Factoring to Solve a ProblemDate: 7/31/96 at 9:21:42 From: Keerthi Kumar Subject: Factoring to Solve a Problem I am a first year algebra student and I have a question involving factoring in problem solving. The instructions are: Solve each problem, rejecting solutions that do not meet the conditions of the problem. Let x = the smaller integer and x + 1 = the larger integer. Solve for x. x (x + 1) = 132 This is how I have been working the problem, but it comes out every time with one or the other of the factors being negative. 2 x (x + 1) = 132 = x * x + x * 1 = 132 = x + x = 132 2 x + x = 132 2 x + x - 132 = 0 (x + 12)(x - 11) = 0 x + 12 = 0 x - 11 = 0 x = -12 x = 11 I know that the answer is 11 and 12, but how am I supposed to get both integers to turn out positive? D.S.K. Date: 7/31/96 at 10:46:2 From: Doctor Jodi Subject: Re: Factoring to Solve a Problem Hi there! Thanks for your question. Since you let x equal the smaller integer, and x + 1 equal the larger integer, the two solutions you get when you solve for x are NOT the smaller integer and the larger integer. Rather they are two possibilities for the smaller integer. Does that make sense? So, if the smaller integer is 11, then the larger integer will be 12. There is still another possibility: if the smaller integer is -12, then the larger integer is -11. Do you see what I mean? Since the question didn't specify positive or negative integers, both of these solutions-- 11 and 12; -11 and -12 -- are correct. Hope this helps. Let us know if you have any more questions. -Doctor Jodi, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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