Completing the Square: Alternate MethodDate: 02/06/98 at 12:30:16 From: Han-Han Wang Subject: New method for solving by completing the square I have come up with a new method for solving quadratic equations by ccompleting the square. Instead of dividing by the coefficient of the quadratic term (and getting complicated fractions), I multiply the equation to get a perfect square coefficient for the quadratic term. I want some feedback. I am sure there are some exceptions to this method. Otherwise schools would teach this, right? Thanks, Hanhan Wang Date: 02/06/98 at 16:22:00 From: Doctor Rob Subject: Re: New method for solving by completing the square This is a valid, alternate method, which is taught in some schools. It works as-is if the linear term has an even coefficient. To make this method fraction-free, if the coefficient of the linear term is odd, you should multiply not by the coefficient of the quadratic term, but by four times it. That will insure that the linear term has an even coefficient, which is necessary to avoid fractions further. In fact, if you always use four times the coefficient of the quadratic term, you will always avoid fractions. Example: 3*x^2 - 7*x - 20 = 0, 9*x^2 - 21*x - 60 = 0, (3*x - 7/2)^2 = 60 + 49/4 = 289/4 = (17/2)^2 See that fractions are still present. If instead of multiplying by 3, we multiply by 4*3, we will still get a perfect square coefficient of the quadratic term: 36*x*2 - 84*x - 240 = 0, (6*x - 7)^2 = 240 + 49 = 289 = 17^2, and you can finish the solution 6*x - 7 = 17 or -17, 6*x = 24 or -10, x = 4 or -5/3. Symbolically, a*x^2 + b*x + c = 0, 4*a^2*x^2 + 4*a*b*x + 4*a*c = 0, (2*a*x + b)^2 = b^2 - 4*a*c and so on. You see, if b is even, then a 4 can be divided out of the last equation, but if it is odd, you can't do that without introducing fractions. The disadvantage of using this method is that the integers involved can get fairly large. You may be faced with taking the square root of a four- or five-digit number. The advantage, as you point out, is that you avoid working with fractions. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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