Cute elementary algebra problem

Discussion in 'Undergraduate Math' started by Stephen J. Herschkorn, Sep 15, 2007.

  1. Came across this problem in the first section of the first chapter of
    the intro calculus text from which I am teaching this semester. I
    thought it was kind of cute.

    Solve x^4 - 4x = 1 exactly.

    Attributed to Murray Klamkin and published in The Mathematics Student
    Journal (1980).
     
    Stephen J. Herschkorn, Sep 15, 2007
    #1
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  2. Stephen J. Herschkorn

    Compendium Guest

    x=-sqrt(2)/2-i·sqrt(sqrt(2)+1/2) or
    x=-sqrt(2)/2+i·sqrt(sqrt(2)+1/2) or
    x=sqrt(2)/2-sqrt(sqrt(2)-1/2) or
    x=sqrt(sqrt(2)-1/2)+sqrt(2)/2
     
    Compendium, Sep 15, 2007
    #2
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  3. This website addresses this issue

    http://www.msc.uky.edu/ken/ma110/text/poly.htm

    My 2nd grade teacher beat this technique into us!

    Brian
     
    Brian VanPelt, Sep 15, 2007
    #3
  4. Stephen J. Herschkorn

    Jeremy Watts Guest

    if by 'exactly' you mean solving it in terms of radicals then the only
    method i know of to do that is 'ferrari's method' , which you'd have to be
    out of your mind to attempt as its hideously complicated.

    is there a solution printed for this?
     
    Jeremy Watts, Sep 15, 2007
    #4
  5. Stephen J. Herschkorn

    ªºª rrock Guest


    It's somewhere (not exactly) around x ~ -0.24903837639842
    but it seems to me it's probably irrational.

     
    ªºª rrock, Sep 15, 2007
    #5
  6. Stephen J. Herschkorn

    Jeremy Watts Guest

    all the roots are irrational, two are real, two complex
     
    Jeremy Watts, Sep 15, 2007
    #6
  7. Stephen J. Herschkorn

    Dana Guest

    Here's cheating...

    x /. Solve[x^4 - 4*x == 1, x][[ 1 ;; 2]]

    {-((Sqrt[-1 + 2*Sqrt[2]] - 1)/Sqrt[2]),
    (Sqrt[-1 + 2*Sqrt[2]] + 1)/Sqrt[2]}

    N[%]

    {x->-0.2490383763983743},
    {x-> 1.663251938771469}
     
    Dana, Sep 15, 2007
    #7
  8. It is possible to do this with only elementary algebra one would learn
    in high school (at the latest) in the U.S. I will post a solution in a
    few days if no one else does.
     
    Stephen J. Herschkorn, Sep 15, 2007
    #8
  9. On Fri, 14 Sep 2007 23:09:24 -0400, "Stephen J. Herschkorn"
    Let s = sqrt(2) and write x^4 - 4x - 1 as the difference of
    two squares:

    x^4 - 4x - 1 =
    (x^4 + 2x^2 + 1) - (2x^2 + 4x + 2) =
    (x^2 + 1)^2 - 2(x + 1)^2 =
    (x^2 + 1 + s(x + 1))(x^2 + 1 - s(x + 1)) =
    (x^2 + sx + (1 + s)) (x^2 - sx + (1 - s))

    Clearly either x^2 + sx + (1+s) = 0 or x^2 - sx + (1-s) = 0.
    But s^2 - 4(1+s) = 2 - 4 - 4s < 0, so only the second factor
    produces real roots. They are (1/2)(s +/- sqrt(-2 + 4s)) =
    (s/2)(1 +/- sqrt(-1 + 2s)). (Barring algebraic errors at
    some point, of course.)


    I actually got this after the fact, by trying to write
    x^4 - 4x - 1 as a product of two monic quadratics.
    Specifically, I divided x^4 - 4x - 1 by x^2 - ax - b; this
    yields a quotient of x^2 + ax + (a^2 + b), where a and b
    must satisfy a^3 + 2ab - 4 = 0 and (a^2 + b)b = 1 in order
    to avoid having a remainder. Note that without loss of
    generality a >= 0.

    Consider the second of these as a quadratic in b and solve
    to get

    b = (1/2)[-a^2 +/- sqrt(a^4 + 4)].

