My Math Forum solve the equation

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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 $\frac{1}{3r^{\small{\frac{3}{2}}}}+\frac{3}{2r^{\s mall{\frac{1}{2}}}}=0$ We see that the least common denominator is $6r^{\small{\frac{3}{2}}}$, so multiply through by this expression to get: $2+9r=0$ $r=-\frac{2}{9}$ 2.) 1/(5t^(1/3)) + 5/(2t^(-2/3))=0 $\frac{1}{5t^{\small{\frac{1}{3}}}}+\frac{5}{2t^{\s mall{-\frac{2}{3}}}}=0$ Rewrite with positive exponents: $\frac{1}{5t^{\small{\frac{1}{3}}}}+\frac{5t^{\smal l{\frac{2}{3}}}}{2}=0$ We see that the least common denominator is $10t^{\small{\frac{1}{3}}}$, so multiply through by this expression to get: $2+25t=0$ $t=-\frac{2}{25}$
 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
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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!!!!

 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 $\frac{1}{3}r^{\small{\frac{3}{2}}}+\frac{3}{2}r^{\ small{\frac{1}{2}}}=0$ Factor: $r^{\small{\frac{1}{2}}}\(\frac{1}{3}r+\frac{3}{2}\ )=0$ If we assume the terms in the original equation represent real values, then we discard the negative root, and are left with: $r=0$ 2.) (1/5)t^(1/3)+(5/2)t^(-2/3)=0 $\frac{1}{5}t^{\small{\frac{1}{3}}}+\frac{5}{2}t^{\ small{-\frac{2}{3}}}=0$ Note: $t\ne0$ hence we may multiply through by $t^{\small{\frac{2}{3}}}$ to get: $\frac{1}{5}t+\frac{5}{2}=0$ $t=-\frac{25}{2}$

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