Results 1 to 3 of 3

Thread: [SOLVED] Inequalities

  1. #1
    Member
    Joined
    Jul 2008
    Posts
    119

    [SOLVED] Inequalities

    This is a little embarassing, but it seems like I do not understand why the answers are what they are for these inequalities.

    1. $\displaystyle 3x^2 - 2x-1 \geq 0 $

    First, I factor the left-hand side.

    $\displaystyle \therefore (3x+1)(x-1) \geq 0 $

    So, $\displaystyle x - 1 \geq 0 $ if we divide both sides by '3x+1'.

    Thus, $\displaystyle x \geq 1 $

    Here's the part I don't understand. Naturally, I think that if we divide 'x-1' from both sides. We would get

    $\displaystyle 3x+1 \geq 0 $

    which yields

    $\displaystyle 3x \geq -1 \therefore x \geq -\frac{1}{3} $ (*)

    But this can not be right. By counterexample, let x = 0; this inequality is false.

    It should be $\displaystyle x \leq -\frac{1}{3} $.

    I believe the reason for this is since for (x-1), if x is a negative number, then (x-1) < 0. So in order for the inequality to be equal then '3x+1' must also be negative (since a negative times a negative equals a positive) which explains why it should be

    $\displaystyle 3x + 1 \leq 0 $ and not $\displaystyle 3x + 1 \geq 0 $

    Hence we solve, we should get $\displaystyle x \leq 1/3 $. What am I doing wrong at (*)?

    =============================
    2. Another confusing inequality.

    $\displaystyle \frac{1}{3-x} < -2 $

    I multiply both sides by '3-x'
    $\displaystyle 1 < -2(3-x) $
    $\displaystyle 1 < -6+2x $
    $\displaystyle 7 < 2x$
    $\displaystyle \frac{7}{2} < x $ (*)

    This answer makes no sense. If x = 4, then the inequality is false.

    Then, I tried a different way. I add both sides by 2

    $\displaystyle \frac{1}{3-x} + 2 < 0 $
    $\displaystyle \frac{2(3-x)+1}{3-x} < 0 $
    $\displaystyle \frac{6-2x+1}{3-x} < 0 $
    $\displaystyle 7 - 2x < 0 $

    So, $\displaystyle x < \frac{7}{2} $. Since, x can not be 3, or else the denominator is zero. I conclude that $\displaystyle 3 < x < \frac{7}{2} $

    What am I doing wrong at (*)?

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    The problem is when you divide both sides by a variable factor! you are losing information about the ineqality.

    After you factor you get

    $\displaystyle (3x+1)(x-1) \ge 0$ from here take your two zero's (this is where the inequality can change signs)

    so we get $\displaystyle x=-\frac{1}{3} \mbox{ or } x=1$

    Now we have three different parts of the number line

    $\displaystyle \left(-\infty,-\frac{1}{3} \right)$

    $\displaystyle \left(-\frac{1}{3},1 \right)$ and

    $\displaystyle (1,\infty)$

    Now we check a point from each interval to see if it is a solution

    $\displaystyle \left(-\infty,-\frac{1}{3} \right)$ check $\displaystyle x=-1$

    In the original and we get $\displaystyle (3(-1)+1)(-1-1)=4 \ge 0$ so the above set is a solution.

    Now we check $\displaystyle \left(-\frac{1}{3},1 \right)$ with $\displaystyle x=0$ and we get $\displaystyle (3(0)+1)(0-1)=-1 \le 0$ so it is false and is not a solution.

    finally we check $\displaystyle (1,\infty)$ with x=2 $\displaystyle (3(2)+1)(2-1)=7 \ge 0$ so this is a solution.

    So the solution to the inequality is

    $\displaystyle \left(-\infty,-\frac{1}{3} \right] \cup [1,\infty)$

    For the rational ineqality get a common denomiantor and get the zero's from both the numerator and denomiator and do the same thing.

    Good luck.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    119
    Thank you TheEmptySet. I was reviewing through my Calculus book and did not know/forgotten how to approach these problems. It was driving me crazy for a while. Your explanation makes more sense. So, I just choose arbitrary numbers within these intervals and check whether the inequality holds... ok. Thanks again. Cheers.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] more inequalities
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Feb 7th 2010, 09:43 AM
  2. [SOLVED] Inequalities
    Posted in the Algebra Forum
    Replies: 9
    Last Post: Jan 22nd 2010, 12:06 PM
  3. [SOLVED] Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Mar 29th 2009, 12:37 AM
  4. [SOLVED] inequalities
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Mar 25th 2009, 07:58 AM
  5. [SOLVED] Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Sep 30th 2008, 08:39 AM

Search Tags


/mathhelpforum @mathhelpforum