    Substitute this into the first condition:

    a^3 - a^3 +/- a sqrt(a^4 + 4) - 4 = 0,
    +/- sqrt(a^4 + 4) = 4/a,
    a^4 + 4 = 16/a^2,
    a^6 + 4a^2 - 16 = 0.

    Let c = a^2, making this c^3 + 4c - 16 = 0; by inspection
    (or the rational root test) this has a solution c = 2.
    Dividing out c - 2 leaves c^2 + 2c + 8, which has no real
    roots, so we take c = 2 and a = sqrt(2) = s; the condition
    a^3 + 2ab - 4 = 0 then becomes 2s + 2sb - 4 = 0, b + 1 = 2/s
    = s, and b = -1 + s. Thus, the quadratic factors are
    x^2 - sx + (1 - s) and x^2 + sx + (1 + s). I then noticed
    that these can be written (x^2 + 1) +/- s(x + 1), which led
    to the slick solution given above.

    [...]

    Brian
     
    Brian M. Scott, Sep 16, 2007
    #9
  10. Why do you exclude the complex roots, easily found from your derivation?

    I got there this way, which is pretty close to Brian's route.

    x^4 - 4x = 1
    x^4 = 4x + 1
    x^4 + 2x^2 + 1 = 2 x^2 + 4x + 2
    (x^2 + 1)^2 = 2 (x + 1)^2
    x^2 + 1 = +/- sqrt(2) (x+1)

    We now have two quadratic equations to solve.

    Any idea if we can express sqrt(sqrt(2) - 1) as a rational linear
    combination of roots of integers? I suspect not.
     
    Stephen J. Herschkorn, Sep 16, 2007
    #10
  11. On Sat, 15 Sep 2007 21:15:57 -0400, "Stephen J. Herschkorn"
    Because of the context that you gave and the later comment
    that the solution can be found using only elementary high
    school algebra: both suggest that real roots are expected,
    though as you say the others are easy enough to find if one
    knows about complex numbers.
    It does seem unlikely, but it's been a *long* time since I
    did anything with field extensions.

    Brian
     
    Brian M. Scott, Sep 16, 2007
    #11
  12. Stephen J. Herschkorn

    Stan Brown Guest

    Sat, 15 Sep 2007 21:15:57 -0400 from Stephen J. Herschkorn
    I used the method at
    http://mathforum.org/library/drmath/view/52657.html
    and unless I made an algebraic error there is no expression
    sqrt(a) ± sqrt(b) for integer a,b that equals sqrt(sqrt(2)-1).
     
    Stan Brown, Sep 16, 2007
    #12
  13. Yeah, if

    sqrt(sqrt(2) - 1) = p sqrt(a) + q sqrt(b)

    for p and q rational numbers, a and b non-square naturals, then

    sqrt(2) - 1 = p^2 a + q^2 b + 2 pq sqrt(ab)

    It seems like -1 would have to equal p^2 a + q^2 b.

    Brian
     
    Brian VanPelt, Sep 16, 2007
    #13
  14. Stephen J. Herschkorn

    Ken Pledger Guest

    No it isn't in cases like this where the reducing cubic has an
    easy factor.


    Only by inspiration. (I've sometimes told students that the best
    method of integrating functions is "integration by inspiration," where
    you can guess the answer and then just check it by differentiating.)

    Ferrari's method leads to the difference of squares

    (x^2 + 1)^2 - 2(x + 1)^2 = 0

    which is easy to factorize. But finding that expression without
    Ferrari's help would involve more inspiration than most of us could hope
    for.

    Ken Pledger.
     
    Ken Pledger, Sep 17, 2007
    #14
  15. Stephen J. Herschkorn

    JillBones Guest

    I get (x^2 + 2*sqrt(x)) * (x^2 - 2*sqrt(x)) = 1 by simple
    factoring. Doesn't solve the problem but raises the question;

    "Does this preclude negative values for x?"

    Bill J
     
    JillBones, Sep 19, 2007
    #15
  16. Stephen J. Herschkorn

    Proginoskes Guest

    That doesn't get you anywhere, since you don't have a 0 on the right-
    hand side.
    Not necessarily; if we're working with complex numbers, square roots
    of negative numbers are okay.

    --- Christopher Heckman
     
    Proginoskes, Sep 19, 2007
    #16
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