Solving a Quartic Equation with SubstitutionsDate: 11/02/2004 at 11:37:03 From: John Subject: How can I solve this out I'm trying to solve y(y+1)(y+2)(y+3) = 7920, which is a problem from my friend's kid. First I multiplied it all out: (y^2 + y)(y + 2)(y + 3) = 7920 (y^3 + 3y^2 + 2y)(y + 3) = 7920 y^4 + 6y^3 + 11y^2 + 6y - 7920 = 0 I tried to solve it as a quadratic equation, but that didn't seem to work since this polynomial has 5 terms. I think it is 3 terms to be a trinomial. Since this equation can't be grouped as a trinomial, I am not able to solve it by factoring. Am I on the right track? I haven't done any maths for more than 10 years, and I think I forgot almost everything I learned! Can you please give me a hint? Date: 11/02/2004 at 13:07:54 From: Doctor Douglas Subject: Re: How can I solve this out Hi John. Your work so far is fine, and you are trying to factor this last equation: y^4 + 6y^3 + 11y^2 + 6y - 7920 = 0. This is not a quadratic equation, because there is a term with y^4. It is a "quartic" equation in y. There exists a formula (similar to the quadratic formula) for extracting the roots, but it is complicated and not often used. You can factor the equation in a number of ways: 1. You can divide through by guesses such as (y-3),(y+6),(y-8),... and see if any of these happen to work. Because this was given as a problem for a schoolkid, I'm guessing that the roots are probably integers, so this not an unreasonable approach. 2. Another way to do this is to realize that the four roots are equally-spaced, because the factors come in a nice arithmetic progression: y,y+1,y+2,y+3. So let's average those and define u = y + 3/2 as the center of this set of four numbers, and the equation becomes (u - 3/2)(u - 1/2)(u + 1/2)(u + 3/2) - 7920 = 0 This is nice, because the factors multiply out such that the cross terms cancel: [(u - 3/2)(u + 3/2)][(u - 1/2)(u + 1/2)] - 7920 = 0 (u^2 - 9/4) (u^2 - 1/4) - 7920 = 0 And if we make one more subsitution, using v = u^2, this IS a quadratic equation in terms of v: (v - 9/4)(v - 1/4) - 7920 = 0 v^2 - 10v/4 + 9/16 - 7920 = 0 16v^2 - 40v + 9 - 16*7920 = 0 16v^2 - 40v - 126711 = 0 Now you can factor this quadratic trinomial using many methods, or you can use the quadratic formula with a = 16, b = -40 and c = 126711: Ask Dr. Math FAQ: Learning to Factor http://mathforum.org/dr.math/faq/faq.learn.factor.html You will find that this quadratic equation factors as follows: 16*(v - 90.25)(v + 87.75) = 0 and has roots of v = 90.25 or -87.75 Since v = u^2, the second root leads to no real solution for u, and we must have v = 90.25 u^2 = 90.25 u = sqrt(90.25) u = 9.5 or -9.5 which means that going back to (u - 3/2)(u - 1/2)(u + 1/2)(u + 3/2) our set of y's is either {8,9,10,11} or {-11,-10,-9,-8}. This is a tough problem because of the substitution steps, so don't feel bad about not being able to do it! - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 11/02/2004 at 14:43:55 From: John Subject: How can I solve this out Thanks for the prompt reply. I followed most of your work, but I'm a little confused by the step where you chose u = y + 3/2. Why do I need to define "u" as the center of the set of numbers, not the beginning or end of the numbers? Is this a maths theory? Thanks. John Date: 11/02/2004 at 14:56:08 From: Doctor Douglas Subject: Re: How can I solve this out Hi John. That's a very good question! Mostly this was simply an inspired guess, guided by our desire to take advantage of the left-right symmetry of the roots. By doing so, we separate the odd (y^3 and y^1) terms from the even (y^4,y^2,y^0) terms, and the latter set is what leads to our quadratic trinomial via the substitution v = y^2. Note that this trick worked only because of the nice progression of factors {y,y+1,y+2,y+3}. It would have been much tougher to work with the set {y,y+1,y+2,y+4}. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 11/02/2004 at 15:49:43 From: John Subject: Thank you (How can I solve this out) Thanks Dr. Math for helping me. Your work was interesting and your comments helpful. I appreciate it! |
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