cdot {y}^{22}} = {\left({15}^{2} \cdot {x}^{18} \cdot {y}^{22}\right)}^{\frac{1}{2}}$
$= {15}^{2 \cdot \frac{1}{2}} \cdot {x}^{18 \cdot \frac{1}{2}} \cdot {y}^{22 \cdot \frac{1}{2}}$
$= 15 \cdot {x}^{9} \cdot {y}^{11}$

Dec 11, 2016

$\sqrt{225 {x}^{18} {y}^{22}} = \left\mid 15 {x}^{9} {y}^{11} \right\mid$

#### Explanation:

Note that:

$225 {x}^{18} {y}^{22} = {15}^{2} \cdot {\left({x}^{9}\right)}^{2} \cdot {\left({y}^{11}\right)}^{2} = {\left(15 {x}^{9} {y}^{11}\right)}^{2}$

So $15 {x}^{9} {y}^{11}$ is a square root of $225 {x}^{18} {y}^{22}$.

We also find:

${\left(- 15 {x}^{9} {y}^{11}\right)}^{2} = 225 {x}^{18} {y}^{22}$

So $- 15 {x}^{9} {y}^{11}$ is also a square root of $225 {x}^{18} {y}^{22}$.

Note that if $a \ge 0$ then $\sqrt{a}$ denotes the non-negative square root.

So if, for example, $x < 0$ and $y > 0$ then $15 {x}^{9} {y}^{11} < 0$ would be the wrong square root.

We can express that we want the non-negative square root by taking the modulus:

$\sqrt{225 {x}^{18} {y}^{22}} = \left\mid 15 {x}^{9} {y}^{11} \right\mid$