Results 1 to 7 of 7

Math Help - Inequality problem

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Inequality problem

    Find the set of values of x for which:

    \frac{x}{x-3} > \frac{1}{x-2}.

    I rearranged it to form:

    \frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0

    Therefore critical points of the denominator are x = 2, 3.

    It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".

    I am not sure then how to work out a set of values for x, they have x<2 and x>3 in the book.

    Thanks in advance for the help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,608
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by craig View Post
    Find the set of values of x for which:
    \frac{x}{x-3} > \frac{1}{x-2}.
    I rearranged it to form:
    \frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0
    Therefore critical points of the denominator are x = 2, 3.
    It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".
    The roots of the numerator are complex numbers.
    The value of the numerator at x=0 is 3. So it is always positive.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Thanks for the reply Plato. Do you have any idea how they worked out their set of values?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by craig View Post
    Find the set of values of x for which:

    \frac{x}{x-3} > \frac{1}{x-2}.

    I rearranged it to form:

    \frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0

    Therefore critical points of the denominator are x = 2, 3.

    It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".
    If the discriminant of a quadratic ax^2+bx+c is negative, then it keeps a constant sign, the same as a.
    So here, since a=1, and the discriminant is 9-12=-3, the above is always positive.

    You can prove it by taking the derivative. For this particular polynomial.

    I am not sure then how to work out a set of values for x, they have x<2 and x>3 in the book.

    Thanks in advance for the help
    Then, for the quotient to be positive, you need the denominator to be positive.

    The product of two terms is positive if and only if both terms have the same sign.
    That is to say [x-2>0 and x-3>0] or [x-2<0 and x-3<0]
    From the first one, you have [x>2 and x>3], which is [x>3]
    From the second one, you have [x<2 and x<3], which is [x<2]

    Does it look clear to you ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Moo View Post
    If the discriminant of a quadratic ax^2+bx+c is negative, then it keeps a constant sign, the same as a.
    That makes perfect sense thank you, if the discriminant is negative - ie two complex roots - then the quadratic keeps the sign of the x^2 constant.

    Never had it explained like this before, but seems much easier.

    Quote Originally Posted by Moo View Post
    Then, for the quotient to be positive, you need the denominator to be positive.

    The product of two terms is positive if and only if both terms have the same sign.
    That is to say [x-2>0 and x-3>0] or [x-2<0 and x-3<0]
    From the first one, you have [x>2 and x>3], which is [x>3]
    From the second one, you have [x<2 and x<3], which is [x<2]

    Does it look clear to you ?
    Yep that's perfectly clear thank you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Sorry just one more question, if the original question had something that was less than zero, would you do the opposite of the above, make one negative and one positive?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie Singaporean's Avatar
    Joined
    May 2009
    From
    Singapore
    Posts
    19
    Not sure if anyone mentioned this but, you can complete the square for the numerator to prove that it is always positive cause i dont understand the discriminant thingy always.

    x^2-3x+3=[x-\frac{3}{2}]^2+\frac{3}{4}

    Since [x-\frac{3}{2}]^2\geq0 for all values of x
    Therefore, [x-\frac{3}{2}]^2+\frac{3}{4}>0 for all values of x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Inequality Problem
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 5th 2011, 08:51 AM
  2. inequality problem =.=
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: May 17th 2010, 01:18 AM
  3. inequality problem
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 19th 2010, 12:28 PM
  4. inequality problem
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 2nd 2010, 11:15 PM
  5. Inequality problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 12th 2008, 09:18 AM

Search Tags


/mathhelpforum @mathhelpforum