4√(5*x^8*y^3)/(27*x^2)

Need to simplify and rationalize the denominator.

Just a little confused where to start, i already seperated the top and bottom of the brackets and each individual thing inside. And the first *4√* being "fourth root"

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- Sep 10th 2009, 03:37 PMDjaevelStuck again :(
4√(5*x^8*y^3)/(27*x^2)

Need to simplify and rationalize the denominator.

Just a little confused where to start, i already seperated the top and bottom of the brackets and each individual thing inside. And the first *4√* being "fourth root" - Sep 10th 2009, 10:40 PMGrandad
Hello DjaevelSince you talk about rationalising the denominator, I assume that you mean $\displaystyle \sqrt[4]{\frac{5x^8y^3}{27x^2}}$ and not $\displaystyle \frac{\sqrt[4]{5x^8y^3}}{27x^2}$ (which is what you've written). So, study what I've done here carefully - I've used the rules of indices (and the fact that $\displaystyle 3^3=27$):

$\displaystyle \sqrt[4]{\frac{5x^8y^3}{27x^2}}=\frac{\sqrt[4]{5}\sqrt[4]{x^8}\sqrt[4]{y^3}}{\sqrt[4]{27}\sqrt[4]{x^2}}$

$\displaystyle =\frac{5^{\frac14}x^2y^{\frac34}}{27^{\frac14}x^{\ frac12}}$

$\displaystyle =\frac{5^{\frac14}x^{\frac32}y^{\frac34}}{3^{\frac 34}}\times\frac{3^{\frac14}}{3^{\frac14}}$

$\displaystyle =\frac{5^{\frac14}3^{\frac14}x^{\frac32}y^{\frac34 }}{3^{\frac34}3^{\frac14}}$

$\displaystyle =\frac{15^{\frac14}x^{\frac32}y^{\frac34}}{3}$

Grandad - Sep 11th 2009, 12:35 PMDjaevel
Alright that makes tons of sense. I just havent done any of this math in a long while and im a whole lot rusty.

Only thing that im stuck on is that its supossed to be X/3 4√15*x^2*y^3

I get where i get the x/3 from, but the x^2 under the 4th root i do not. And i get where the 4th root comes from. - Sep 11th 2009, 09:54 PMGrandad
Hello DjaevelContinuing from the answer I gave:

$\displaystyle \frac{15^{\frac14}x^{\frac32}y^{\frac34}}{3}$

$\displaystyle = \frac{\sqrt[4]{15\,x^{(4\times\frac32)}\,y^{(4\times\frac34)}}}{ 3}$

$\displaystyle = \frac{\sqrt[4]{15\,x^6\,y^3}}{3}$

$\displaystyle = \frac{\sqrt[4]{15\,x^4\,x^2\,y^3}}{3}$

$\displaystyle = \frac{x}{3}\sqrt[4]{15\,x^2\,y^3}$

Grandad