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How to find x in -2≤x^2≤4?

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data

    -2≤x2≤4

    How to find .....<x<.....?
    Show I take square root of everything? What about the negative -2?
     
  2. jcsd
  3. Sep 5, 2010 #2
    Draw a graph:
    y = x^2
    y =-2
    y = -4

    You would see -2≤x2≤4 same as x2≤4
     
  4. Sep 6, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, since [itex]x^2[/itex] is never negative, [itex]-2\le x^2\le 4[/itex] is exactly the same as [itex]0\le x^2\le 4[/itex]. But also note that both positive and negative x will give a positive square.

    You could also attempt this as two separate inequalities: You should immediately see that [itex]-2\le x^2[/itex] is true for all x. What about [itex]x^2\le 4[/itex]? I recommend solving inequalities like this by first solving the related equality. What are the two solutions to [itex]x^2= 4[/itex]? Those two numbers (lets call them a and b with a< b) divide the set of all real numbers into 3 intervals: x< a, a< x< b, and b< x. In each of those we have either [itex]x^2< 4[/itex] or [itex]x^2> 4[/itex]. You could choose one point in each interval to determine which is true for all points in that interval.
     
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