# Sqrt 128x^9y^16/[16x^2y]? Simplify

Sep 3, 2017

$\frac{\sqrt{2 {x}^{5}} {y}^{7}}{8}$

#### Explanation:

We have: $\frac{\sqrt{128 {x}^{9} {y}^{16}}}{16 {x}^{2} y}$

$= \frac{\sqrt{128} \cdot \sqrt{{x}^{9}} \cdot \sqrt{{y}^{16}}}{16 {x}^{2} y}$

$= \frac{8 \sqrt{2} \cdot \sqrt{{x}^{9}} \cdot {y}^{8}}{16 {x}^{2} y}$

$= \frac{8 \sqrt{2} {y}^{8} \cdot \sqrt{{x}^{9}}}{16 {x}^{2} y}$

$= \frac{\sqrt{2} {y}^{7}}{8} \cdot \left({x}^{\frac{9}{2} - 2}\right)$

$= \frac{\sqrt{2} {y}^{7}}{8} \cdot \left({x}^{\frac{5}{2}}\right)$

$= \frac{\sqrt{2} {y}^{7}}{8} \cdot \sqrt{{x}^{5}}$

$= \frac{\sqrt{2 {x}^{5}} {y}^{7}}{8}$

Sep 3, 2017

I different interpretation of the question to that by Tazwar Sikder

$2 {x}^{3} {y}^{7} \sqrt{2 x y} \textcolor{w h i t e}{\mathrm{dd} d}$ Answer checked and confirmed

#### Explanation:

Assumption: the question is meant to read:

sqrt((128x^9y^16)/(16x^2y)

If you are ever not sure about the roots of a large number do a rough sketch of a prime factor tree.

Write as:

$\frac{\sqrt{128 {x}^{9} {y}^{16}}}{\sqrt{16 {x}^{2} y}}$

This is the same as

(sqrt((2^2)^3xx2xx (x^2)^4xx x xx(y^2)^8))/(sqrt(4^2xx x^2xxy)
$\textcolor{w h i t e}{\text{d}}$

$\frac{{2}^{3} {x}^{4} {y}^{8} \sqrt{2 x}}{4 x \sqrt{y}}$

Cancel out the $4 x$ in the denominator

$\frac{2 {x}^{3} {y}^{8} \sqrt{2 x}}{\sqrt{y}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{y}}{\sqrt{y}}$

$\frac{2 {x}^{3} {y}^{8} \sqrt{2 x y}}{y}$

Cancel out the $y$ in the denominator

$2 {x}^{3} {y}^{7} \sqrt{2 x y}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check by substituting arbitrary values for $x \mathmr{and} y$

I choose x=2; y=2

sqrt((128x^9y^16)/(16x^2y)=5792.61875.....#

$2 {x}^{3} {y}^{7} \sqrt{2 x y} = 5792.61875 \ldots . \textcolor{red}{\leftarrow \text{ It works}}$