Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

When the Rational Root Theorem Fails

  1. Sep 10, 2012 #1
    On a math test, one of the questions was to solve [itex]-\sqrt{7-x}=-\frac{x^2}{2}+12x-10[/itex]. I solved graphically with a calculator, but later tried to solve algebraically, when I had more time. The equation is equivalent (with extraneous solutions) to [itex]x^4 + 48x^3 +536x^2 -956x + 372=0[/itex]. This quartic has no rational roots (the rational roots theorem gives ±1, ±2, ±3, ±4, ±6, ±12, ±31, ±62, ±93, ±124, ±186, ±372, but none of these are zeros. I have shown that all real zeros lie in (-62, 2), but this is as far as I have gotten. What is the next step after the Rational Roots Theorem (RRT) has failed to find a root?

    Thanks!
     
  2. jcsd
  3. Sep 10, 2012 #2
    The quartic formula! Didn't you learn it in school? :p
     
  4. Sep 11, 2012 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    For many purposes, the best way to represent a root in this situation is something of the form:

    r is the root of [itex]x^4 + 48x^3 +536x^2 -956x + 372=0[/itex] between 0.6 and 0.7​

    Incidentally, I plugged your original equation and your quartic equation into wolframalpha and the solutions don't agree -- you've made an error in arithmetic or in transcription. (although correcting your arithmetic won't make things any easier)
     
  5. Sep 11, 2012 #4
    The original equation, [itex]-\sqrt{7-x} = -\frac{x^2}{2} + 12x - 10 [/itex] has two solutions, at x1≈0.996 and x2≈-25.242, but the quartic has two extra solutions: at x3≈0.607 and at x4≈-24.361. These last two are extraneous solutions, but x1 and x2 are also solutions of the quartic.
     
  6. Sep 11, 2012 #5

    uart

    User Avatar
    Science Advisor

    I'm pretty sure that the original equation has just one real solution. Write it as:
    [tex]\sqrt{7-x} = \frac{x^2}{2} - 12x + 10 [/tex]
    and it's clear that the LHS is the upper half of a sideways parabola, with domain [itex]x \leq 7[/itex], while the RHS is a vertical parabola with axis of symmetry [itex]x=12[/itex]. Therefore, where the domains overlap both curves are monotonic. So there is only one real solution.

    BTW. Fixed point iteration of [itex]x = \frac{1}{12} (0.5*x^2 + 10 - \sqrt{7-x} )[/itex] gives this solution at about 0.640259457069546.
     
  7. Sep 11, 2012 #6
    Your comment about one solution threw me off. We had gone over the problem (graphically, of course) in class, and there were two solutions. I then noticed that I had typed the wrong equation, and simply copied it from my first post. I switched the signs:

    [tex]\sqrt{7-x} = \frac{x^2}{2} + 12x - 10 [/tex]

    should be correct. Wolfram Alpha confirms that there are two zeros here. Regardless of the signs, how would I find these solutions algebraically, without relying on the quartic formula?

    Thanks!
     
    Last edited: Sep 11, 2012
  8. Sep 11, 2012 #7

    uart

    User Avatar
    Science Advisor

    Yes obviously that changes things. The axis of symmetry of that parabola is x = -12, so that it is now possible for the [itex]\sqrt{7-x}[/itex] curve to intersect with both "sides" of the parabola (I haven't checked that it actually does, but will take your word for it). With the previously given equation however, this was simply not possible. :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: When the Rational Root Theorem Fails
  1. Rational root (Replies: 5)

  2. Rational roots (Replies: 2)

Loading...