Do Irrational Roots Always Occur in Conjugate Pairs?Date: 03/13/2002 at 08:51:29 From: Libby Lerner Subject: Do irrational roots always occur in conjugate pairs? A colleague and I have been puzzled by the equation x^3+4x^2+3x+1 = 0. Upon graphing, there appear to be two imaginary solutions and one real solution. According to the rational root theorem, if the root is rational, then it should be either 1 or -1. It isn't. If the root is irrational, then its conjugate should also be a root. It isn't, since there is only one real root. What are we missing? Thanks for your help. Date: 03/13/2002 at 11:17:20 From: Doctor Paul Subject: Re: Do irrational roots always occur in conjugate pairs? The irrational conjugate roots theorem says: Let p(x) be any polynomial with rational coefficients. If a + b*sqrt(c) is a root of p(x), where sqrt(c) is irrational and a and b are rational, then another root is a - b*sqrt(c). The problem is that you have no guarantee that the root of a generic cubic can be written in the form x = a + b*sqrt(c). It is in fact the case that whenever a cubic has only one real irrational root, this irrational root will never be expressable in the form x = a + b*sqrt(c), where a and b are rational and sqrt(c) is irrational. If this were not the case, then the irrational conjugate roots theorem would tell us that x = a - b*sqrt(c) would be another irrational root, and we would have problems (we are assuming there is only one real root) unless we could conlude that a + b*sqrt(c) = a - b*sqrt(c) (so that we in fact had only one real root). But if a + b*sqrt(c) = a - b*sqrt(c), then 2*b*sqrt(c) = 0, which forces b = 0 (since sqrt(c) is irrational, c cannot be zero). But b = 0 means that the root is just x = a, where a is rational. And this cannot be, since we assumed that the root was irrational. Thus it must be the case that whenever a cubic has only one real irrational root, this irrational root will never be expressable in the form x = a + b*sqrt(c) where a and b are rational and sqrt(c) is irrational. Notice that if the cubic has one real rational root, say x = p/q then choose a = p/q, b = 0, and c = any nonsquare to satisfy the irrational conjugate roots theorem. In this case, a + b*sqrt(c) = a = a - b*sqrt(c), and things work out nicely. Let's look more specifically at what's going on with p(x) = x^3+4x^2+3x+1 We need to solve p(x) = 0, and I'll use Cardano's formula, as described in our FAQ: Cubic and Quartic Equations http://mathforum.com/dr.math/faq/faq.cubic.equations.html Make the substitution x = (y - 4/3) to obtain: y^3- 7/3*y + 47/27 (verify this) so if we have y^3 + p*y + q = 0 then it must be the case that p = -7/3 and q = 47/27 Now I deviate from the FAQ a bit (I like my method better), but the ideas are similar. Let D = -4*p^3 - 27*q^2 (this is the discriminant of the cubic) Then D = -4 * (-7/3)^3 - 27 * (47/27)^2 = -31 Then we let [ -27 * q 3 * sqrt(-3 * D) ] A = cuberoot |--------- + ----------------- | [ 2 2 ] and [ -27 * q 3 * sqrt(-3 * D) ] B = cuberoot |--------- - ----------------- | [ 2 2 ] So in this case, A = [-47/2 + 3/2*sqrt(93)]^(1/3) and B = [-47/2 - 3/2*sqrt(93)]^(1/3) Then L = (A+B)/3 M = (t^2*A + t*B)/3 and N = (t*A + t^2*B)/3 are the three roots, where t is the first primitive third root of unity: t = -1/2 + i*sqrt(3)/2 Thus we have: 1 [ -47 3*sqrt(93) ] L = - * cuberoot |----- + ---------- | + 3 [ 2 2 ] 1 [ -47 3*sqrt(93) ] - * cuberoot |----- - ---------- | 3 [ 2 2 ] It is a simple exercise to verify that y^3- 7/3*y + 47/27 has only one real root. Inasmuch as L ~= -1.814565703 is real, we have found the only real root of y^3- 7/3*y + 47/27 = 0. (I am not computing M and N because they're messy and they're not really needed to talk about what we're discussing) Thus L' = L - (4/3) ~= -3.14789903633 (remember we made the substitution x = y - 4/3) will satisfy the original cubic p(x). Thus we have: 1 [ -47 3*sqrt(93) ] L' = - * cuberoot |----- + ---------- | + 3 [ 2 2 ] 1 [ -47 3*sqrt(93) ] 4 - * cuberoot |----- - ---------- | - - 3 [ 2 2 ] 3 with p(L') = 0. As determined earlier, L' will not be expressible in the form a + b*sqrt(c), and so the irrational conjugate roots theorem does not apply. I suppose there is a more direct way to prove that L' will not be expressible in the form a + b*sqrt(c) without relying on the irrational conjugate roots theorem, but it's not immediately clear to me how this would be done. Here's another example I think you might like. Consider the polynomial: p(x) = x^3 + x^2 - 2 Plainly, x = 1 is a root and dividing p(x) by (x-1) allows us to use the quadratic formula to find out that the other two roots are x = -1 +- i But using Cardano's formula as described above gives us the following real root: 1/3 * [ cuberoot(26 + 15*sqrt(3)) + cuberoot(26 - 15*sqrt(3)) - 1] Unless we want to start being contradictory, this real root had better be equal to one. And that is in fact the case, but seeing this requires us to be a bit sneaky. Notice that: cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3) and that cuberoot(26 - 15*sqrt(3)) = 2 - sqrt(3) thus cuberoot(26 + 15*sqrt(3)) + cuberoot(26 - 15*sqrt(3)) = 2 + sqrt(3) + 2 - sqrt(3) = 4 and the root is 1/3 * (4 - 1) = 1. How do we see that cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3)? Start with something you know to be true: 26 + 15*sqrt(3) = 26 + 15*sqrt(3) = 8 + 12*sqrt(3) + 18 + 3*sqrt(3) = 2^3 + 3 * 2^2 * sqrt(3)^1 + 3 * 2^1 * sqrt(3)^2 + sqrt(3)^3 = (2 + sqrt(3))^3 taking cuberoots on both sides yields the desired result: cuberoot(26 + 15*sqrt(3)) = 2 + sqrt(3) So in this case the real root of p(x) was rational and we were able to express our cuberoot terms in the proper form. I think that if you investigate this further, you'll find that this is not a coincidence. I hope this helps. Please write back if you'd like to talk about this some more or if anything is unclear. I've left out some steps to save time. If you have trouble reconstructing my work, just let me know and I'll go back and fill in the details. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/