1. Not finding help here? Sign up for a free 30min 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!

Irrational Equation - I end up in a dead end

  1. May 29, 2013 #1
    1. The problem statement, all variables and given/known data
    This is the equation:
    2/(2 - x) + 6/(x^2 - x - 2) = 1


    2. Relevant equations
    sqrt= square root
    ^ = to the power of

    3. The attempt at a solution
    First thing that comes to mind is to turn it into this:
    2x^2 - 2x - 2 + 12 - 6x = (2 - 2)(x^2 -x -2)

    Then it gets real ugly:
    x^3 - x^2 - 8x + 14 = 0

    If only I could somehow get rid of that +14, I would then divide it all by x and get a normal quadric equation...
    x(x^2 - x - 8) + 14 = 0
     
    Last edited: May 29, 2013
  2. jcsd
  3. May 29, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    That should be
    $$
    2x^2 - 2x -4 +12 -6x = (2-x)(x^2-x-2)
    $$

    But have you tried dividing ##x^2-x-2## by ##2-x## first?
     
  4. May 29, 2013 #3

    AGNuke

    User Avatar
    Gold Member

    [tex]\frac{2}{2-x}+\frac{6}{x^2-x-2}=1[/tex]
    [tex]\frac{2x^2-2x-4+ 12-6x}{(2-x)(x^2-x-2)}=1[/tex]

    You made an error in the first step itself. That's why you are stuck. Try correcting it.
     
    Last edited: May 29, 2013
  5. May 29, 2013 #4

    DrClaude

    User Avatar

    Staff: Mentor

    You've made an error yourself :wink:

    $$
    \frac{2x^2-2x-4+ 12-6x}{(2-x)(x^2-x-2)}=1
    $$
     
  6. May 29, 2013 #5
    thanks for the help, I eventually end up with
    (2x-4)/(x^2 - x - 2) = 1

    which I turn into
    x^2 - x + 2 = 0
    and turns out that D < 0, so no real roots.
     
  7. May 29, 2013 #6

    DrClaude

    User Avatar

    Staff: Mentor

    That is incorrect. Please show your work.
     
  8. Jun 4, 2013 #7

    Curious3141

    User Avatar
    Homework Helper

    Don't expand the LHS. Factorise. Same for the RHS.

    After factoring both sides, you can group terms on the LHS.

    You'll find that ##(2-x)^2## is common on both sides.
     
  9. Jun 4, 2013 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In the original form of the problem, simplify the left-hand-side as much as possible before doing anything else.
     
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: Irrational Equation - I end up in a dead end
Loading...