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

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

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Answer 1 · 2017-09-03T09:25:48

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

Explanation:

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

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

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

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

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

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

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

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

Answer 2 · 2017-09-03T09:49:07

I different interpretation of the question to that by Tazwar Sikder

$2 x^{3} y^{7} \sqrt{2 x y} d d d$ $2x^3y^7sqrt(2xy) color(white)(ddd)$ Answer checked and confirmed

Explanation:

Assumption: the question is meant to read:

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

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

Write as:

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

This is the same as

$\frac{\sqrt{{(2^{2})}^{3} \times 2 \times {(x^{2})}^{4} \times x \times {(y^{2})}^{8}}}{\sqrt{4^{2} \times x^{2} \times y}}$ $(sqrt((2^2)^3xx2xx (x^2)^4xx x xx(y^2)^8))/(sqrt(4^2xx x^2xxy)$
$d$ $color(white)("d")$

$\frac{2^{3} x^{4} y^{8} \sqrt{2 x}}{4 x \sqrt{y}}$ $(2^3x^4y^8sqrt(2x))/( 4xsqrt(y))$

Cancel out the $4 x$ $4x$ in the denominator

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

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

$\frac{2 x^{3} y^{8} \sqrt{2 x y}}{y}$ $(2x^3y^8sqrt(2xy))/y$

Cancel out the $y$ $y$ in the denominator

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

I choose $x = 2; y = 2$ $x=2; y=2$

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

$2x^3y^7sqrt(2xy) =5792.61875.... color(red)(larr" It works")$

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

2 Answers

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