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Math Help - strange quadratic

  1. #1
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    strange quadratic

    a)101.22= 6/1+y + 106/(1+y)^2

    b) 101.22= 112+6y/(1+y)^2

    c) 101.22y^2 +202.44y+101.22= 112+6y


    and i was wondering how they got there from a to c
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  2. #2
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    Quote Originally Posted by matlondon View Post
    a)101.22= 6/1+y + 106/(1+y)^2

    b) 101.22= 112+6y/(1+y)^2

    c) 101.22y^2 +202.44y+101.22= 112+6y


    and i was wondering how they got there from a to c
    Could you put the appropriate parenthesis in here? I can make out what b) is supposed to be, but only because I see how to go from b) to c). (Multiply both sides of the equation by (1 + y)^2 and expand.) But I can't make any sense out of how to go from a) to b). There are too many ambiguities in a).

    b) 101.22= (112+6y)/(1+y)^2

    c) 101.22y^2 +202.44y+101.22= 112+6y

    -Dan

    PS Oh I get a) now. It's supposed to be
    a) 101.22= 6/(1+y) + 106/(1+y)^2

    Add the two fractions:
    \displaystyle  \frac{6}{1+y} + \frac{106}{(1+y)^2}

    \displaystyle = \frac{6(1 + y)}{(1 + y)^2} + \frac{106}{(1+y)^2}

    \displaystyle \frac{6(1 + y) + 106}{(1 + y)^2}

    etc.
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  3. #3
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    e^(i*pi)'s Avatar
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    I suspect a) is meant to read 101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2}
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by matlondon View Post
    a)101.22= 6/1+y + 106/(1+y)^2

    b) 101.22= 112+6y/(1+y)^2

    c) 101.22y^2 +202.44y+101.22= 112+6y


    and i was wondering how they got there from a to c
    Assuming e^(i*pi) is correct, to get your answer simply multiply through by (1+y)^2, expand, and then you'll have something more familiar. (If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)
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  5. #5
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    101.22 = \dfrac{6}{1+y} + \dfrac{6}{(1+y)^2} + \dfrac{106}{(1+y)^3}
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  6. #6
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    Quote Originally Posted by Swlabr View Post
    Assuming e^(i*pi) is correct, to get your answer simply multiply through by (1+y)^2, expand, and then you'll have something more familiar. (If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)
    how would i do that "(If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)"
    or if you can't show me could you direct me to a site that does.
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  7. #7
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    You need brackets on your first term on the RHS. At the moment it reads 6 + y whereas I suspect you mean \dfrac{6}{1+y}

    It's the same principle as Swlabr stated in post 4. However, this time, you need to multiply through by (1+y)^3 and then expand


    how would i do that "(If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)"
    or if you can't show me could you direct me to a site that does.
    u = 1+y \implies y = u-1

    101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2} \implies 101.22 = \dfrac{6}{u} + \dfrac{106}{u^2}

    when you multiply by u^2 it gives 101.22u^2 = 6u + 106 \implies 101.22u^2-6u-106 = 0

    Solve for u (I suggest the quadratic formula) and then back substitute to find y. (Remember that -1 is not in the domain)

    Personally I think it's easier than expanding!
    Last edited by e^(i*pi); February 18th 2011 at 04:25 AM. Reason: expanding on post
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  8. #8
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    Quote Originally Posted by matlondon View Post
    thanks guys so what if it was 101.22= 6/1+y + 6/(1+y)^2+106/(1+y)^3

    101.22=\displaystyle\frac{6}{1+y}+\frac{6}{(1+y)^2  }+\frac{106}{(1+y)^3}

    \displaystyle\ 101.22=\frac{6(1+y)^2}{(1+y)^3}+\frac{6(1+y)}{(1+y  )^3}+\frac{106}{(1+y)^3}

    Now we are adding the same fractions

    \displaystyle\ 101.22=\frac{6(1+y)^2+6(1+y)+106}{(1+y)^3}

    101.22(1+y)^3=6(1+y)^2+6(1+y)+106

    and multiply out...
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  9. #9
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    Quote Originally Posted by e^(i*pi) View Post
    You need brackets on your first term on the RHS. At the moment it reads 6 + y whereas I suspect you mean \dfrac{6}{1+y}

    It's the same principle as Swlabr stated in post 4. However, this time, you need to multiply through by (1+y)^3 and then expand




    u = 1+y \implies y = u-1

    101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2} \implies 101.22 = \dfrac{6}{u} + \dfrac{106}{u^2}

    when you multiply by u^2 it gives 101.22u^2 = 6u + 106 \implies 101.22u^2-6u-106 = 0

    Solve for u (I suggest the quadratic formula) and then back substitute to find y. (Remember that -1 is not in the domain)

    Personally I think it's easier than expanding!
    U=1+y/2
    [MATH\]dfrac{6}{u} + dfrac{6}{u^2}\+dfrac{6}{u^3}dfrac{106}{u^4}[/tex]
    how would i do that , is there any sites that can help me with it
    Last edited by matlondon; February 18th 2011 at 04:54 AM. Reason: messed up
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  10. #10
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    Or, let x= \frac{1}{y+ 1} so that [tex]101.22= \frac{6}{1+y}+ \frac{106}{(1+y)^2}[\math] becomes
    101.22= 6x+ 106x^2

    Solve that for x, then y= \frac{1}{x}- 1

    For [tex]101.22= \frac{6}{1+y}+ \frac{6}{(1+y)^2}+ \frac{106}{(1+ y)^3}[tex]
    the same substitution give
    101.22= 6x+ 6x^2+ 106x^3
    Of course, cubics are considerably harder to solve than quadratics.
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