Results 1 to 8 of 8

Math Help - How to solve this inequality ?

  1. #1
    Junior Member silvercats's Avatar
    Joined
    Mar 2011
    Posts
    67

    How to solve this inequality ?

    [ (x-1)-(x+5) ] / (x+5) < 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1490
    \displaystyle \frac{(x - 1)^2 - (x + 5)}{x + 5} < 0

    First note that \displaystyle x\neq -5 since this would give a zero denominator.

    Now you need to test two separate cases, one where the denominator is negative, and one where the denominator is positive (because multiplying or dividing by a negative number changes the inequality sign).

    Case 1: \displaystyle x + 5 < 0 \implies x < -5

    \displaystyle \begin{align*} \frac{(x - 1)^2 - (x + 5)}{x + 5} &< 0 \\ (x - 1)^2 - (x + 5) &> 0 \\ x^2 - 2x + 1 - x - 5 &> 0\\ x^2 - 3x - 4 &> 0 \\ x^2 - 3x + \left(-\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 - 4 &> 0 \\ \left(x - \frac{3}{2}\right)^2 - \frac{9}{4} - \frac{16}{4} &> 0 \\ \left(x - \frac{3}{2}\right)^2 - \frac{25}{4} &> 0 \\ \left(x - \frac{3}{2}\right)^2 &> \frac{25}{4} \\ \left|x - \frac{3}{2}\right| &> \frac{5}{2}\end{align*}

    So \displaystyle x - \frac{3}{2} < -\frac{5}{2} \implies x < -1 or \displaystyle x - \frac{3}{2} > \frac{5}{2} \implies x > 4. Since in this case, we already said that \displaystyle x < -5, that means the solution set for this case is \displaystyle x < -5.


    Now you try Case 2, where \displaystyle x + 5 > 0 \implies x > -5.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by silvercats View Post
    [ (x-1)-(x+5) ] / (x+5) < 0
    As an alternative,
    you either have "positive numerator" and "negative denominator"
    or "negative numerator" and "positive denominator".
    Those are the only ways the fraction can be negative.

    x<-5\Rightarrow\ (x-1)^2-(x+5)>0\Rightarrow\ x^2-2x+1-x-5>0

    \Rightarrow\ x^2-3x-4>0\Rightarrow\ (x-4)(x+1)>0

    Both of these factors must be positive or both must be negative.

    Both are positive for x>4 but then we cannot have x>4 and x<-5 at the same time.

    Both are negative for x<-1 and for x<-1 and x<-5 at the same time, then x<-5



    x>-5\Rightarrow\ (x-1)^2-(x+5)<0\Rightarrow\ (x-4)(x+1)<0

    These factors must have opposite sign.
    For the first positive and the second negative...
    x>4 and x<-1 and the second conflicts with the first.

    For the first negative and the second positive...
    x<4 and x>-1 which is consistent with x>-5
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member silvercats's Avatar
    Joined
    Mar 2011
    Posts
    67
    Thanks you both!But I thought we can't multiply both sides by (x+5) ,If it is a inequality .That is what I've heard before
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,884
    Thanks
    673
    Quote Originally Posted by silvercats View Post
    [ (x-1)-(x+5) ] / (x+5) < 0
    find critical values ...

    look for the values of x that make the numerator 0 ... x = 4 , x = -1

    look for the values of x that make the denominator 0 ... x = -5

    these three x-values break the number line into four regions ... since the rational expression is continuous in each region, pick any single value of x in each region and determine if that x-value makes the expression positive or negative.

    the region(s) where the expression is negative is your solution set ...

    x < -5 , -1 < x < 4
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by silvercats View Post
    Thanks you both!But I thought we can't multiply both sides by (x+5) ,If it is a inequality .That is what I've heard before
    You need to bear in mind that 2>1 but -2<-1.
    This scenario can occur if you multiply or divide both sides of an inequality by a negative number.
    Hence, for a variable (such as x) in the denomimator, you must account for the possibility
    of the denominator being negative.
    Then, if you multiply both sides by the negative denominator, you must reverse the inequality sign.
    If the denominator is positive, no reversal is necessary.
    But you have to account for these cases.

    The method I gave avoids this.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1490
    Quote Originally Posted by Archie Meade View Post
    You need to bear in mind that 2>1 but -2<-1.
    This scenario can occur if you multiply or divide both sides of an inequality by a negative number.
    Hence, for a variable (such as x) in the denomimator, you must account for the possibility
    of the denominator being negative.
    Then, if you multiply both sides by the negative denominator, you must reverse the inequality sign.
    If the denominator is positive, no reversal is necessary.
    But you have to account for these cases.

    The method I gave avoids this.
    So did the method I gave - i.e. to treat the solution process as two separate cases, one where the denominator is negative, the other where the denominator is positive.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member silvercats's Avatar
    Joined
    Mar 2011
    Posts
    67
    oh I see....... I can understand now.Thanks Archie Meade,Prove It and skeeter for the great answers
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solve inequality.
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 1st 2011, 04:06 AM
  2. solve the inequality
    Posted in the Algebra Forum
    Replies: 7
    Last Post: April 30th 2009, 05:29 PM
  3. Solve the inequality...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 18th 2009, 07:18 PM
  4. solve the inequality x^2+2x<3
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 8th 2009, 09:04 AM
  5. solve inequality
    Posted in the Algebra Forum
    Replies: 4
    Last Post: December 22nd 2007, 11:35 AM

Search Tags


/mathhelpforum @mathhelpforum