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Absolute value question

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    how do I solve this? I am confused...
    |2x - 1| = x^2

    2. Relevant equations

    3. The attempt at a solution
    when i tried it I ended up with a solution set of (1,-1). But the official answer is quite different so I am confused!!
  2. jcsd
  3. May 28, 2010 #2


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    Homework Helper

    The absolute value function is defined as follows:

    [tex]|x| = x, x\geq 0[/tex]

    [tex]|x|= -x, x< 0[/tex]

    But you have 2x-1 so replace the x in the absolute value for 2x-1. Try this and then solve both equations [itex]2x-1=x^2[/itex] and [itex]-(2x-1)=x^2[/itex]. Once you find your solutions, remember to be sure to scrap the solutions that aren't valid. That is, if you get an answer of -10 for [itex]2x-1=x^2[/itex] then you know it's not valid because we are assuming 2x-1>0, or, x>1/2.
  4. May 28, 2010 #3
    You could also approach this graphically.

    Imagine the line y=2x-1, that's a line with a positive slope and intercepts the y-axis at (0,-1). Now, |2x-1| refers to the absolute values of y (i.e. |y|) so you have to reflect what is below the x-axis in the x-axis itself to get your |y|=|2x-1| which looks like a 'V' shaped graph. The function of the graph that you obtain can be defined as follows:

    [tex] y = 2x-1 for x >= \frac{1}{2} [/tex]

    [tex] y = 1 - 2x for x <= \frac{1}{2} [/tex]

    Now all you have to do is sketch the graph of [tex] x^2 [/tex] on top of that and simply determine where your V-shaped graph (y=|2x-1|) is equal to the graph of y=x^2 by finding the points of intersection.
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