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Thread: modulus equation

  1. #1
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    modulus equation

    am trying to solve for x in this equation

    $\displaystyle x^2-\mid x \mid - 12 < 0$
    $\displaystyle \mid x \mid <12-x^2$
    $\displaystyle \mid x \mid^2 >{(12-x^2})^2$ squaring both sides

    i end up with $\displaystyle x^4 - 24x^2+144=0$ can i get some helf solving for x. that if what i did was correct.
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  2. #2
    MHF Contributor
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    Hi

    I suggest you to separate the 2 cases $\displaystyle x\leq 0$ and $\displaystyle x \geq 0$

    In the first case you get $\displaystyle x^2 + x - 12 < 0$

    $\displaystyle (x+4)(x-3) < 0$

    Find the solutions and take the ones that are $\displaystyle \leq 0$

    Do the same for the second case and finally join the 2 sets of solutions
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  3. #3
    MHF Contributor
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    Quote Originally Posted by sigma1 View Post
    am trying to solve for x in this equation

    $\displaystyle x^2-\mid x \mid - 12 < 0$
    $\displaystyle \mid x \mid <12-x^2$
    $\displaystyle \mid x \mid^2 >{(12-x^2})^2$ squaring both sides

    i end up with $\displaystyle x^4 - 24x^2+144=0$ can i get some helf solving for x. that if what i did was correct.
    Hi sigma1,

    you need c-ordination here,

    you have made errors with polarity above.

    $\displaystyle x^2-|x|-12<0$

    If x is positive, this is

    $\displaystyle x^2-x-12<0$

    $\displaystyle (x-4)(x+3)<0$

    this is negative if one factor is positive and the other negative.
    If -3<x<4 the left factor is negative and the right one positive.
    But remember, x must be positive here.
    If x>4, they are both positive.
    If x<-3 they are both negative.

    If x is negative, this is

    $\displaystyle x^2+x-12<0$

    $\displaystyle (x-3)(x+4)<0$

    This can be evaluated in the same way as above.
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