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Math Help - Show this inequality

  1. #1
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    Show this inequality

    Show that the expression

    (6+y)(4-y)(y+4)(y-2)-65<0

    for all values of y are real numbers .

    I tried expanding the whole thing .. but no idea how to show it ..

    (-y^2-2y+24)(y^2+2y-8)-65<0
    -y^4-2y^3+8y^2-2y^3-4y^2+16y+24y^2+48y-192-65<0
    -y^4-4y^3+28y^2+64y-257<0

    Wonder what i can from here ..
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  2. #2
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    if you plot the expanded function on a graph and show that it is always less than zero then that should work. is this any help??
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  3. #3
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    Re :

    Quote Originally Posted by hmmmm View Post
    if you plot the expanded function on a graph and show that it is always less than zero then that should work. is this any help??

    THanks for your reply but besides plotting it on a graph to show , is that any other method such as showing it in the equation itself ?
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  4. #4
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    not that i can see apart from differentiating to show maximum value that is less than 257 sorry
    Last edited by hmmmm; January 11th 2009 at 07:48 AM.
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  5. #5
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    Quote Originally Posted by mathaddict View Post
    Show that the expression

    (6+y)(4-y)(y+4)(y-2)-65<0

    for all values of y are real numbers .
    What you want to show is that

    (1)\;\;\;-65 < (y+6)(y+4)(y-2)(y-4)

    You can actually show that

    -64 < (y+6)(y+4)(y-2)(y-4)

    or

    0 < (y+6)(y+4)(y-2)(y-4) +64

    Factoring the RHS gives

    0 < (y^2 + 2y - 16)^2

    which is clearly true which makes (1) true.
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  6. #6
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    Re :

    Thanks Danny .

    By the way , how did you factorise 0 < (y+6)(y+4)(y-2)(y-4) +65 to 0 < (y^2 + 2y - 16)^2 .
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  7. #7
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    Quote Originally Posted by mathaddict View Post
    Thanks Danny .

    By the way , how did you factorise 0 < (y+6)(y+4)(y-2)(y-4) +65 to 0 < (y^2 + 2y - 16)^2 .
    When I looked at the factors

    (y+6)(y+4)(y-2)(y-4) I saw (y+6)(y-4)(y+4)(y-2)

    and grouping them in pairs I saw a difference of 2 in y

    (y^2+2y-24)(y^2+2y-8)

    then

    (y^2+2y-16-8)(y^2+2y-16+8)

    If we let u = y^2 +2y -16 then (u-8)(u+8) = u^2-64

    so (u-8)(u+8)+64 = u^2-64+64 = u^2 > 0 giving

    (y^2 +2y -16) > 0.

    Hope it helps.
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