# Math Help - Stuck again :(

1. ## Stuck 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"

2. Hello Djaevel
Originally Posted by Djaevel
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"
Since you talk about rationalising the denominator, I assume that you mean $\sqrt[4]{\frac{5x^8y^3}{27x^2}}$ and not $\frac{\sqrt[4]{5x^8y^3}}{27x^2}$ (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 $3^3=27$):

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

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

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

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

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

3. 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.

4. Hello Djaevel
Originally Posted by Djaevel
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.
Continuing from the answer I gave:

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

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

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

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

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