Page 1 of 2 12 LastLast
Results 1 to 15 of 30

Math Help - Help with solving inequalities

  1. #1
    Junior Member
    Joined
    Feb 2007
    Posts
    53

    Help with solving inequalities

    Please help with solving the following inequalities (2):

    1) x^2 + 4x - 32 < 0

    I read in the book that you're supposed to factor and then plug in some "key" numbers that will cause the equation to equal zero (if I'm understanding this correctly). The only thing that I could come up with was:

    (x+8) (x-4), but plugging in an '8' in the equation doesn't equal zero. How would I solve this? Thanks in advance.

    2) This is an inequality that's just one variable (not sure how to proceed):

    \frac{x+1}{2x-3} ≥ 0

    Thanks for your help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,649
    Thanks
    1597
    Awards
    1
    Here is the idea. If x<-8 then the factor (x+8) is negative and for x>-8 that factor is positive.
    If x<4 then the factor (x-4) is negative and for x>4 that factor is positive.

    Now if -8<x<4 we have one factor positive and the other negative, hence their product is negative. So have we found the complete solution?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Plato View Post
    Here is the idea. If x<-8 then the factor (x+8) is negative and for x>-8 that factor is positive.
    If x<4 then the factor (x-4) is negative and for x>4 that factor is positive.

    Now if -8<x<4 we have one factor positive and the other negative, hence their product is negative. So have we found the complete solution?
    Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by lilrhino View Post
    Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...
    I need some guidance on this please...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    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 lilrhino View Post
    Plato thanks for your response. I'm still trying to fathom this, but from what I understand 8 does not satisfy the inequality. So, I'd say we haven't found the complete solution. I'm reading my book and holding my breath (at the same time)...
    -8 is the point you should be looking at. Plato's solution is complete. Look at the graph.
    Attached Thumbnails Attached Thumbnails Help with solving inequalities-quad.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Jhevon View Post
    -8 is the point you should be looking at. Plato's solution is complete. Look at the graph.
    Thanks Plato & Jhevon!

    It helps to graph it first (I see). What approach should I use in addition to graphing to solve these sort of inequalities? Do I just use a test number from each?

    Also, how should I proceed to solve my 2nd inequality?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by lilrhino View Post
    1) x^2 + 4x - 32 < 0
    First, note that x^2+4x-32<0\iff(x-4)(x+8)<0

    Now you have to analyze two possibilities:

    \blacktriangleright~(x-4)>0~~\wedge~~(x+8)<0

    \blacktriangleright~(x-4)<0~~\wedge~~(x+8)>0

    After, find the intersection of the first one and do the same for the second one. Finally, the intervals that you'll have, join them and that's the answer.

    Apply the same idea with the next exercise, but remember that you cannot take the denominator equal to 0.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Krizalid View Post
    First, note that x^2+4x-32<0\iff(x-4)(x+8)<0

    Now you have to analyze two possibilities:

    \blacktriangleright~(x-4)>0~~\wedge~~(x+8)<0

    \blacktriangleright~(x-4)<0~~\wedge~~(x+8)>0

    After, find the intersection of the first one and do the same for the second one. Finally, the intervals that you'll have, join them and that's the answer.

    Apply the same idea with the next exercise, but remember that you cannot take the denominator equal to 0.
    I'll give it a try. Thanks for the explanation.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    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 lilrhino View Post
    Thanks Plato & Jhevon!

    It helps to graph it first (I see). What approach should I use in addition to graphing to solve these sort of inequalities? Do I just use a test number from each?
    i usually use a number-line approach. write the roots on the number line, in this case that would be -8 and +4, then test numbers in the different regions separated by the roots. so you would test like -9, 0 and 5 by plugging them into the original equation to see where the function would be negative or positive. you would realize that it is negative in the region between -8 and +4, which is the answer.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    By the way lilrhino, use \ge to generate a\ge0, for example
    Follow Math Help Forum on Facebook and Google+

  11. #11
    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 Krizalid View Post
    By the way lilrhino, use \ge to generate a\ge0, for example
    i always use \geq, that works as well.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Jhevon View Post
    i usually use a number-line approach. write the roots on the number line, in this case that would be -8 and +4, then test numbers in the different regions separated by the roots. so you would test like -9, 0 and 5 by plugging them into the original equation to see where the function would be negative or positive. you would realize that it is negative in the region between -8 and +4, which is the answer.
    I'll try the number line approach because using a table is lengthy. Thanks for the explanation Jhevon.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by Jhevon View Post
    i always use \geq, that works as well.
    Haha

    If you prefer too, use \geqslant.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Feb 2007
    Posts
    53
    Quote Originally Posted by Jhevon View Post
    i always use \geq, that works as well.
    Quote Originally Posted by Jhevon View Post
    haha, it wasn't a preference. i learnt LaTex from the tutorial here, in the tutorial \geq is what they have
    I guess I'd better review the tutorial because I haven't a clue what you two are talking about
    Follow Math Help Forum on Facebook and Google+

  15. #15
    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 lilrhino View Post
    I guess I'd better review the tutorial because I haven't a clue what you two are talking about
    the tutorial is here

    click on the pdf file in the first post
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Solving Inequalities
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 8th 2009, 04:57 AM
  2. Solving Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 11th 2008, 01:01 PM
  3. Solving Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2008, 02:00 AM
  4. Solving Inequalities
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 13th 2008, 08:21 PM
  5. Solving inequalities
    Posted in the Algebra Forum
    Replies: 4
    Last Post: February 17th 2007, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum