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Math Help - How close?

  1. #1
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    How close?

    Hello,

    Can you please help me figure this out?

    Problem: How close to -3 do we have to take x so that

    1/(x+3)^4 > 10,000
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  2. #2
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    (x+3)^4 is non-negative, so we can just move some stuff around:


    \frac{1}{(x+3)^4} > 10,000

    \implies~ \frac{1}{10,000} > (x+3)^4

    \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|

    \implies~ \frac{1}{10} > |x+3|
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  3. #3
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    Quote Originally Posted by drumist View Post
    (x+3)^4 is non-negative, so we can just move some stuff around:


    \frac{1}{(x+3)^4} > 10,000

    \implies~ \frac{1}{10,000} > (x+3)^4

    \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|

    \implies~ \frac{1}{10} > |x+3|
    Great thanks!

    I just have a couple of rookie questions.

    If it was negative, you couldn't move stuff around?

    and why did the (x+3) become an absolute value?
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  4. #4
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    Quote Originally Posted by l flipboi l View Post
    Great thanks!

    I just have a couple of rookie questions.

    If it was negative, you couldn't move stuff around?

    and why did the (x+3) become an absolute value?
    Say for example we considered something like \frac{1}{(x+3)^3} > 1000
    . It's possible for (x+3)^3 to be a negative number, but if it was negative, the value on the left-hand side would have been negative, and a negative number can't be greater than 1000. So we just need to make a side note that we are only considering values of x such that x>-3. From then we can continue the problem.

    As for your second question: Any time you take an even-powered root of a value, you must take its absolute value. In other words:

    a^2 = 25 ~\implies~ \sqrt{a^2}=\sqrt{25} ~\implies~ |a|=5 ~\implies~ a=\pm5

    Notice that had we not done so, we would not have found the solution a=-5, but clearly (-5)^2=25.

    You do not need to do this when taking odd-powered roots.
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