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Math Help - how do u find the value of x for these:

  1. #1
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    how do u find the value of x for these:

    i know these are easy questions, but i am just a bit confused:


    please indicate which question you are answering:

    a) y=6x^5 + 192

    b) 48-3x^2

    c) 6x^-1 + 5

    d) y= (1-x^2)(x^-1)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by juiicycouture View Post
    i know these are easy questions, but i am just a bit confused:


    please indicate which question you are answering:

    a) y=6x^5 + 192

    b) 48-3x^2

    c) 6x^-1 + 5

    d) y= (1-x^2)(x^-1)
    you mean the x-values as in the roots right?

    a) y=6x^5 + 192
    for the roots


    6x^5 + 192 = 0
    => 6x^5 = -192
    => x^5 = -192/6 = -32
    => x = -32^(1/5) = -2


    b) 48-3x^2
    for roots:

    48-3x^2 = 0
    => 3x^2 = 48
    => x^2 = 48/3 = 16
    => x = +/- sqrt(16) = +/- 4

    c) 6x^-1 + 5
    for roots:

    6x^-1 + 5 = 0
    => 6x^-1 = -5
    => x^-1 = -5/6
    => x = -6/5 ............i just took the inverse

    d) y= (1-x^2)(x^-1)
    for roots:

    (1-x^2)(x^-1) = 0

    => (1 - x^2) = 0 or (x^-1)=0

    since x^-1 is never 0, we must have

    (1 - x^2) = 0 ...........this is the difference of 2 squares
    => (1 + x)(1 - x) = 0
    => x = -1 or x = 1
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