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  1. #1
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    Please help

    Im not sure how to get to the answers here.

    $\displaystyle \alpha+\beta=-4$

    $\displaystyle \alpha\beta=3$

    $\displaystyle \alpha=$
    $\displaystyle \beta=$

    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.

    $\displaystyle \alpha+\beta=-4$

    $\displaystyle \alpha\beta=3$

    $\displaystyle \alpha=$
    $\displaystyle \beta=$

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

    $\displaystyle \alpha = -\beta - 4$.


    Substituting into equation 2 gives

    $\displaystyle (-\beta - 4)\beta = 3$

    $\displaystyle -\beta^2 - 4\beta = 3$

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

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

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


    Now since $\displaystyle \alpha = -\beta - 4$

    Case 1:

    $\displaystyle \alpha = -(-3) - 4$

    $\displaystyle \alpha = -1$.


    Case 2:

    $\displaystyle \alpha = -(-1) - 4$

    $\displaystyle \alpha = -3$.



    So the solutions are $\displaystyle \alpha = -1, \beta = -3$ and $\displaystyle \alpha = -3, \beta = -1$.
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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.

    $\displaystyle \alpha+\beta=-4$

    $\displaystyle \alpha\beta=3$

    $\displaystyle \alpha=$
    $\displaystyle \beta=$

    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. $\displaystyle \alpha(-\alpha-4)=3$

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

    $\displaystyle \alpha+\beta=-4$

    $\displaystyle \alpha\beta=3$

    $\displaystyle \alpha=$
    $\displaystyle \beta=$

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

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

    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,
    $\displaystyle x^2+4x+3$
    oh I see, $\displaystyle x^2+(\alpha+\beta)x+\alpha\beta$

    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 $\displaystyle \alpha^2+\beta^2=32,\alpha\beta=-8$

    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 $\displaystyle (\alpha+ \beta)^2= \alpha^2+ 2\alpha\beta+ \beta^2$ will help a lot here!

    That will tell you what $\displaystyle \alpha+ \beta$ 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 $\displaystyle \alpha^2+\beta^2=32,\alpha\beta=-8$

    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

    $\displaystyle \alpha^2\beta^2 = 64$.

    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 $\displaystyle \alpha^2+\beta^2=32,\alpha\beta=-8$

    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?
    $\displaystyle (\alpha +\beta )^2-2\alpha \beta =32$
    $\displaystyle (\alpha +\beta )^2=16$
    $\displaystyle \alpha +\beta=4, \text{or}, \alpha +\beta=-4$
    now,
    solve either $\displaystyle x^2-4x-8$ or $\displaystyle x^2+4x-8$.
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  10. #10
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    Quote Originally Posted by Raoh View Post
    solve either [...] or $\displaystyle x^2+4x-8$.
    ok, I used the quadratic equation on one and it basically turned out like this...

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

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

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

    $\displaystyle =-2+2\sqrt{32}$

    Is this correct?

    hmmm... maybe more like

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

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

    $\displaystyle =-2+\sqrt{12}$
    Last edited by davidman; Jan 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...

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

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

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

    $\displaystyle =-2+2\sqrt{32}$

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

    I get it now!

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