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Math Help - quadratic

  1. #1
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    quadratic

    trying to find the possible values of x to 2DP, ok so far i have done this

    M=wx(1-x)
    2
    original question but with the values i had given put in

    70=1.5x(20-x)
    2

    70X2=1.5x(20-x)

    140=1.5x(20-x)

    140=x(20-x)
    1.5

    93.3=x(20-x)

    93.3=20x-x2 (squared)

    93.3=x-x2
    20

    4.665=x-x2


    is this right, and where do i go to next?
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  2. #2
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    hope this helps im sure some one else would be better at this than me.

    you started off right but as your doing a quadratic you need to get your formula into a form that looks like
    0 = ax^2+bx+c

    so multiplying by 2 worked but then you needed to expand the bracket on the right hand side and rearrange everything to one side so you have 0 =

    once thats done you just use the quadratic formula giving you two results for x
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  3. #3
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    still dont really get it
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  4. #4
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    Quote Originally Posted by hoikey View Post
    trying to find the possible values of x to 2DP, ok so far i have done this

    M=wx(1-x)
    2
    original question but with the values i had given put in

    70=1.5x(20-x)
    2

    70X2=1.5x(20-x)

    140=1.5x(20-x)

    140=x(20-x)
    1.5

    93.3=x(20-x)

    93.3=20x-x2 (squared)

    93.3=x-x2 (<- Can't do that. If you divide by 20 both side, you would have gotten \frac{93.3}{20}=x-\frac{1}{20}x^2 instead)
    20

    4.665=x-x2


    is this right, and where do i go to next?
    There are many ways to find the approximated "x". These are some that came up in my mind,

    You can do it algebraically (as p3achy has mentioned):
    1). Get the equation to the form: y=ax^2+bx+c and solve it with the quadratic formula.
    2). Or, Completing the square.

    Or graphically:
    Graph the two lines/curves: y_1 = 20-x^2 and y_2=93.3, then find the intersection.
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  5. #5
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    knew there would be a few different ways to do it really depends which way your most comfortable with.
    Last edited by p3achy; January 4th 2009 at 02:03 PM. Reason: better reply:-)
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  6. #6
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    no, thats why it all went wrong haha
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  7. #7
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    have you been asked to use the quadratic formula to solve it?
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  8. #8
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    Quote Originally Posted by hoikey View Post
    still confused as hell. could someone do it and explain it then i will be able to see what your on about lol. havent done this in ages so kinda struggling with it
    Here is one way using the quadratic formula:
    70=\frac{1.5x(20-x)}{2}
    ...
    140=1.5x(20-x)
    140=30x-1.5x^2
    1.5x^2-30x+140=0 (In the form: ax^2+bx+c=0)
    Then a=1.5, b=-30, c=140, Then using the quadratic formula:
    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},
    we have
    x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1.5)(140)}}{2(1.5)}.
    Thus x=\frac{30\pm \sqrt{60}}{3}
    So x \approx 12.58 or x \approx 7.42
    Last edited by chabmgph; January 4th 2009 at 02:15 PM. Reason: typo
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  9. #9
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    if ya want to use the quadratic formula i can run you throught putting it in the right format..
    so you were at
    140=1.5x(20-x)
    so going dead simple you do 1.5x times 20 and then 1.5 x times -x
    giving you
    140=30x - 1.5x^2
    any easier yet?

    darn wasnt gonna give it away
    Last edited by p3achy; January 4th 2009 at 02:16 PM. Reason: beaten to it
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  10. #10
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    may sound stupid but no. how would i get the x2 to just a normal x?
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  11. #11
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    Quote Originally Posted by p3achy View Post
    if ya want to use the quadratic formula i can run you throught putting it in the right format..
    so you were at
    140=1.5x(20-x)
    so going dead simple you do 1.5x times 20 and then 1.5 x times -x
    giving you
    140=30x - 1.5x^2
    any easier yet?

    darn wasnt gonna give it away
    oops. Sorry.
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  12. #12
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    Quote Originally Posted by chabmgph View Post
    Here is one way using the quadratic formula:
    70=\frac{1.5x(20-x)}{2}
    ...
    140=1.5x(20-x)
    140=30x-1.5x^2
    1.5x^2-30x+140=0 (In the form: ax^2+bx+c=0)
    Then a=1.5, b=-30, c=140, Then using the quadratic formula:
    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a},
    we have
    x=\frac{-(-30)\pm \sqrt{(-30)^2-4(1.5)(140)}}{2(1.5)}.
    Thus x=\frac{30\pm \sqrt{60}}{3}
    So x \approx 12.58 or x \approx 7.42
    oh yeah, my tutor said something about that quadratic formula thing, i just couldnt see the different parts of it. might be cause im trying to complete a whole assignment in the last minute and im proper stressed. thanks mate
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  13. #13
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    oops. Sorry.
    lol only the blonde in me that was making them work for it
    nothing like leaving assignments to the last minute to increase motivation
    any other help needed im sure there is someone with bigger brains than me that can help
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  14. #14
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    Quote Originally Posted by p3achy View Post
    lol only the blonde in me that was making them work for it
    nothing like leaving assignments to the last minute to increase motivation
    any other help needed im sure there is someone with bigger brains than me that can help
    its just that over the time ive had to do it ive lost a mate in a car accident so i havent really felt too motivated to do maths lol
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