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June 7th, 2012, 04:33 PM   #1
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solve the equation

1/3r^(3/2)+3/2r^(1/2)=0

and

1/5t^(1/3)+5/2t^(-2/3)=0

I need help with these 2 questions. Thanks.
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June 7th, 2012, 05:31 PM   #2
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Re: solve the equation

I could guess as to what the equations look like, but won't.

Could you either:
1) Visit the link in my profile for LATEX (preferred), or
2) at least add some parenthesis to clear things up?
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June 7th, 2012, 06:19 PM   #3
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Re: solve the equation

I feel lucky...I'll guess you mean:

1.) 1/(3r^(3/2)) + 3/(2r^(1/2))=0



We see that the least common denominator is , so multiply through by this expression to get:





2.) 1/(5t^(1/3)) + 5/(2t^(-2/3))=0



Rewrite with positive exponents:



We see that the least common denominator is , so multiply through by this expression to get:



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June 7th, 2012, 06:48 PM   #4
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For the questions as posted, r = 0 and t = -25/2.
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June 7th, 2012, 07:14 PM   #5
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Re: solve the equation

Quote:
Originally Posted by ahereicome
1/3r^(3/2)+3/2r^(1/2)=0

and

1/5t^(1/3)+5/2t^(-2/3)=0

I need help with these 2 questions. Thanks.
I'm sorry guys; next time, I will add parentheses, and the LATEX is kinda hard to use lol, but here it's after I edited:
1) (1/3r^(3/2))+(3/2r^(1/2))=0

2) (1/5t^(1/3))+(5/2t^(-2/3))=0

skipjack actually got the answers for me; it's the same as on the textbook, but I want the steps, thanks!!!!
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June 7th, 2012, 07:37 PM   #6
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Re: solve the equation

The parentheses you added are unnecessary as they do not change the interpretation of the equations. To make it more clear, you could type:

1.) (1/3)r^(3/2)+(3/2)r^(1/2)=0



Factor:



If we assume the terms in the original equation represent real values, then we discard the negative root, and are left with:



2.) (1/5)t^(1/3)+(5/2)t^(-2/3)=0



Note: hence we may multiply through by to get:



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