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Homework Help: Incredibly small equation but cant find answer

  1. Mar 10, 2006 #1
    cuberoot(x) = x-6

    I know the answer is 8, but I dont know how to solve it... how embarrassing.
    Taking Ap chem and not knowing how to find this out. im just in Algebra 2.
    :cry:
     
  2. jcsd
  3. Mar 10, 2006 #2
    Cube both sides:
    [tex]x=(x-6)^3=x^3-18x^2+108x-216[/tex]

    So
    [tex]0=x^3-18x^2+107x-216[/tex].

    I think the best, low level, way to try to solve this is to use the "Rational Roots Test." It's a simple enough method once you've learned it. Basically, take any polynomial equation of the form [tex]0=ax^n+bx^{n-1}+...+cx+d[/tex]. Look at the leading coefficient (a), in this case it's 1. Factor 1 into all possibilities: [tex]1= \pm 1[/tex]. Next look at the constant term (d) and factor that: [tex]-216=\pm (1, 2, 3, 4, 6, 8, 9, 12, ...)[/tex]. If any rational roots to the equation exist, they must be of the form: x = (factor of constant term)/(factor of leading term). So make your list and try all the possibilities. If one (or more) works, then you've found a solution. If none of them do then the roots are not rational numbers and you need a more advanced method.

    -Dan
     
  4. Mar 10, 2006 #3
    [tex]0=x^3-18x^2+107x-216[/tex]
    tahts pretty much where i got to...except i moved 216 to the other side, took out one X from the right side, but noticed the equation was nonfactorable.


    well... from the way you said it, it could have just been easier to plug in numbers into X to find a solution... which is what i did to find that X=8.
    Obviously, there must be an equation to do this.


    If you know how to do this, could you solve it for me? I mean its not a homework it was just a classwork that we turned in to a substitute teacher... but no one knew how to do it.

    I just want to learn step by step.
     
    Last edited: Mar 10, 2006
  5. Mar 11, 2006 #4

    lightgrav

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    Homework Helper

    Once you get the first root, just use "long division" :
    divide the " x^3 - 18x^2 +107x -216 " , treating the x-powers like place value,
    by the " x-8 " root . I get " x^2 - 10x + 27 " .
    So, now what are the other roots?
    They're not integers, since -3 + -9 is not -10 ... (-3*-9 = 27) but not far off.
     
  6. Mar 11, 2006 #5
    Actually, there isn't an equation. Well, for cubic and quartic equations there sort of is, but they're rather nastier than the quadratic equation. And if the degree of the polynomial is 5 or larger, you're pretty much stuck with numerical approximation in general.

    If you're feeling up to some algebra, look up "Cardano's Method" for solving cubic equations. There's a ton of algebra involved, and the solutions usually need quite a bit of tidying before they simplify for even simple problems, but the process isn't that hard to understand.

    -Dan
     
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