Solving Equations that Equal ZeroDate: 02/26/97 at 21:58:17 From: Thomas G. Morton, Jr. Subject: Help with Algebra Dear Dr. Math: My son is in the 8th grade. I am trying to help him with his algebra but I really need some pointers. For example: (1) (5x+10)(7x-2) = 0 How do I solve it for x? (2) 3x(2x-1)(x+2) = 0 In solving for x, are there 3 answers or 2? (3) How do I find the solution set of 2x^2 - 4x + 2 = 0? These are problems my son missed on his test. I can't solve them either. Any ideas would really be appreciated. Date: 02/26/97 at 23:33:35 From: Doctor Mike Subject: Re: Help with Algebra Hi Mr. Morton, There is a key fact that should help you here. If a product of 2 numbers (2 numbers multiplied together) is equal to zero, then at least one of them must be zero. In the case of your first problem this means that either 5x+10 = 0 or 7x-2 = 0. The solutions to these 2 equations are x = -2 and x = 2/7, respectively, so these are the solutions to the original equation. Your second problem can be solved in a similar fashion. For problems like (3) where the equation is not already factored as in your first 2 problems, it is best to make the equation as simple as possible before we start to factor. In this case, since all the terms on the left side have an even number as their coefficient, we can factor 2 out of everything to get: 2(x^2-2x+1)=0 This factors as 2(x-1)(x-1)=0 Since both of these factors are the same there is only one answer, namely x = 1. Factoring expressions like this is not a simple thing. There are some techniques which are often useful, but some of these expressions cannot be factored no matter how hard you try (unless you go to "complex numbers" which I think you can put off for a while). There is not an easy way to pick this up. It just needs to be studied and practiced. If you encounter another specific exercise that stubbornly resists your efforts to figure it out, please write back and we may be able to help. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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