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Math Help - (Algebraic Inequalities) Can somebody show me the first step to these 3 equations?

  1. #1
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    Wink (Algebraic Inequalities) Can somebody show me the first step to these 3 equations?

    Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it).

    1.

    {{(x-1)(x-2)(x-3)}\over{(x+1)(x+2)(x+3)}}>1

    2.

    a^4+a^3-a-1<0

    3.

    x^6-9x^3+8>0

    Thank you sooo much!

    Nate
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  2. #2
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    Quote Originally Posted by nathan02079 View Post
    Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it).

    1.

    {{(x-1)(x-2)(x-3)}\over{(x+1)(x+2)(x+3)}}>1
    If (x+1)(x+2)(x+3)>0 multiplying through by it gives:

    (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)

    which you expand and simplify and draw conclusions.

    But if (x+1)(x+2)(x+3)<0 we get instead:

    (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)

    which you expand and simplify and draw conclusions.

    .
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  3. #3
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    Quote Originally Posted by nathan02079 View Post
    Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it).

    3.

    x^6-9x^3+8>0

    x^6-9x^3+8 is a quadratic in y=x^3

    .
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  4. #4
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    Quote Originally Posted by nathan02079 View Post
    Hello, I'm absolutely stuck. :S My teacher gave me these 'challenge' problems and I have no idea how to get started. Can somebody explain how to start each equation? (Not solving it would be okay. I would love to get the answer...just don't know how to start it).

    2.

    a^4+a^3-a-1<0
    Does the left hand side have any obvious roots or factors?

    .
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  5. #5
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    Quote Originally Posted by Constatine11 View Post
    If (x+1)(x+2)(x+3)>0 multiplying through by it gives:

    (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)

    which you expand and simplify and draw conclusions.

    But if (x+1)(x+2)(x+3)<0 we get instead:

    (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)

    which you expand and simplify and draw conclusions.

    .
    (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)
    X^3-3x^2+2x-3x^2+9x-6>x^3+3x^2+2x+3x^2+9x+6
    -12x^2-12>0
    -12(x+1)(x-1)>0

    and


    (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)
    X^3-3x^2+2x-3x^2+9x-6<x^3+3x^2+2x+3x^2+9x+6
    -12x^2-12<0
    -12(x+1)(x-1)<0

    I have yet to get to the answer

    Quote Originally Posted by Constatine11 View Post
    x^6-9x^3+8 is a quadratic in y=x^3

    .
    Sorry, what does this mean exactly?

    Quote Originally Posted by Constatine11 View Post
    Does the left hand side have any obvious roots or factors?

    .
    a(a^3+a^2-1)-1<0

    Is it?
    Last edited by nathan02079; January 28th 2009 at 10:58 PM.
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  6. #6
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    Quote Originally Posted by nathan02079 View Post
    (x-1)(x-2)(x-3)>(x+1)(x+2)(x+3)
    X^3-3x^2+2x-3x^2+9x-6>x^3+3x^2+2x+3x^2+9x+6
    -12x^2-12>0
    -12(x+1)(x-1)>0
    -12x^2-12>0 \implies x^2<-1

    which is impossible for real x

    (x-1)(x-2)(x-3)<(x+1)(x+2)(x+3)
    X^3-3x^2+2x-3x^2+9x-6<x^3+3x^2+2x+3x^2+9x+6
    -12x^2-12<0
    -12(x+1)(x-1)<0
    -12x^2-12<0 \implies x^2>-1

    which is always true for real x. So we conclude that:

    (x+1)(x+2)(x+3)<0

    or that: x \in (-\infty,-3) \cup (-2,-1)

    .
    Last edited by Constatine11; January 28th 2009 at 11:32 PM.
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  7. #7
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    Quote Originally Posted by nathan02079
    Quote Originally Posted by Constatine11 View Post
    x^6-9x^3+8 is a quadratic in y=x^3

    .
    Sorry, what does this mean exactly?
    It means that if you make the substitution y=x^3 you get the quadratic:

     <br />
y^2-9y+8<br />

    and so:

    y^2-9y+8>0

    is satisfied on (-\infty, y_1) \cup (y_2, \infty) , where y_1 \le y_2 are the roots of y^2-9y+8=0, which now needs to be translated into something to do with x

    .
    Last edited by Constatine11; January 28th 2009 at 11:31 PM. Reason: Fixed the quote tags
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  8. #8
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    Quote Originally Posted by nathan02079 View Post
    a(a^3+a^2-1)-1<0

    Is it?


    Consider:

    a^4+a^3-a-1=0

    The rational root theorem suggests that if this has rational roots then they are either a=1 or a=-1, test these. If one or other of these are roots then (a-1) and/or (a+1) are factors of a^4+a^3-a-1, so take out the factors and see what that transforms:

    a^4+a^3-a-1<0

    into, and draw any conclusions.

    (what this should do, if you get it right, is lead you to find all the real roots of a^4+a^3-a-1 then the inequality will be satisfied between appropriate consecutive pairs of these roots)

    .
    Last edited by Constatine11; January 28th 2009 at 11:41 PM.
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