Results 1 to 8 of 8

Math Help - How would I set these two inequalities up to solve?

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    113

    How would I set these two inequalities up to solve?

    Would I set both of them equal to 0 and solve using the quadratic formula? Those that help me will be thanked!


    Solve each inequality:


    1. 4x^2 - 5x - 6 <= 0



    2. 2x^2 + 9x + 4 < 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by bobbyboy1111 View Post
    Would I set both of them equal to 0 and solve using the quadratic formula? Those that help me will be thanked!



    Solve each inequality:


    1. 4x^2 - 5x - 6 <= 0



    2. 2x^2 + 9x + 4 < 0
    Factorize the LHS of both the inequalities by splitting the middle term.

    Spoiler:
    1. 4x^2 - 5x - 6=4x^2-8x+3x-6=4x(x-2)+3(x-2)=(x-2)(4x+3)

    2. 2x^2 + 9x + 4=2x^2+8x+x+4=2x(x+4)+(x+4)=(x+4)(2x+1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2009
    Posts
    27
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2009
    Posts
    113
    Quote Originally Posted by alexmahone View Post
    Factorize the LHS of both the inequalities by splitting the middle term.

    Spoiler:
    1. 4x^2 - 5x - 6=4x^2-8x+3x-6=4x(x-2)+3(x-2)=(x-2)(4x+3)

    2. 2x^2 + 9x + 4=2x^2+8x+x+4=2x(x+4)+(x+4)=(x+4)(2x+1)

    Whoa, what does "LHS" mean, how do I split the middle term, and how do I factorize?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Aug 2009
    Posts
    29
    Quote Originally Posted by bobbyboy1111 View Post
    Whoa, what is the LHS mean, how do I split the middle term, and how do I factorize?
    LHS = Left Hand Side

    You need to decompose the polynomials on the LHS into a product of factors which when multiplied give the original form.

    It should be clear that (x-2)(4x-3) = 0 for x=2 and x = -\frac{3}{4}. It can easily be checked that for values between -3/4 and 2 it is negative.

    See also http://www.wolframalpha.com/input/?i=4+x^2+-+5+x+-+6+<%3D+0
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2009
    Posts
    27
    Dont you know how to factorise quadratice equations ? & did you viewed the link I offered .
    Last edited by Aladdin; August 9th 2009 at 08:29 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2009
    Posts
    113
    Quote Originally Posted by Aladdin View Post
    Dont you know how to factorise quadratice equations ? & did you viewed the link I offered .
    Yes, I did, but there aren't any truly similar examples.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    The examples, like "x2 + 2x 8 < 0" and "x2 + 6x 9 > 0" are quadratic, they have "zero" on the other side of the inequality, and they involve factoring, just as you've been advised to proceed here. How are those examples "truly" different from what you are doing?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 18th 2011, 08:54 PM
  2. Please help solve inequalities
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 30th 2008, 10:24 AM
  3. Replies: 1
    Last Post: November 27th 2008, 11:28 AM
  4. please help me solve inequalities
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: May 12th 2007, 06:40 AM
  5. solve and Graph the inequalities
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 27th 2007, 01:20 AM

Search Tags


/mathhelpforum @mathhelpforum