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Equation question

  1. Jan 16, 2005 #1
    hi ,

    I wanted to find out all the possible ways to solve :

    (x^2 + 4x) / 3 + 84 / (x^2 +4x) = 11


    Please could you show me in each case the method of working out .


    Thanks a lot.


    Roger
     
  2. jcsd
  3. Jan 16, 2005 #2

    arildno

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    There are infinite ways of "solving" an equation.
    For example, you may always insert a "solving step" by adding a-a=0 to one side of an equation (where a is some number).

    Hence, we are more interested in finding EFFICIENT ways of solving an equation, rather than POSSIBLE ways.

    In your case, it seems the simplest way of solving this, is to introduce the variable:
    [tex]y=x^{2}+4x[/tex]
    Multiplying your original equation by "y", and rearranging, gives:
    [tex]\frac{1}{3}y^{2}-11y+84=0[/tex]

    Use this in your further calculations.
     
  4. Jan 16, 2005 #3
    Might be me being stupid but does this equation mean [tex]\frac{(x^2 + 4x)}{\frac{3 + 84}{(x^2 +4x)}} = 11[/tex] ???

    The Bob (2004 ©)
     
  5. Jan 16, 2005 #4
    That's what I thought.
     
  6. Jan 16, 2005 #5

    arildno

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    roger's notation is quite consistent with how you usually plug in formulas in a computer program, in which case the expression is to be read as:
    [tex]\frac{x^{2}+4x}{3}+\frac{84}{x^{2}+4x}}=11[/tex]
     
  7. Jan 16, 2005 #6
    Thanks, but is there any other method without substituting for y ?

    I'm assuming substituting for y, is what you meant by...'' introduce the variable '' ?



    Roger
     
  8. Jan 16, 2005 #7

    dextercioby

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    Yes.That's what he meant.I hope u solved the eq.One more thing though,please learn how to edit formulas with Tex.People might get confused due to your inability of getting them clear with what u want...

    Daniel.
     
  9. Jan 16, 2005 #8
    [tex]\frac{x^{2}+4x}{3}+\frac{84}{x^{2}+4x}}=11[/tex]

    => [tex]\frac{[(x^2 + 4x)(x^2 + 4x)]+(84\times3)}{3x^2 + 12x}=11[/tex]

    It is another way to solve it up it would take a while and would not be benifical at all.

    The Bob (2004 ©)
     
  10. Jan 16, 2005 #9

    dextercioby

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    Forget it,Bob.It's 4-th order.It would take a couple of hours to solve it...

    Daniel.
     
  11. Jan 16, 2005 #10
    I know. That is why I said it was unbenifical because it would take so long to solve.

    [tex]\frac{x^4 + 8x^3 + 16x^2 +252}{3x^2 + 12x} = 11[/tex]

    The Bob (2004 ©)
     
  12. Jan 16, 2005 #11
    Well this has made me curious.

    How would the 4th order equation be solved or any other similar equation of same or higher order ?

    Roger
     
  13. Jan 16, 2005 #12

    dextercioby

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    No higher order than 4 general algebraic equation can be solved.This was shown round 200 yrs ago by Abel and Ruffini.

    Check out the wolfram site.Search for "quartic".

    Daniel.
     
  14. Jan 17, 2005 #13

    arildno

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    Another solution procedure is the guess&try method.
    Occasionally, it works surprisingly well.
     
  15. Jan 17, 2005 #14
    Or you could do a numeric solution with a computer.... but i dont think thats how this praticular problem was meant to be solved. But its true no equation can be solved by traditional algebraic methods in orders higher than 4th.
     
  16. Jan 17, 2005 #15

    HallsofIvy

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    There cannot exist a formula using only roots because there exist solutions to polynomial equations of degree 5 or higher that cannot be written in terms of roots.

    I would not use the phrase "cannot be solved"!
     
  17. Jan 17, 2005 #16

    dextercioby

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    Sorry,you're right...I went too far with the generalization and used inappropriate words.

    It won't happen... :wink:

    Daniel.
     
  18. Jan 17, 2005 #17
    [tex]\frac{x^4 + 8x^3 + 16x^2 +252}{3x^2 + 12x} = 11[/tex]

    [tex]x^4 + 8x^3 + 16x^2 + 252 = 33x^2 + 132x[/tex]

    [tex]f(x) = x^4 + 8x^3 - 17x^2 -132x + 252 = 0[/tex]

    f(3) = 0

    f(2) = 0

    [tex](x - 3)(x - 2)( x^2 +13x + 42) = 0[/tex]

    Problem is that the answer to [tex]x^2 +13x + 42[/tex] is x = 6 or 7 which doesn't work so it works to a point.

    The Bob (2004 ©)
     
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