Rational Root Rigging
Date: 12/17/2013 at 22:39:08 From: Roger Subject: submitted earlier about rational equations Consider this equation: x/(x + 4) = -3/(x - 2) + 18/((x + 4)(x - 2). There are 2 solutions: x = -3 x = 2 But x = 2 is extraneous because it makes a denominator equal 0. With rational equations, sometimes you can get 2 solutions, both of which work; and sometimes neither works. Is there a way to knowingly generate rational equations so as to result in a pre-determined set of outcomes? I cannot find a reliable algorithm. Thanks.
Date: 12/17/2013 at 23:50:07 From: Doctor Peterson Subject: Re: submitted earlier about rational equations Hi, Roger. It's technically wrong to say that there are two solutions; it is the quadratic equation you obtained in the process of solving it that has two solutions. Your equation has ONE solution. Here's a problem I just made up: x 3 2 ----- + ----- = ------- x - 1 x + 1 x^2 - 1 The standard way to solve this is to multiply by the least common denominator (LCD), x^2 - 1, yielding x(x + 1) + 3(x - 1) = 2 The solution for this is x^2 + 4x - 5 = 0 (x + 5)(x - 1) = 0 x = -5 or 1 Of these, 1 is extraneous because the factor (x - 1) is found in the denominator of the original equation. So the only actual solution is -5. Here's how I came up with that equation. I wanted the equation to have the general form A C E --- + --- = --- B D BD Here, A, B, C, D, and E are either monomials or binomials. Multiplying by the LCD, I get AD + BC = E I want one solution to be a zero of B or D, let's say B, so that it will be extraneous. This means that B must be a factor of AD + BC - E = (AD - E) + BC Since B is already a factor of BC, it has to be a factor of the polynomial AD - E. Now I pick specific values for A, B and D, arbitrarily. I choose A = x B = x - 1 D = x + 1 (Interestingly, C doesn't come into this at all, so I will be free to choose it at will later!) Now I want to choose E to make B a factor of AD - E; that is, I can choose E = AD - BP, where P can be whatever makes things look good: E = AD - BP = x(x + 1) - P(x - 1) I want E to have degree less than 2, so I'd like P to have the form (x + p) in order to cancel out the x^2 term: E = x(x + 1) - (x + p)(x - 1) = x^2 + x - (x^2 + (p - 1)x - p) = x^2 + x - x^2 - (p - 1)x + p = -(p - 2)x + p To make it simple, I'll take p = 2 so that E is just a constant: E = -(2 - 2)x + 2 = 2 Now my equation (with C still unchosen) is x C 2 ----- + ----- = ------- x - 1 x + 1 x^2 - 1 Now I can choose C in order to make the solution a nice number. I'll make C a constant c rather than a binomial. Our equation after multiplying by the LCD is AD + BC - E = 0 (AD - E) + BC = 0 BP + BC = 0 B(P + C) = 0 So the non-extraneous solution will be the solution of P + C = 0 x + 2 + C = 0 x = -(C + 2) I can pick any integer value for c and I'll get a nice solution. I picked 3, so the other solution is -5. And there's my equation: x 3 2 ----- + ----- = ------- x - 1 x + 1 x^2 - 1 The whole process is similar in some ways to partial fractions. (I tried doing that explicitly, but I found it gave me less control over the results.) But there may well be a way to adapt and improve upon that technique ... I'll try to make another equation with the same general pattern, to see if my easy success that time was a fluke. I'll pick A = x + 1 B = x - 2 D = x + 3 Again, letting P = x + p to cancel x^2, we have E = AD - BP = (x + 1)(x + 3) - (x + p)(x - 2) = x^2 + 4x + 3 - x^2 + 2x - px + 2p = (6 - p)x + (2p + 3) I'll take p = -1 and P = x - 1, so E = 7x + 1 The other solution is the solution of P + C = 0 I'll make C a binomial x + c this time, just to be different: x - 1 + x + c = 0 2x = 1 - c I'll take c = 2, so the solution to the equation will be -1/2 (which is not extraneous as it would be for c = -3 or -5). The equation is x + 1 x + 2 7x + 1 ----- + ----- = ----------- x - 2 x + 3 x^2 + x - 6 Let's solve it and see that it works: (x + 1)(x + 3) + (x + 2)(x - 2) = 7x + 1 x^2 + 4x + 3 + x^2 - 4 = 7x + 1 2x^2 - 3x - 2 = 0 (2x + 1)(x - 2) = 0 As expected, the solutions of this are x = -1/2 and 2, but the latter is extraneous. Looks like the technique works. I've had to do this before in making tests, but hadn't worked out a routine method, so this has been an interesting and useful exercise. Still, I'll leave this question open in case anyone else has a better method. