Solving for a Variable in One EquationDate: 3/18/96 at 20:52:4 From: Anonymous Subject: Quadratic Equations Please help! I need to write the following in standard form and solve for the indicated variable: (2r+3)(2r-1) = -(3r+1) Thank you, Robert Date: 3/20/96 at 10:56:55 From: Doctor Patrick Subject: Re: Quadratic Equations Hi! To solve for r we are going to have to set one side of the equation to 0. To do this we must add the -(3r+1) to both sides. This gives us (2r+3)(2r-1) + (3r+1) = 0 . Unfortunately, we can't do anything with the problem until we clear the parentheses. Once we do this we get (4r^2 + 4r - 3) + (3r+1) = 0. Remove the parentheses: 4r^2 + 4r - 3 + 3r + 1 = 0. Now we add like terms and get 4r^2 + 7r - 2 = 0. Now all that is left to do is to factor the problem and solve each half separately. I'll let you try this part on your own, but write back if you need more help. Good luck! -Doctor Patrick, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/