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Math Help - Quadratic equations

  1. #1
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    Quadratic equations

    I am working with some quadratic equations(from my notes) and I am having trouble understanding what is going on.

    For one I have....
    1.
    3y^2-6y-2=0

    then it goes to

    2.
    y=[6+√(36+24)]/6 and I am not sure what happened there. EDIT:woops forgot about the quadratic formula :P

    then

    3.
    y=6+√(4*15) ->I don't know what happened to the 6 that was on the bottom.


    4.
    y=(3+√15)/3 ->where did the 4 go? where did these 3's come from?

    5.
    y=1+[3/(3+√15)] ->still confused

    6.(3+√15+3)/(3+√15) ->I actually know what happened here

    7.
    y=[(6+√15)(3-√15)]/[(3+√15)(3-√15)] ->What??!?! No clue how we got from step 6 to 7.

    8.
    y=(18+3√15-6√15-15)/(9-15) ->I can see what happened here

    9.y=3-3√15 ->what happened to the 9-15 on the bottom


    ps. I am pretty sure I copied the notes down correctly but maybe I am wrong.
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    Quote Originally Posted by ehpoc View Post
    I am working with some quadratic equations(from my notes) and I am having trouble understanding what is going on.

    For one I have....
    1.
    3y^2-6y-2=0

    then it goes to

    2.
    y=[6+√(36+24)]/6 and I am not sure what happened there. EDIT:woops forgot about the quadratic formula :P
    Yep!
    Note that this is only one of the 2 solutions (the "+" one).


    Quote Originally Posted by ehpoc View Post
    then

    3.
    y=6+√(4*15) ->I don't know what happened to the 6 that was on the bottom.
    Yep, someone dropped the 6 by accident.


    Quote Originally Posted by ehpoc View Post
    4.
    y=(3+√15)/3 ->where did the 4 go? where did these 3's come from?
    The 4 was factored out of the square root.
    √(4*15) = √4 √15 = 2 √15

    At the same time both nominator and denomitor were divided by 2.
    As you can see the lost 6 in the denominator magically reappeared here.


    Quote Originally Posted by ehpoc View Post
    5.
    y=1+[3/(3+√15)] ->still confused
    I'm afraid his step is incorrect.
    So what comes after is also incorrect.
    y=3-3√15 is not a solution to the quadratic equation.


    Quote Originally Posted by ehpoc View Post
    6.(3+√15+3)/(3+√15) ->I actually know what happened here

    7.
    y=[(6+√15)(3-√15)]/[(3+√15)(3-√15)] ->What??!?! No clue how we got from step 6 to 7.

    8.
    y=(18+3√15-6√15-15)/(9-15) ->I can see what happened here

    9.y=3-3√15 ->what happened to the 9-15 on the bottom


    ps. I am pretty sure I copied the notes down correctly but maybe I am wrong.
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  3. #3
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    Re: Quadratic equations

    Here is what I did, tell me if this is still technically correct.

    I had

    3x^2-6x-2=0

    I divided by 3

    x^2-2x-2/3=0

    Then did the quadratic

    x=[2√(4+8/3)]/2

    x=√20/8
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    You appear to have left out part of the quadratic formula.

    The solution of
    ax^2+bx+c=0

    is
    x={-b \pm \sqrt{b^2 - 4ac} \over 2a}
    What did you do with (-b)?
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  5. #5
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    Re: Quadratic equations

    It is there....

    the 2?

    but the 2 on top and 2 on bottom cancel out right?
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  6. #6
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    Aha!
    In that case, what did you do with the \pm?
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  7. #7
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    Re: Quadratic equations

    Oh sorry....I am using this as applied to solving continued fractions, and thus in a continued fraction it is positive.

    But that aside does it look correct?
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  8. #8
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    There should be a plus (or a minus) between the 2 and the square root.
    Effectively this means that the 2 in the nominator does not simply cancel against the 2 in the denominator.

    Perhaps you can add a plus in between and redo you calculation?
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  9. #9
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    Re: Quadratic equations

    won't I end up with the exact same answer but with a plus/minus symbol infront?
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  10. #10
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    Huh?

    With a plus in between, it's an addition.
    Without anything in between, it's an implied multiplication.

    How would that be able to yield the same answer?
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  11. #11
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    Re: Quadratic equations

    woops ya never thought about that.

    so it would be in this case (because it is a continued fraction it is plus)

    x=(2+√20/8)/2

    Is this correct? Is the √(20/8) correct?
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  12. #12
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    Quote Originally Posted by ehpoc View Post
    woops ya never thought about that.

    so it would be in this case (because it is a continued fraction it is plus)

    x=(2+√20/8)/2

    Is this correct? Is the √(20/8) correct?
    First off, I don't understand your reference to a continued fraction.
    That is not what this is.

    But I'm afraid √(20/8) is not correct.
    You had √(4 + 8/3).
    This comes down to adding fractions:

    4 + {8 \over 3} = {4 \cdot 3 \over 3} + {8 \over 3} = {4 \cdot 3 + 8 \over 3}
    Last edited by ILikeSerena; December 17th 2011 at 11:32 AM.
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  13. #13
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    Re: Quadratic equations

    This quadratic equation is being using in solving for x in a continued fraction. If you don't know what that is it does not really matter. All it means is that the plus/minus changes to a plus.

    Anyways that was a typo

    I meant

    x=(2+√20/3)/2
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  14. #14
    Super Member ILikeSerena's Avatar
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    Re: Quadratic equations

    Yep! That is correct.
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  15. #15
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    Re: Quadratic equations

    Quote Originally Posted by ehpoc View Post
    I meant
    x=(2+√20/3)/2
    ...which is SAME as x = [3 + sqrt(15)] / 3
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