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Finding x for an equation

  1. Sep 2, 2011 #1
    So how would I find the x value when y = 2.2 ?



    Thanks. I'm on mobile can't use latex
  2. jcsd
  3. Sep 2, 2011 #2


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    Well plug in y= 2.2 and get 2.2 = x^3-5x+6. You will need to solve using a numerical method most likely.
  4. Sep 2, 2011 #3
    I got x^3-5x+3.8=0 but I don't know what to do next.
  5. Sep 3, 2011 #4
  6. Sep 3, 2011 #5


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    What methods for root finding have you learnt? (i.e. what methods for finding 'x' do you know about? Simple algebra, Newton-Raphson method, etc.)
  7. Sep 3, 2011 #6
    -3.8 = x(x**2 - 5)

    See if you can stumble upon a haphazard number x that satisfies this eqn ( EG: 1 is very near and satisfactory )

    then divide the original eqn by (x- xhaphazard )

    Obtain second degree Polynomial where x1,2 =( (-b + sqrt(b**2-4ac)) / 2a and the other root where the sign before sqrt term is a negative )

    Good luck.
  8. Sep 3, 2011 #7


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    Well, I am not sure I would call it "very close. However, Crossfit415, you can use that to get a better solution: if x= 0, [itex]x^3-5x+3.8= 3.8[/itex] and if x= 1, itex]x^3-5x+3.8= -.2[/itex]. one valule is positive and the other negative so the function must be 0 between them- that means there must be a root between 0 and 1. .5 is halfway between and [itex](.5)^3- 5(.5)+ 3.8= 1.425. That is positive so there must be a root half way between .5 and 1. Halfway between those is .75. (.75)^3- 5(.75)+ 3.8= 0.471875. Again, that is positive so there must be a root between .75 and 1. Halfway between those is 0.875, etc. Carry that out to whatever accuracy you want and, as stallionx says, divide the polynomial by x- a to get a quadratic equation for the other two roots.

    You might suspect, right from the start, that since -.2 is much closer to 0 than 3.8, the root should be closer to 1 than to 0. That's correct and a little faster numerical method would be the "method of secants". Instead of picking exactly half way between possible x values, we calculate the equation of the line between the given points. The line between (0, 3.8) and (1, -.2) is y= -4x+ 3.8. That will be 0 when x= 3.8/4= 0.95. Faster "convergence" to the correct solution but more calculation required at each step.

    "Newton's method", replacing the curve by a tangent line at each step, would be even faster but requires Calculus.
  9. Sep 3, 2011 #8


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    You could multiply both sides by 5 and perform the rational root test. If that fails though, iterative methods like previous posters have mentioned is the best way to go.
  10. Sep 3, 2011 #9


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    Since it's already a depressed cubic you can transform it into a quadratic (in z^3) with the substitution:

    [tex]x = z + \frac{5}{3z}[/tex]

    However it requires a bit of complex arithmetic and you need to take all three branches of the complex cubed root to get all three (in this case real) solutions.
  11. Sep 3, 2011 #10
    Once you find one root by means of be it Newton-Raphson or a computer program, the other 2 are ready to be found by the means I tried to describe above.
  12. Sep 5, 2011 #11


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    Just as there is a formula for solving a quadratic (x=-b +/- sqrt(b2-4ac) etc...) then there is also a formula for solving a cubic. It does look a bit formidable at first glance, but will simplify in your case because your coefficient of x2 is zero.

    See http://en.wikipedia.org/wiki/Cubic_function" [Broken]
    Last edited by a moderator: May 5, 2017
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