Annoying math question

• aiglosicicle

  25 Dec 06, 19:56

  there is this inequalities math question that appears pretty easy but i cant seem to solve it.
  
  its really bugging me and i have to solve it! haha
  
  hope some people here can help!
  
  (x+1)/(x-1) [smaller or equals to] x
  
  
• hegu

  25 Dec 06, 20:17

  (x+1)/(x-1) - x < 0
  
  [(x+1) - x(x-1)]/(x-1) < 0
  
  (1+2x-x^2)/(x+1) < 0
  
  sry lazy to factorise further
  
  but you know how to continue right? (find the roots of the LHS...)
• yuko-ogura

  25 Dec 06, 20:33

  hegu,ur back.
  
  long time no see.
• hegu

  25 Dec 06, 20:52

  Originally posted by yuko-ogura:

  hegu,ur back.
  
  long time no see.

  haha, im always here just that i dun post that often
• ^tamago^

  25 Dec 06, 21:01

  1-sqrt(2) <= x < 1
  
  OR
  
  x >= 1+sqrt(2)
• hegu

  25 Dec 06, 21:09

  Originally posted by ^tamago^:

  1-sqrt(2) <= x < 1
  
  OR
  
  x >= 1+sqrt(2)

  tamago kor kor comes to the rescue!
• ^tamago^

  25 Dec 06, 21:11

  Multiply both sides by (x-1)²:
  (x+1)(x-1) <= x(x-1)², x is not 1
  x²-1 <= x³-2x²+x
  x³-3x²+x+1 >= 0
  (x-1)(x²-2x-1) >= 0
  (x-1)(x-[1+sqrt(2)])(x-[1-sqrt(2)]) >= 0
  
• ^tamago^

  25 Dec 06, 21:12

  Originally posted by hegu:

  tamago kor kor comes to the rescue!

  
• Orange*Asterisk

  1 Jan 07, 00:45

  Originally posted by hegu:

  tamago kor kor comes to the rescue!

  shud be jie jie ba
• mahawarrior

  1 Jan 07, 11:23

  Multiply both sides by the square of the denominator lor. Then the rest should be pretty easy bah...
• 798

  1 Jan 07, 23:31

  Originally posted by aiglosicicle:
  there is this inequalities math question that appears pretty easy but i cant seem to solve it.
  
  its really bugging me and i have to solve it! haha
  
  hope some people here can help!
  
  [b](x+1)/(x-1) [smaller or equals to] x
  
  [/b]

  (x+1)/(x-1) <= x
  x+1 <= x(x-1)
  x+1 <= x^2-x
  0 <= x^2-x-x-1
  x^2-2x-1 >= 0
  
  using quad equation:
  x >= 2.414 or -0.4142
  
  as tis is an inequalities question, hence
  x must be more than or equal to 2.414 or (1+sqrt(2))
• hegu

  2 Jan 07, 13:03

  Originally posted by 798:

  (x+1)/(x-1) <= x
  x+1 <= x(x-1)
  x+1 <= x^2-x
  0 <= x^2-x-x-1
  x^2-2x-1 >= 0
  
  using quad equation:
  x >= 2.414 or -0.4142
  
  as tis is an inequalities question, hence
  x must be more than or equal to 2.414 or (1+sqrt(2))

  haha, your first step to second step is alr wrong but anyway tamago kor...i mean jie jie has alr solved it
• 798

  2 Jan 07, 22:46

  Originally posted by hegu:

  haha, your first step to second step is alr wrong but anyway tamago kor...i mean jie jie has alr solved it

  wrong meh? anyway still can get correct ans wah.
  
  tamago's method involve many working, mine very simple working.
• hisoka

  2 Jan 07, 22:49

  multiply te denominator out lor the solve as pernormal jus tthat you need to make 3 cases one for the denominator <0 and one for denominator >0 and lastly no solution for the denominator = 0
• 798

  2 Jan 07, 22:52

  Originally posted by hisoka:

  multiply te denominator out lor the solve as pernormal jus tthat you need to make 3 cases one for the denominator <0 and one for denominator >0 and lastly no solution for the denominator = 0

  i get ur point, if x = 1
  
  LHS is infinite whereas RHS is 1, this is not right... -_-|
• hisoka

  2 Jan 07, 22:57

  Originally posted by 798:

  i get ur point, if x = 1
  
  LHS is infinite whereas RHS is 1, this is not right... -_-|

  actualy might still be correct de.
  
  cos if you take limits hor and assume its continous the LHS at limx->1 is 1 also i think. but in this case should be undefined
• 798

  2 Jan 07, 23:04

  Originally posted by hisoka:

  actualy might still be correct de.
  
  cos if you take limits hor and assume its continous the LHS at limx->1 is 1 also i think. but in this case should be undefined

  maybe the correct answer should be:
  
  x is within the range of:
  -0.414 to 1.414 & x is not equal to 1