Roots of an EquationDate: 4/16/96 at 22:42:39 From: Jonathan Stone Subject: Solving Quadratic Equations I'm trying to help my daughter with these. Example 6x^2+x-2 = 0. We can factor it to (3x+2)(2X-1), I think. We're not sure what to do now to solve it to two numbers. A good method of factoring and solving these with an example or two would be greatly appreciated. Thanks again. Jonathan Stone Date: 4/17/96 at 13:6:35 From: Doctor Patrick Subject: Re: Solving Quadratic Equations Hi! You're right so far - the factoring seems to be correct, and you do have to solve it for two answers. Whenever you have problems with exponents, there will be as many answers as the highest power of the variable you are solving for, so in a problem with x^2 there will be two seperate answers. The reason for needing these two answers comes from the fact that x can me either positive or negative, but when it is squared it will be positive. For example both 2^2 and -2^2 equals 4. Since the x^2 in our equation can be either, we need to solve for both cases. We do this be first making sure that the equation is set to 0, then factoring it as you did. Then, we solve for each have of the equation separately. So your (3x+2)(2X-1) = 0 becomes: 3x+2=0 and 2x-1=0. The justification for doing this is that since the two terms, when multiplied together, equal 0, one of the terms must also be equal to 0, since you only get 0 as a product of multiplication when one of the terms is also equal to 0. So since (3x+2)(2X-1) = 0 either 3x+2 = 0, or 2x-1 = 0. We then solve for both possiblities. I'll let you try to solve these two equations on your own. Good luck! Write back to us if you need more help with this, or other problems. -Doctor Patrick, The Math Forum |
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