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June 7th, 2012, 04:33 PM  #1 
Newbie Joined: May 2012 Posts: 21 Thanks: 0  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. 
June 7th, 2012, 05:31 PM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  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? 
June 7th, 2012, 06:19 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  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: 
June 7th, 2012, 06:48 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2207 
For the questions as posted, r = 0 and t = 25/2.

June 7th, 2012, 07:14 PM  #5  
Newbie Joined: May 2012 Posts: 21 Thanks: 0  Re: solve the equation Quote:
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!!!!  
June 7th, 2012, 07:37 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  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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