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Inequality Question

  1. Sep 25, 2015 #1
    How do you solve x for the below inequality?

    ##\frac{a}{x^2} < -b##

    My attempt is:

    ##\frac{a}{x^2} + b < 0##

    ##\frac{a + bx^2}{x^2} < 0##
     
  2. jcsd
  3. Sep 25, 2015 #2

    andrewkirk

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    No. If it asks you to solve for x, it means you need to get x by itself on one side of the (in)equation. How would you go about doing that?
     
  4. Sep 25, 2015 #3
    Do you mean it has no solution?

    Can I do such as below?

    ##\frac{a}{x^2} < -b##

    ##a < -bx^2##

    ##-\frac{a}{b} > x^2##

    ##x < ±\sqrt{-\frac{a}{b}}##

    ##x < ±i\sqrt{\frac{a}{b}}##

    ##x < ±i\frac{\sqrt{ab}}{b}##
     
  5. Sep 25, 2015 #4
    You have to be more careful than that. Split in into 4 cases according to whether a and b are positive or negative. And a fifth case if b=0.
     
  6. Sep 25, 2015 #5

    Mark44

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    The four cases can be reduced to two: ab > 0 and ab < 0.
     
  7. Sep 26, 2015 #6

    HallsofIvy

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    No, they can't. For example if a and b are both equal to -1, the inequality is [itex]-\frac{1}{x^2}< 1[/itex]. Multiplying both sides by the positive [itex]x^2[/itex], [itex]-1< x^2[/itex] which is true for all x. But if a and b are both equal to 1, we have [itex]\frac{1}{x^2}< -1[/itex] so that [itex]1< -x^2[/itex] and multiplying both sides by negative 1, [itex]-1> x^2[/itex] which is never true.
     
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