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Math Help - Some more problems..(solving for x)

  1. #1
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    Some more problems..(solving for x)

    I don't know why I just can't get these stupid problems..it seems that every problem has something different about it and I just don't get what I'm doing wrong...


    Okay, first one..I worked it out all the way but it just looks really wrong and none of my answers work when I plug them in, so I don't know what I did wrong:

    8(m-4)^4 - 10(m-4)^2 + 3 = 0

    I used substitution..and made y=(m-4)^2 and y^2 = (m-4)^4

    8y^2 - 10y + 3 = 0
    8y^2 - 4y - 6y + 3 = 0
    4y(2y - 1) - 3 (2y-1) = 0
    4 y - 3 = 0
    y = 3/4 and 2y - 1 = 0 ---> y = 1/2

    Then I plugged the y's in

    y = (m-4)^2
    1/2 = (m-4)^2
    1/sqroot(2) = m - 4
    m = 4 + 1/sqroot(2)
    simplified to: 4 + or - 1/sqrt(2)

    and

    3/4 = (m-4)^2
    sqrt(3)/4 = m-4
    m = sqrt(3)/2 + 4
    m = 4sqrt(3) + 3

    Yeah, it looks confusing =/ but the point is that neither of my answers worked out..Can anyone point out what I did wrong? If it's not too hard to read?

    2. 4(x+1)^1/2 - 5(x+1)^2 + (x+1)^5/2 = 0

    I used substitution again

    x = (x+1)^1/2
    x^4 = (x+1)^2
    x^5 = (x+1)^5

    4x - 5x^4 + x^5 = 0
    x(4-5x^3 + x^4) = 0

    and once I get here I'm pretty lost as to what to do next =/

    3. (x^2 - 9) / (x^2 - 2x - 3) = 3/2

    So I assume I should factor everything first:

    (x-3)(x+3) / (x-3)(x+1) = 3/2
    (x+3) / (x+1) = 3/2
    then..I guess I multiply by...the denominators..to ..get rid of them?

    and got

    2(x+3) = 3(x+1)
    2x+6 = 3x+3
    x=3
    but when I plug it in it doesn't work...and everything it doesn't work I just feel like I did something wrong

    Sorry for the long post..appreciate any help I can get
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  2. #2
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    8(m-4)^4 - 10(m-4)^2 + 3 = 0
    => 8(m-4)^4 - 6(m-4)^2 - 4(m-4)^2 + 3 = 0
    =>(4(m-4)^2 -3)(2(m-4)^2-1) = 0
    =>((m-4)^2 -3/4)((m-4)^2-1/2) = 0
    =>(m-4-sqrt(3)/2)(m-4+sqrt(3)/2)(m-4-1/sqrt(2))(m-4+1/sqrt(2))=0
    => m = 4+sqrt(3)/2, 4-sqrt(3)/2, 4+1/sqrt(2), 4-1/sqrt(2)
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  3. #3
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    (x^2 - 9) / (x^2 - 2x - 3) = 3/2
    => 2(x-3)(x+3) = 3(x-3)(x+1)
    => 2(x-3)(x+3) - 3(x-3)(x+1) = 0
    => (x-3)(2x + 6 - 3x -3) = 0
    => (x-3)(-x + 3) = 0
    => (x-3)^2 = 0 ;
    => x = 3
    But this can not be the solution to the equation as the function (x^2 - 9) / (x^2 - 2x - 3) is not defined at x = 3.


    Therefor no solution.
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  4. #4
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    Thank you! I guess I had done it correctly

    Can anyone help me out with the second one?

    2. 4(x+1)^1/2 - 5(x+1)^2 + (x+1)^5/2 = 0

    I used substitution again

    x = (x+1)^1/2
    x^4 = (x+1)^2
    x^5 = (x+1)^5

    4x - 5x^4 + x^5 = 0
    x(4-5x^3 + x^4) = 0

    and once I get here I'm pretty lost as to what to do next =
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