1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Explanation for increase of degree of equation

  1. Sep 21, 2016 #1
    So say that I have the quadratic equation ##x^2 - 2x +1 = 0##. If I multiply each side by ##x##, a valid operation, I get the new equation ##x^3 - 2x^2 + x = 0##, which has a different solution set as the first, namely ##\left\{ {0, 1}\right\}##. If I make the substitution ##x^2 = 2x -1 = 0##, then I now have ##x^3 - 2(2x - 1) + x = x^3 - 3x + 2##, which has the solution set ##\left\{ {-2,1}\right\}##. I could repeat this process ad infinitum, but I am just wondering what is going on. If I am just doing valid operations and making valid substitutions, then why is the solution set constantly changing from what is originally was, i.e, ##\left\{ {1}\right\}##?

    Also, given that I transformed ##x^2 - 2x + 1 = 0## to ##x^3 - 3x + 2 = 0##, is there any way to go in the reverse direction, that is, to start with ##x^3 - 3x + 2 = 0## and go to ##x^2 - 2x + 1 = 0##?
     
  2. jcsd
  3. Sep 21, 2016 #2

    Mark44

    Staff: Mentor

    The only change you're making is to enlarge the solution set.

    If you leave your equations in factored form, it should be more obvious.
    ##(x - 1)^2 = 0## Solution {1}
    ##x(x - 1)^2 = 0## Solution {0, 1}
    In your third equation, what you've done seems murky as you have written it, but if ##x^2 = 2x - 1##, then ##x^2 - 2x + 1 = 0## or ##(x - 1)^2 = 0##. Multiplying both sides by ##(x - 1)^2## raises the degree of the polynomial to 5, but there are no new values in the solution set.
    Factor it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Explanation for increase of degree of equation
Loading...