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Solving a Quartic Equation with Substitutions

Date: 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!
Associated Topics:
High School Basic Algebra
High School Polynomials

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