Cubic CuriosityDate: 09/21/2013 at 09:48:16 From: atwan Subject: whats wrong with this solution? equation Hi, Dr. Math. I'm trying to find out where is the wrong step below: x^2 + x + 1 = 0 x(x + 1) + 1 = 0 [1] -x^2 = x + 1 [2] -x^3 = -1 x = 1 is not a solution for either equation [1] or [2]. Date: 09/21/2013 at 10:23:04 From: Doctor Peterson Subject: Re: whats wrong with this solution? equation Hi, Atwan. It's hard to follow your thinking, because you didn't explain each step you took. I'll restate this more fully, and with slight changes in the equations to make the result more dramatic. Suppose that x is a solution of x^2 + x + 1 = 0 [0] Then we can write x^2 + x = -1 Now factor this and obtain x(x + 1) = -1 [1] But starting from [0], we can also obtain x + 1 = -x^2 [2] When we substitute [2] into [1], we get x(-x^2) = -1 Simplifying, -x^3 = -1 x^3 = 1 [3] So any solution of [0] is a cube root of 1. And this is true! The solutions of [0] are the NON-REAL cube roots of 1, namely (-1 +/- i sqrt(3))/2. In fact, we can see this by factoring [3]: x^3 - 1 = 0 (x - 1)(x^2 + x + 1) = 0 Your question is, why isn't this reversible, so that ANY solution of [3], including x = 1, is a solution of [0]? And, in fact, why isn't 1 a solution of [1] or [2]? The answer is that the procedure you followed is not reversible. Combining [1] and [2] to make [3] can introduce solutions of [3] that are not solutions of [1] or [2]. In particular, reversing the process would require combining, say, [2] and [3] to get [1], but 1 is not a solution of [2]. In general, when we add two equations a = b and c = d to obtain a new equation a + c = b + d, we can say that if the first two are true, so is the last. But we can't say that if the last is true, so are the first two. For example, when we solve a system of two linear equations by addition, we sometimes obtain the equation 0 = 0, which is true for all x and for all y. But this does not mean that each of the equations in the system is true for all x and all y; each is only true for certain ordered pairs. Thanks for an interesting question! I haven't see this particular "paradox," which is a good reminder that addition of equations has to be stated carefully to avoid the implication that it is fully reversible. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/21/2013 at 10:39:41 From: atwan Subject: Thank you (whats wrong with this solution? equation) Thank you, doctor. |
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