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Solving for x (quadratic)

  1. Nov 4, 2009 #1
    How do you explicitly solve for x in terms of y using this quadratic formula?

    y = 2x2 + x

    I need to solve for x in order to find the inverse fxn. I know how to factor, but after that I'm stuck.
    Any suggestions?

    ~Jules~


    * And what about polynomials of a higher order, like x3 + 2x2 + x, or something similar?
     
  2. jcsd
  3. Nov 5, 2009 #2

    Andrew Mason

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    There is no inverse function for y = 2x^2 + x. For some values of y there are more than one value of x (eg. y = 0 or y = 1). So you cannot express x in terms of y.

    AM
     
  4. Nov 5, 2009 #3

    HallsofIvy

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    Are you not aware of the quadratic formula? If [itex]ax^2+ bx+ c= 0[/itex] then [itex]x= (-b\pm\sqrt{b^2- 4ac})/(2a)[/itex].

    In particular, if [itex]y= 2x^2+ x[/itex] then [itex]2x^2+ x- y= 0[/itex] so
    [tex]\frac{-1\pm\sqrt{1+8y}}{4}[/tex]

    It is that [itex]\pm[/itex] that prevents this from being a "true" inverse, as Andrew Mason said. We could divide y into two functions, restricting the domain:

    If [itex]f_1(x)= 2x^2+ x[/itex] for [itex]-\infty< x\le -1/4[/itex] then
    [tex]f_1^{-1}(x)= \frac{-1-\sqrt{1+8x}{4}[/tex]

    If [tex]f_2(x)= 2x^2+ x[/itex] for [itex]-1/4\le x< \infty[/itex] then
    [tex]f_2^{-1}(x)= \frac{-1+ \sqrt{1+8x}{4}[/tex]

    There is a general formula for cubics and quartics but they are extremely complicated. It can be shown that cannot be a general formula solve polynomial equations of degree higher than four using only algebraic functions.
     
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