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$\sqrt[5]{(-5)^5}$

Simplify the root:

$\sqrt[4]{x^4}$

$-\sqrt[3]{x^9}$

$-\sqrt{x^{18}}$

Thank you to everyone who replies <3

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By index $4 \sqrt{81}$ do you mean the fourth root of $81, \sqrt[4]{81}$ To write roots in $\LaTeX$, which are easier to read, you enclose in dollar signs \sqrt[index]{number to take the root of} so to get $\sqrt[4]{81}$ I wrote \sqrt[4]{81} –  Ross Millikan May 1 '13 at 21:37
    
I believe that I do. I'm sorry, I must seem like an invalid. –  user75437 May 1 '13 at 21:38
    
It said "index" in my book, so I assumed that was what it was called. –  user75437 May 1 '13 at 21:39
    
oh sweet. I only learned the symbol for square root from imgur. lol. Thank you. –  user75437 May 1 '13 at 21:40
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No, many people don't know $\LaTeX$, but it makes math much easier to read. Here is a place to start. The point about right-clicking and picking Show Math as -> TeX Commands is a great way to learn. –  Ross Millikan May 1 '13 at 21:41

2 Answers 2

Hint: Do you know that powers and roots are inverse functions, so $\sqrt [a]b=b^{\frac 1a}$? Then do you know a law of exponents about $(a^b)^c$? These will take you far.

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The teacher that I have isn't the greatest. When I was attempting to learn factoring and the distributive prop. he told us NOT to use FOIL. It made life pretty hard, as that's how I learned from the start. He just has a different teaching style I guess. –  user75437 May 1 '13 at 21:52

Brace yourselves, for square roots are coming..

Assuming that the question is wether

$$ \sqrt[5]{(-5)^5}$$

is a real number or not, note that

$$(-5)^5=(-1\cdot5)^5=(-1)^5 5^5=-5^5$$

$$\sqrt[5]{(-5)^5}=\sqrt[5]{-1}\sqrt[5]{5^5} $$

so it is not real.

For the simplifications note that

$$ \sqrt[n]{a}=(a)^{1/n}$$

and

$$ (a^{n})^{m}=a^{nm}$$

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So any negatives are not real numbers? these square roots are dark and full of terrors –  user75437 May 1 '13 at 23:48
2  
@Nivalth $(-1)$ doesn't have a real fifth root?! You may want to look at what you wrote again... –  RecklessReckoner May 2 '13 at 3:19

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