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 12/20/2013 at 22:11:59 From: Doctor Peterson Subject: Re: submitted earlier about rational equations Hi, Roger. No one else has added any better ideas, but I've thought a little more about how to make this easier to carry out in practice. Here is another example, worked out in a way that is easier to follow, without needing to refer to A, B, ... and magical formulas. I start by choosing my denominators; they must have a common factor, so I'll choose 2x - 1, x + 2, and (2x - 1)(x + 2) I'll put all terms on the left for now. I have to choose the extraneous solution of the equation I'm going to write (which will be the zero of one of my factors). I'll pick 1/2. I'll hold off on deciding what the actual solution will be until I have more to go on. I can pick anything for the first numerator; I'll take x: x A B ------ + ----- + --------------- = 0 2x - 1 x + 2 (2x - 1)(x + 2) Now in order for the factor 2x - 1 to yield an extraneous solution, it must be a factor of the new numerator, x(x + 2) + A(2x - 1) + B This implies that x(x + 2) + B must be a multiple of (2x - 1); let's do that by choosing B in this way: B = (2x - 1)(...) - x(x + 2) I can pick anything I like for that missing factor; if I wanted B to be linear, I'd want it to start with x/2 so the x^2 terms would cancel. But let's go ahead and let B be quadratic, since I didn't do one like that before. I'll take this: B = (2x - 1)(x - 1) - x(x + 2) = (2x^2 - 3x + 1) - (x^2 + 2x) = x^2 - 5x + 1 So, x(x + 2) + B = (2x - 1)(x - 1) Now the left side above will factor as x(x + 2) + A(2x - 1) + B = [x(x + 2) + B] + A(2x - 1) = (2x - 1)(x - 1) + A(2x - 1) = (2x - 1)(x - 1 + A) Again, I can pick anything I want for A (not necessarily a constant), with the goal of making the root of this a nice number (since that will be the actual solution of my equation): x - 1 + A = 0 Let's arbitrarily take A = 2x + a, so the equation becomes 3x - 1 + a = 0 This must be satisfied by the actual solution x; given that x, I'll take a = 1 - 3x I'll pick x = 2, so a = -5, and A = 2x - 5. Here's my equation: x^2 - 5x + 1 x 2x - 5 --------------- + ------ + ------ = 0 (2x - 1)(x + 2) 2x - 1 x + 2 For presentation to students, I might rearrange it this way: x 2x - 5 -x^2 + 5x - 1 ------ + ------ = ------------- 2x - 1 x + 2 2x^2 + 3x - 2 I'll "leave it as an exercise for the reader" to solve this and confirm that it does what I wanted, with the only solution being x = 2. Note that in this process, we arbitrarily specified the factors of the denominators, one numerator, the extraneous root, and the genuine root, and also made a partially arbitrary choice of the complexity of the other two numerators, though the specifics were then forced on us. I think this is a reasonable procedure for writing this kind of equation. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 12/20/2013 at 22:25:21 From: Roger Subject: Thank you (submitted earlier about rational equations) THANK YOU for a great solution. I hope you will submit this to NCTM's "Mathematics Teacher" journal, because a lot of people could use it. Your explanations are great and easy to follow. Have a Merry Christmas and happy holiday season. - rg
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