1. ## Roots Need Help

This section talks about how you can simplify roots by combining two properties, ill refer to as A & B.

in$\displaystyle {^n}\sqrt{a^m}$ if m is at least as large as n then this can be simplified using

Property A. $\displaystyle ({^n}\sqrt{ab})={^n}\sqrt{a} {^n\sqrt{b}}$
&
Property B. $\displaystyle {^n}\sqrt{a^n} = a$

then goes on to list this as an example.

$\displaystyle \sqrt{32x^3}=\sqrt{2^22^22^1x^2x^1}= \sqrt{2^2} \sqrt{2^2} \sqrt{x^2} \sqrt{2x} = 2 * 2x\sqrt{2x} = 4x\sqrt{2x}$

I have no idea whats going on here. If someone could please walk me through this it would be greatly appreciated. Many thanks.

2. Originally Posted by cmf0106
This section talks about how you can simplify roots by combining two properties, ill refer to as A & B.

in$\displaystyle {^n}\sqrt{a^m}$ if m is at least as large as n then this can be simplified using

Property A. $\displaystyle ({^n}\sqrt{ab})={^n}\sqrt{a} {^n\sqrt{b}}$
&
Property B. $\displaystyle {^n}\sqrt{a^n} = a$

then goes on to list this as an example.

$\displaystyle \sqrt{32x^3}=\sqrt{2^22^22^1x^2x^1}= \sqrt{2^2} \sqrt{2^2} \sqrt{x^2} \sqrt{2x} = 2 * 2x\sqrt{2x} = 4x\sqrt{2x}$

I have no idea whats going on here. If someone could please walk me through this it would be greatly appreciated. Many thanks.
Let me show another way to look at it. See if you like it.

$\displaystyle \sqrt{32x^3}$

First thing I do is change everything in the radicand to exponents:

$\displaystyle \sqrt{2^5x^3}$

Now divide the each exponent of each factor in the radicand by the index of the radical, in this case (2):

$\displaystyle \frac{5}{2}=2 \ \ remainder \ \ 1$

$\displaystyle \frac{3}{2}=1 \ \ remainder \ \ 1$

This means that a $\displaystyle 2^2$ is extracted from the radicand and a $\displaystyle 2^1$ remains. $\displaystyle x^1$ is extracted from the radicand and $\displaystyle x^1$ remains. What's extracted is 4x and what's left in the radicand is 2x.

$\displaystyle \sqrt{2^5x^3}=2^2x\sqrt{2x}=4x\sqrt{2x}$

3. EDIT: Double checked, I think that makes sense. Let me try your method on an additional problem see if I have the right idea.

$\displaystyle {^3}\sqrt{8y}^3$

1. Change To Exponents $\displaystyle \sqrt{2^3y^3}$

2. Divide each exponent of each factor by the index of the radical in this case 3

$\displaystyle 3\div3=1$no remains
$\displaystyle 3\div3=1$no remains

&

$\displaystyle {^3}\sqrt{16x^7y^5}$

1. Change to Exponents
$\displaystyle \sqrt{2^4x^7y^5}$

2.
(2^4) = $\displaystyle 4\div3$= 1 & 1 remains
(x^7) = 2 and 1 remains
(y^5) = 1 and 2 remains

3. $\displaystyle 2x^2y^3$ $\displaystyle {^3}\sqrt{2xy^2}$

Do these look right?
And it seems that two criteria always apply
1. M is at least as large as N
2. in your final answer if there is a radicand it will always be to the originals index

4. Originally Posted by cmf0106
EDIT: Double checked, That makes more sense up until where you state 2^2 is extracted from the radicand. Shouldn't it be 2^1 since 5/2= 2 remainder 1?
No.

$\displaystyle \sqrt{2^5x^3}=2^{\frac{5}{2}}x^{\frac{3}{2}}=2^2 \cdot 2^{\frac{1}{2}} \cdot x^1 \cdot x^{\frac{1}{2}}=4x \cdot 2^{\frac{1}{2}} \cdot x^{\frac{1}{2}}=4x\sqrt{2x}$

5. Originally Posted by cmf0106
EDIT: Double checked, I think that makes sense. Let me try your method on an additional problem see if I have the right idea.

$\displaystyle {^3}\sqrt{8y}^3$

1. Change To Exponents $\displaystyle \sqrt{2^3y^3}$

2. Divide each exponent of each factor by the index of the radical in this case 3

$\displaystyle 3\div3=1$no remains
$\displaystyle 3\div3=1$no remains