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# 1.) Rewrite in simplest radical form:

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1.) Rewrite in simplest radical form: Feb 26, 2019

#1
+2

I find it easiest to put these in expnential form  and then simplify

√  27x^11y^6  =

(3^3)^(1/2) * (x^11)^(1/2) * (y^6)^(1/2)  =

(3)^(3/2) * (x)^11/2 * (y)^3

Now.....we can  find out how many powers of  each thing come out of the radical and how many powers stay in by this procedure :

Divide 3/2  and express as a mixed number =  1 + 1/2

The whole number tells how many powers of 3  come out of the radical =  3^1

The numerator tells how many powers of 3 stay in the radical = 3^1

Similarly

11/2 =   5 + 1/2

5 powers of x come out......1 power stays in

And

3  =  3

We have no mixed number.....so 3 powers of  y come out   and there remain 0 that stay in

So....we have

3^1* x^5 * y^3 √  [ 3^1 * x^1 ]  =

3x^5y^3 √ [ 3x ]   Feb 26, 2019
#2
+2

√ [8x^3y^8 ]  =

√[ 2^3 * x^3 * y^8] =

(2^3)^(1/2) * (x^3)^(1/2) * (y^8)^(1/2) =

(2)^(3/2) * (x)^(3/2) * y^4

As before

3/2 = 1 + 1/2         1 power of 2 comes out 1 stays in

3/2   = 1 + 1/2        1 power of x comes out, 1 stays in

4 = 4                     4 powers of y come out......0 remain in

So we have

2xy^4√ [ 2x ]   Feb 26, 2019
#3
+2

√72x^9y^4  =  √2 * √36 *x^9 * y^4

We can write

√2* (36)^(1/2) * (x^9)^(1/2) * (y^4)^(1/2 )  =

√2 *  6   * x^(9/2) * (y)^(2)

The only thing here is to find out how many x's stay in and how many come out

9/2 = 4 + 1/2        4 powers come out and 1 power stays in......so we have

6x^4y^2 √ [ 2 x ]   Feb 26, 2019