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January 23, 2017
Posted by **Ben** on Tuesday, April 1, 2008 at 9:09pm.

3 times the square root of 11x^3y (the three is an exponet) over -2 times the square root of 12x^4y (the 4 is an exponet)

- Algebra 2 -
**DrBob222**, Tuesday, April 1, 2008 at 9:52pmsquare top and bottom.

3*sqrt(11x^3y)/-2*sqrt(12x^4y)=

9(11x^3y)/4(12x^4y) =

y cancels

x^3 cancels leaving x on the bottom.

9(11)/4(12x) =

Divide 9 and 12 by 3

3(11)/(4*4x) =

33/16x

Check my work carefully. - Algebra 2 -
**Ben**, Tuesday, April 1, 2008 at 10:30pmActually, I figured it out.....

Because of what it says in the directions: Simplify Each Expression. Rationalize all denominators. Assume that all variables are positive, I have to rationalize all denominators, which means you can't have a radical in the denominator so you have to multiply the numerator and the denominator both by the denominator so......

3*sqrt of 11x^3 y (it is actually 11x cubed then the y follows but i did it this way to keep from confusing anybody) over -2*sqrt of 12x^4 y equals both the top and bottom times -2*sqrt of 12x^4 y which gives you negative 6*sqrt132x^7 y^2 over 4*sqrt24x^8 y^2 which simplified is 12x^4 y, then you multiply 4 by 12x^4 y which gives you -6*sqrt132x^7 y^2 over 48x^4, you then factor down as much as possible of what is under the square root, in this case, 132 can be broken down, using the factor tree to 2*sqrt of 33, this is because we are squaring so there has to be 2 of a number before you can bring it out like there were two 2s so i brought them out and this makes it were 2 is going to be multiplied by -6 12on the outside of the sqrt which gives you -12*sqrt of 33x^7 y^12 over 48x^4, then you break down the "x" and the "y", 2 goes into 7 three times so u bring a x^3 on the outside, the 2 goes into 12 six times so you bring y^6 on the outside which gives you -12x^3 y^6*sqrt of 33x over 48x^4 y, then the y^6 on top can be divided by the y on the bottom which gives you y^5 on top for the final answer of -12x^3 y^5*sqrt of 33x over 48x^4!

Thanks for the help anyway though!