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Math Help - How to Solve this type of problem for practice SAT?

  1. #1
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    How to Solve this type of problem for practice SAT?

    Hello members,

    I am working on my SAT practice and i came across this problem.

    x^-2=16

    I dont remember learning this. Any help would be appreciated. Thanks.

    PS. That is supposed to be X to the -2 power.
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  2. #2
    A Plied Mathematician
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    This equation is the same as

    \dfrac{1}{x^{2}}=16.

    You can flip both sides to obtain

    x^{2}=\dfrac{1}{16}.

    Does that give you any ideas?
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  3. #3
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    Yes, Thank you for the help!
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  4. #4
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    \displaystyle \frac{1}{x^{2}}=16 \implies 1 = 16x^2 \implies \dfrac{1}{16}=x^2 \implies x = \sqrt{\frac{1}{16}}
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  5. #5
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    Quote Originally Posted by pickslides View Post
    \displaystyle \frac{1}{x^{2}}=16 \implies 1 = 16x^2 \implies \dfrac{1}{16}=x^2 \implies x = \sqrt{\frac{1}{16}}
    \displaystyle x = \mathbf{\pm}\sqrt{\frac{1}{16}}
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  6. #6
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    Here is a quick way to get the answer on the SAT:

    x^{-2}=16

    Simply raise both sides to the reciprocal power:

    (x^{-2})^{-\frac{1}{2}}=16^{-\frac{1}{2}}=.25

    So grid in .25


    Notes:

    (1) To solve this you only need to type the following into your calculator: 16^{-.5}

    (2) If this were a multiple choice question you could simply plug the answer choices in for x in the original question. You should almost always start with choice (C) when using this strategy. (Of course the solution I have given is better, but it requires some mathematical knowledge, whereas this one does not.)

    (3) Note that this problem is incomplete as you have stated it since -.25 is also a solution. So either you haven't written the question precisely or this isn't an actual SAT practice problem. In any case, the solution to a grid-in can't be negative.

    (4) The above posters have given a better solution to this problem, but on the SAT you don't generally want the most complete solution. You want to get the answer as quickly and efficiently as possible.
    Last edited by DrSteve; January 5th 2011 at 04:11 AM.
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