# Thread:

1. ## 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?

2. 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?
$\displaystyle \frac{\sqrt[5]{9^4}}{9} = \frac{\sqrt[5]{3^8}}{9} = \frac{\sqrt[5]{3^5 \cdot 3^3}}{9} = \frac{3\sqrt[5]{3^3}}{9} = \frac{\sqrt[5]{3^3}}{3}$

3. How did you get the root sign to type

4. 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?

$\displaystyle\frac{1}{\sqrt[5]{9}}=\frac{1}{\sqrt[5]{3^2}}=\frac{1}{3^{\frac{2}{5}}}=\frac{3^{\frac{3} {5}}}{3^{\frac{2}{5}}\;3^{\frac{3}{5}}}$

5. 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:

$\sqrt{(3+x)(3+x)+(-3+x)(-3+x)} = \sqrt{9+6x+x^2+9-6x+x^2}<br /> = \sqrt{2x^2+18} =$
$\sqrt{2(X^2+9)} = ?$
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 $2\sqrt{3x}$

6. ## another way

by fractional exponents (probably not any easier tho)

$<br /> \frac{1}{9^{\frac{1}{5}}} \rightarrow \frac{1}{3^{\frac{2}{5}}}<br />$

$<br /> \frac{1}{3^{\frac{2}{5}}}<br /> \times<br /> \frac<br /> {3^<br /> {\frac{3}{5}}<br /> }<br /> {3^<br /> {\frac{3}{5}}<br /> }<br /> \rightarrow<br /> \frac<br /> {3^{\frac{3}{5}}}<br /> {3^{\frac{5}{5}}}<br /> \rightarrow<br /> \frac<br /> {27^<br /> {\frac{1}{5}}<br /> }<br /> {3}<br />$

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

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

$\displaystyle = \sqrt{12x}$

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

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

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

8. 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?

9. $<br /> \sqrt{9 +6x+x^2-9+6x-x^2}<br /> \rightarrow<br /> \sqrt{12x}<br /> \rightarrow<br /> \sqrt{3}\sqrt{4}\sqrt{x}<br /> \rightarrow<br /> 2\sqrt{3x}<br />$

10. 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 ...

$-(3-x)^2 = -(9 - 6x + x^2) = -9 + 6x - x^2$

11. Oh...it's always something simple..

Thanks for the help
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9698947668075562, "perplexity_flag": "middle"}