# Simplifying roots

• Dec 14th 2010, 04:57 PM
Kieth89
Simplifying roots
Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

1 over the 5th root of 9

I am told to rationalize the denominator. I got to this:

5th root of 9 to the 4th power over 9

This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

5th root of 27 over 3

Help?
• Dec 14th 2010, 05:02 PM
skeeter
Quote:

Originally Posted by Kieth89
Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

1 over the 5th root of 9

I am told to rationalize the denominator. I got to this:

5th root of 9 to the 4th power over 9

This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

5th root of 27 over 3

Help?

$\displaystyle \frac{\sqrt[5]{9^4}}{9} = \frac{\sqrt[5]{3^8}}{9} = \frac{\sqrt[5]{3^5 \cdot 3^3}}{9} = \frac{3\sqrt[5]{3^3}}{9} = \frac{\sqrt[5]{3^3}}{3}$
• Dec 14th 2010, 05:06 PM
Kieth89
How did you get the root sign to type
• Dec 14th 2010, 05:07 PM
Quote:

Originally Posted by Kieth89
Alright, I've got a final in College Algebra tomorrow but can't figure out how my teacher got this answer. Here was the original problem (as best I can type it):

1 over the 5th root of 9

I am told to rationalize the denominator. I got to this:

5th root of 9 to the 4th power over 9

This works as the answer, but it can simplify even further apparently. I can't figure out how it simplified further into:

5th root of 27 over 3

Help?

$\displaystyle\frac{1}{\sqrt[5]{9}}=\frac{1}{\sqrt[5]{3^2}}=\frac{1}{3^{\frac{2}{5}}}=\frac{3^{\frac{3} {5}}}{3^{\frac{2}{5}}\;3^{\frac{3}{5}}}$
• Dec 14th 2010, 05:29 PM
Kieth89
Could you help me with this one, too?

$\sqrt{(3+x)^2-(3-x)^2}$

Alright, I know that I can't take the things out of the root until they are being multiplied, so what I did at first was just what the problem says:

$\sqrt{(3+x)(3+x)+(-3+x)(-3+x)} = \sqrt{9+6x+x^2+9-6x+x^2}
= \sqrt{2x^2+18} =$

$\sqrt{2(X^2+9)} = ?$
I don't know if I messed up somewhere, or am not seeing the key, but I'm not getting any further than that. The correct answer is supposed to be $2\sqrt{3x}$
• Dec 14th 2010, 05:31 PM
bigwave
another way
by fractional exponents (probably not any easier tho)

$
\frac{1}{9^{\frac{1}{5}}} \rightarrow \frac{1}{3^{\frac{2}{5}}}
$

$
\frac{1}{3^{\frac{2}{5}}}
\times
\frac
{3^
{\frac{3}{5}}
}
{3^
{\frac{3}{5}}
}
\rightarrow
\frac
{3^{\frac{3}{5}}}
{3^{\frac{5}{5}}}
\rightarrow
\frac
{27^
{\frac{1}{5}}
}
{3}
$
• Dec 14th 2010, 05:35 PM
Prove It
$\displaystyle \sqrt{(3+x)^2 - (3 - x)^2} = \sqrt{9 + 6x + x^2 - (9 - 6x + x^2)}$

$\displaystyle = \sqrt{9 + 6x + x^2 - 9 + 6x - x^2}$

$\displaystyle = \sqrt{12x}$

$\displaystyle = \sqrt{4\cdot 3x}$

$\displaystyle =\sqrt{4}\cdot \sqrt{3x}$

$\displaystyle = 2\sqrt{3x}$
• Dec 14th 2010, 05:37 PM
Kieth89
Quote:

Originally Posted by Prove It
$\displaystyle \sqrt{(3+x)^2 - (3 - x)^2} = \sqrt{9 + 6x + x^2 - (9 - 6x + x^2)}$

$\displaystyle = \sqrt{9 + 6x + x^2 - 9 + 6x - x^2}$

$\displaystyle = \sqrt{12x}$

$\displaystyle = \sqrt{4\cdot 3x}$

$\displaystyle =\sqrt{4}\cdot \sqrt{3x}$

$\displaystyle = 2\sqrt{3x}$

So did distributing the minus first mess it all up for me?
• Dec 14th 2010, 05:38 PM
bigwave
$
\sqrt{9 +6x+x^2-9+6x-x^2}
\rightarrow
\sqrt{12x}
\rightarrow
\sqrt{3}\sqrt{4}\sqrt{x}
\rightarrow
2\sqrt{3x}
$
• Dec 14th 2010, 05:48 PM
skeeter
Quote:

Originally Posted by Kieth89
So did distributing the minus first mess it all up for me?

order of operations, remember? exponents first, then multiply by -1 ...

$-(3-x)^2 = -(9 - 6x + x^2) = -9 + 6x - x^2$
• Dec 14th 2010, 05:54 PM
Kieth89
Oh...it's always something simple..

Thanks for the help :)