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Math Help - re-arrange

  1. #1
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    re-arrange

    this is an common algebra error I make. Just want confirm which is the correct answer.




    when using the sqrt do I reduce the power on the x also?
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  2. #2
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    In the Real Numbers, you follow something like this stream of logic.

    Right-Hand Side
    x^4 is positive
    8 is positive
    1 is positive
    Negative sign - The right hand side is negative. It cannot be positive. It cannot be zero.

    Left-Hand Side
    y^2 is positive

    How many positive numbers can you name that are equal to a negative number?

    Note: When did 2 get to be the square root of 8? Cube root, perhaps.
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  3. #3
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    It should be obvious that since x can not be 0 and x^4 is always nonnegative, that -1/(8x^4) is always negative. You can not square any number to get a negative number. So you can not solve for y.
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  4. #4
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    Remember that the square root of negative one is imaginary.
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