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Math Help - Quadratics further solving

  1. #1
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    Quadratics further solving

    Hi Guys

    Im stuck on the following question

    3x^2+16x-12=0 hence solve 3(y-(7/y))^(2)+16(y-(7/y))-12=0

    I can get the factors (3x-2) and (x+6) for the first one and so x=2/3 x= -6

    its the second part im stuck on

    any help would be great please

    Thanks
    Pete
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  2. #2
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    Substitue

    Follow above:
    In the second equation, let x=y-(7/y)
    so u get the first equation by substitution of x
    ..which u have solved already..
    so
    y-(7/y)=2/3
    and y-(7/y)= -6
    two more quadratics for you to solve!
    (multiply throughout by y)
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  3. #3
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    im still confused about the 2nd part
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  4. #4
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    Quote Originally Posted by zeeshahmad View Post
    y-(7/y)=2/3
    and y-(7/y)= -6
    you have been given the answer here, have a go at solving these equations.
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  5. #5
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    so the bottom one would be y^(2)-7y=-6y? and then y^(2)-7y+6y=0? and then y^(2)-y=0?

    not sure about the 2/3 part though
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  6. #6
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    Multiply y through both sides you get

    y^2-\frac{2y}{3}-7

    and

    y^2+6y-7
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  7. #7
    Junior Member RHandford's Avatar
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    I think the point is that y-(7/y) is the same as x in the first equation, therefore the results are the same - you have solved the first one, so by default you haver also solved the second
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  8. #8
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    hmmm ok thanks for that

    where did i go wrong on the one above (post 5)
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  9. #9
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    so if y-(7/y) = x do i still have to solve that part of equation? or just go into 3(2/3)^(2)+16(-6)-12=0?
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  10. #10
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    Clarity

    Ok... now lets try to make things clear:
    first of all there were two equations, first one which u already solved and found two values of x.

    The second equation ie the one that contains y, requires a bit of common sense.

    If you write both equations on paper in maths form, you will notice that the second one
    just has y-(7/y) instead of x.

    So... your first step will be to solve the second equation for y-(7/y).
    Look, if you change x for z in the first equation, doesn't matter, does it?

    So similarly u r just using y-(7/y) instead of x. But you already did that in the first equation.
    And thats where you get the two new equations
    y-(7/y)=2/3
    and y-(7/y)= -6

    But your objective of the second equation is NOT to find y-(7/y) which is x. It is to find y. It is these two new equations that u now need to solve, forget the original second equation, because these two equations are parts of the second one.
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  11. #11
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    ah now i see

    ahh thanks zeeshahmad now i kinda feel a little silly but at least i think i understand now.

    your a star, ill post back what i get when i work those equations out.

    Thanks again

    Pete
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  12. #12
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    so now i get y-(7/y) = -6, so y^(2) -7 = -6y, so y^(2)+6y-7=0, so (y+7)(y-1) =0 so y=-7 y=1

    y-(7/y)= 2/3, y^(2)-7=2y/3 so multiply out by 3 gives 3y^(2)-21=2y, so 3y^(2)-2y-21=0 so (3y+7) (y-3)=0 so y = -(7/3) y=3?
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  13. #13
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    Exactly

    Yup
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  14. #14
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    your my hero

    Thanks for your help
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