Results 1 to 7 of 7

Math Help - inequalities: 2 methods

  1. #1
    Member
    Joined
    May 2009
    Posts
    91

    Unhappy inequalities: 2 methods

    Hi guys,

    I am trying to find the values of x for which

     x - 3 < \frac{x - 4}{x}

    Attached is a diagram from which it is easy to see the answer x < 0! i.e these are the values of x for which the straight line x - 3 is below the curve (which I have expressed as:

    1 - \frac{4}{x}

    But, as we all know, inequalities can be solved in three ways: by sketching, by calculation and by completing the square. So I have tried it by calculation - here goes:


    x - 3 < (x - 4) / x

    x^2 - 3x < x - 4

    x^2 - 4x + 4 < 0

    (x - 2) (x - 2) < 0

    (x - 2)^2 < 0


    Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!
    Attached Thumbnails Attached Thumbnails inequalities: 2 methods-m14c15.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by s_ingram View Post
    ...So I have tried it by calculation - here goes:


    x - 3 < (x - 4) / x

    x^2 - 3x < x - 4

    x^2 - 4x + 4 < 0

    (x - 2) (x - 2) < 0

    (x - 2)^2 < 0


    ... Can anyone see what I am doing wrong!
    You assumed that x \geq 0. If you do the calculations for x < 0 you'll get the result which you already know.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by s_ingram View Post
    x - 3 < (x - 4) / x

    x^2 - 3x < x - 4

    x^2 - 4x + 4 < 0

    (x - 2) (x - 2) < 0

    (x - 2)^2 < 0


    Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!
    I wouldn't multiply both sides by x. Instead, I would do this:
    \begin{aligned}<br />
x - 3 &< \frac{x - 4}{x} \\<br />
x - 3 - \frac{x - 4}{x} &< 0 \\<br />
\frac{x^2 - 3x}{x} - \frac{x - 4}{x} &< 0 \\<br />
\frac{x^2 - 4x + 4}{x} &< 0 \\<br />
\frac{(x - 2)^2}{x} &< 0<br />
\end{aligned}

    At this point I would make a sign chart. First, I need "critical points." These are values which (1) the fraction equals 0, and (2) the fraction is undefined. I find (1) by setting the numerator equal to zero:
    \begin{aligned}<br />
(x - 2)^2 &= 0 \\<br />
x - 2 &= 0 \\<br />
x &= 2<br />
\end{aligned}
    I find (2) by setting the denominator equal to zero: x = 0.

    Then I draw my sign chart like this:
    Code:
             und              0
    ----------+---------------+-------------
              0               2
    (und = undefined)

    Now take the fraction
    \frac{(x - 2)^2}{x} and test a value less than 0. You will see that the fraction is negative, so I make a notation on my sign chart.
    Code:
      (-)^2
      -----
       (-)   und              0
    ----------+---------------+-------------
       neg    0               2
    Pick a number between 0 and 2 and test the fraction, and then pick a number greater than 2 and test the fraction. When you're done, you should get this:
    Code:
      (-)^2          (-)^2          (+)^2
      -----          -----          -----
       (-)    und     (+)      0     (+)
     ----------+---------------+-------------
        neg    0      pos      2     pos
    We want a range of numbers where the fraction is less than 0, or negative. So the solutions are x < 0.


    01
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2009
    Posts
    91
    Sorry, where do I assume that x \geq 0? If x is less than zero say -5, then (x - 2)^2 gives (-7)^2 = 49 and this is not less than 0.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2009
    Posts
    91
    Thanks again yeongil. I can see your method works and I have used something like it when checking the ranges of functions for which they are real or unreal, but I don't see why you would not multiply by x. Is it wrong to do so and if so why?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    May 2009
    Posts
    527
    To answer both of your posts, you assumed that x \geq 0 when you multiplied both sides by x. Remember that when you multiply both sides of an inequality by a negative number, you flip the sign. When you went from this step:
    x - 3\;{\color{red}<}\;\frac{x - 4}{x}
    to this step:
    x^2 - 3x\;{\color{red}<}\;x - 4
    you multiplied both sides by x, and since you didn't change the sign, you assumed that x \geq 0.

    The problem was to solve for x, and if we pretend that we do not know what x is beforehand, we shouldn't multiply both sides by x; instead, subtract the fraction from both sides and proceed like I did earlier.


    01
    Last edited by yeongil; July 9th 2009 at 02:14 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2009
    Posts
    91
    Excellent answer! Now I see. I suspected Earboth was on to something but I didn't understand what he meant.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof methods
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: September 17th 2011, 07:15 AM
  2. 2 Methods
    Posted in the Algebra Forum
    Replies: 6
    Last Post: September 16th 2011, 11:58 AM
  3. dw/dt how to solve by two methods
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 4th 2010, 05:49 PM
  4. Numerical methods
    Posted in the Calculus Forum
    Replies: 3
    Last Post: January 12th 2008, 06:07 PM
  5. analytical methods
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 19th 2007, 05:18 AM

Search Tags


/mathhelpforum @mathhelpforum