Results 1 to 6 of 6

Math Help - inequality 2

  1. #1
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    inequality 2

    I couldn't solve the following questions.
    1)|x^2+x-5|>2X+1
    2)|X-2|-|X+1|-X+2<0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,569
    Thanks
    1428

    Re: inequality 2

    In the first, square both sides, then solve the resulting polynomial.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    Re: inequality 2

    You don't have to square both sides.
    x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,493
    Thanks
    966

    Re: inequality 2

    Quote Originally Posted by kastamonu View Post
    You don't have to square both sides.
    x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.
    true you don't have to, but doing so gives you the same answer and is easier to solve than the absolute value equations.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    2,493
    Thanks
    966

    Re: inequality 2

    Quote Originally Posted by kastamonu View Post
    You don't have to square both sides.
    x^2+x-5>2x+1 and -x^2-x+5<2x+1 and I solved both equations. But couldn't get the answer.
    looking at the first one and applying Prove It's method.

    $|x^2+x-5|> 2x+1 \Rightarrow (x^2+x-5)^2> (2x+1)^2$

    $x^4+2 x^3-9 x^2-10 x+25>4 x^2+4 x+1$

    $x^4+2 x^3-13 x^2-14 x+24>0$

    whittling this down by checking for factors of the form $(x\pm r)$ where r divides 24 we eventually reduce this to

    $(x-3) (x-1) (x+2) (x+4)>0$

    checking signs from $-\infty \to \infty$ we get

    $(\infty, -4) \rightarrow +$

    $(-4, -2) \rightarrow -$

    $(-2, 1) \rightarrow +$

    $(1, 3) \rightarrow -$

    $(3,\infty) \rightarrow +$

    and thus our solution is

    $(\infty, -4) \cup (-2, 1) \cup (3, \infty)$

    Solving that is much more methodical than trying to solve the absolute value equation directly.

    Apply this to the second problem and see if you can work it out.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2012
    From
    Istanbul
    Posts
    181
    Thanks
    1

    Re: inequality 2

    Second one can be solved by this way:
    1)If -1<x<2 then x>1
    2)if x<-1 then
    -x+2+x+1-x+2<0
    x>5
    According to my book answer is x>1.Then it is correct.
    Last edited by kastamonu; May 10th 2014 at 03:39 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 26th 2013, 02:10 AM
  2. Replies: 2
    Last Post: January 11th 2011, 08:20 PM
  3. Replies: 3
    Last Post: December 12th 2010, 01:16 PM
  4. Inequality
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 24th 2010, 11:32 AM
  5. Inequality
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: October 5th 2008, 10:27 AM

Search Tags


/mathhelpforum @mathhelpforum