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Math Help - Finding the values of x for a boundary on a graph

  1. #1
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    Finding the values of x for a boundary on a graph

    Hi I need to find the values of x for which

    \frac{x^{3}+5x-12}{x-3} > 4

    Now I could - 4(x-3)/(x-3) from the RHS to the LHS, which is what the markscheme does, but why cant I do this:

    x^{3}+5x-12> 4(x-3)
    x^{3}+5x-12> 4x-12
    x^{3}+5x> 4x

    x^{3}+x

    that only gives me x = 0 and i as opposed to the other way which gives me 0,3,i

    Why doesn't the stupid math work?

    Thanks
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by thomas49th View Post
    Hi I need to find the values of x for which

    \frac{x^{3}+5x-12}{x-3} > 4

    Now I could - 4(x-3)/(x-3) from the RHS to the LHS, which is what the markscheme does, but why cant I do this:

    x^{3}+5x-12> 4(x-3)
    x^{3}+5x-12> 4x-12
    x^{3}+5x> 4x

    x^{3}+x

    that only gives me x = 0 and i as opposed to the other way which gives me 0,3,i

    Why doesn't the stupid math work?

    Thanks
    move the 4 to the left side:
    \frac{x^{3}+5x-12}{x-3} - 4 > 0
    \frac{x^{3}+5x-12-4(x-3)}{x-3} > 0
    \frac{x^{3}+5x-12-4x+12}{x-3} > 0
    \frac{x^{3}+x}{x-3} > 0
    \frac{x(x^{2}+1)}{x-3} > 0


    Can you finish the rest?

    Good luck!!
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  3. #3
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    Sorry I didn't make myself clear The method what you posted above is what works , but I understand why the method I posted earlier doesn't work either. I'm multiplying x-3 to the other side of the inequality but following it through makes me lose x=3 root. I don't see why

    Thanks
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  4. #4
    Super Member craig's Avatar
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    Quote Originally Posted by thomas49th View Post
    I'm multiplying x-3 to the other side of the inequality but following it through makes me lose x=3 root.
    Don't multiply through by x-3. This could be a minus number, and you should know that if you multiply an inequality by a minus you need to swap the sign around.

    The method to use is to either subtract like in the above example, or multiply through by (x-3)^2 as you know this is a positive number.
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