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Math Help - Please help

  1. #1
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    Please help

    Im not sure how to get to the answers here.

    \alpha+\beta=-4

    \alpha\beta=3

    \alpha=
    \beta=

    would be great to get some help. Thanks.
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  2. #2
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    Quote Originally Posted by davidman View Post
    Im not sure how to get to the answers here.

    \alpha+\beta=-4

    \alpha\beta=3

    \alpha=
    \beta=

    would be great to get some help. Thanks.
    From equation 1 we can see

    \alpha = -\beta - 4.


    Substituting into equation 2 gives

    (-\beta - 4)\beta = 3

    -\beta^2 - 4\beta = 3

    0 = \beta^2 + 4\beta + 3

    0 = (\beta + 3)(\beta + 1)

    \beta = -3 or \beta = -1.


    Now since \alpha = -\beta - 4

    Case 1:

    \alpha = -(-3) - 4

    \alpha = -1.


    Case 2:

    \alpha = -(-1) - 4

    \alpha = -3.



    So the solutions are \alpha = -1, \beta = -3 and \alpha = -3, \beta = -1.
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  3. #3
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    Quote Originally Posted by davidman View Post
    Im not sure how to get to the answers here.

    \alpha+\beta=-4

    \alpha\beta=3

    \alpha=
    \beta=

    would be great to get some help. Thanks.
    I assume that this is a system of simultaneous equations. If so:

    1. Calculate from the first equation and plug in this term into the 2nd one:

    2. \alpha(-\alpha-4)=3

    This is a quadratic equation in \alpha. Use the quadratic formula to solve this equation.
    Spoiler:
    \alpha = -3~\vee~ \alpha = -1 . Now calculate .


    EDIT: Too late ...as usual
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  4. #4
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    Quote Originally Posted by davidman View Post
    Im not sure how to get to the answers here.

    \alpha+\beta=-4

    \alpha\beta=3

    \alpha=
    \beta=

    would be great to get some help. Thanks.
    hello,
    solve the equation,
    x^2+4x+3
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  5. #5
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    thanks for the really quick input.

    I have a similar question where \alpha^2+\beta^2=32,\alpha\beta=-8

    and it completely throws me off that they're squared all of a sudden.

    Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
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  6. #6
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    Quote Originally Posted by Raoh View Post
    hello,
    solve the equation,
    x^2+4x+3
    oh I see, x^2+(\alpha+\beta)x+\alpha\beta

    thanks
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  7. #7
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    Quote Originally Posted by davidman View Post
    thanks for the really quick input.

    I have a similar question where \alpha^2+\beta^2=32,\alpha\beta=-8

    and it completely throws me off that they're squared all of a sudden.

    Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
    I think the fact that (\alpha+ \beta)^2= \alpha^2+ 2\alpha\beta+ \beta^2 will help a lot here!

    That will tell you what \alpha+ \beta is (actually, there are two answers to that) and then it becomes just like the previous problem.
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  8. #8
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    Quote Originally Posted by davidman View Post
    thanks for the really quick input.

    I have a similar question where \alpha^2+\beta^2=32,\alpha\beta=-8

    and it completely throws me off that they're squared all of a sudden.

    Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
    Square the second equation

    \alpha^2\beta^2 = 64.

    Now solve using a similar method to the first question.
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  9. #9
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    Quote Originally Posted by davidman View Post
    thanks for the really quick input.

    I have a similar question where \alpha^2+\beta^2=32,\alpha\beta=-8

    and it completely throws me off that they're squared all of a sudden.

    Do you solve this the same way by taking the square root of LHS and solving the simultaneous equation?
    (\alpha +\beta )^2-2\alpha \beta =32
    (\alpha +\beta )^2=16
    \alpha +\beta=4, \text{or}, \alpha +\beta=-4
    now,
    solve either x^2-4x-8 or x^2+4x-8.
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  10. #10
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    Quote Originally Posted by Raoh View Post
    solve either [...] or x^2+4x-8.
    ok, I used the quadratic equation on one and it basically turned out like this...

    x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, a=1, b=4, c=-8

    x_1=\frac{-4+\sqrt{16+32}}{2}

    =\frac{-4+4\sqrt{32}}{2}

    =-2+2\sqrt{32}

    Is this correct?

    hmmm... maybe more like

    =\frac{-4+\sqrt{48}}{2}

    =\frac{-4+\sqrt{4\times{12}}}{2}

    =-2+\sqrt{12}
    Last edited by davidman; January 21st 2010 at 05:26 AM.
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  11. #11
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    Quote Originally Posted by davidman View Post
    ok, I used the quadratic equation on one and it basically turned out like this...

    x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, a=1, b=4, c=-8

    x_1=\frac{-4+\sqrt{16+32}}{2}

    =\frac{-4+4\sqrt{32}}{2}

    =-2+2\sqrt{32}

    Is this correct?
    check your work again,
    x_1=-2(1+\sqrt{3})
    and x_2=2(\sqrt{3}-1).
    P.S : \sqrt{16+32}=4\sqrt{3}.
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  12. #12
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    omg, THANK YOU

    I get it now!

    \sqrt{16+32}=\sqrt{48}=\sqrt{4\times4\times3}
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