Results 1 to 4 of 4

Math Help - inequality

  1. #1
    Newbie
    Joined
    Feb 2007
    Posts
    23

    inequality

    (x-2)*(x+5) < 0

    do you just make the < an = and solve there?

    I tried expanding doing:

    x^2 + 3x - 10

    I know (x + 2)(x + 1) = x^2 + 3x + 2 which is similar, so then

    (x + 2)(x + 1) - 8 < 0 .. but that doesn't really help me

    hmm
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by flash101 View Post
    (x-2)*(x+5) < 0

    do you just make the < an = and solve there?

    I tried expanding doing:

    x^2 + 3x - 10

    I know (x + 2)(x + 1) = x^2 + 3x + 2 which is similar, so then

    (x + 2)(x + 1) - 8 < 0 .. but that doesn't really help me

    hmm
    you treat the < as an equal sign, then test your results at the end

    (x - 2)(x + 5) < 0

    => x - 2 < 0 or x + 5 < 0

    => x < 2 or x < -5

    so now we basically think of, or draw a number line, placing the numbers -5 and 2 in there relative positions, and test what regions the original equation holds true for
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, flash!

    (x-2)(x+5) \:< \:0
    Leave it in factored form . . .

    We have: . (x - 2)(x + 5) \:<\:0

    It says: the product of two numbers is negative.
    So one factor must be positive and the other must be negative.
    . . There are two possible cases.


    [1] . (x - 2) is positive and (x + 5) is negative.

    So we have: . \begin{array}{ccc}x - 2 \:>\:0 & \Rightarrow & x \:> \:2 \\ x+5 \:<\:0 & \Rightarrow & x \:<\:\text{-}5\end{array}

    Hence, x is a number which greater than 2 and less than -5.
    . . This is clearly impossible.


    [2] . (x-2) is negative and (x+5) is positive.

    So we have: . \begin{array}{ccc}x - 2 \:<\:0 & \Rightarrow & x \:<\:2 \\ x + 5 \:>\:0 & \Rightarrow & x \:>\:\text{-}5\end{array}

    Hence, x is a number which is less than 2 and greater than -5.

    This is possible for x between -5 and +2: . \boxed{-5 \:<\:x\:<\:2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    Meh
    Posts
    1,630
    Thanks
    6
    Flash, remember when you multiply with a negative number, the  > will become a  < and vice versa.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2011, 09:20 PM
  2. Replies: 3
    Last Post: December 12th 2010, 02:16 PM
  3. inequality
    Posted in the Math Challenge Problems Forum
    Replies: 7
    Last Post: July 25th 2010, 07:11 PM
  4. Inequality help
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: July 8th 2010, 07:24 AM
  5. Inequality :\
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 12th 2009, 02:57 PM

Search Tags


/mathhelpforum @mathhelpforum