Incredibly small equation but cant find answer

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Homework Help Overview

The discussion revolves around solving the equation cuberoot(x) = x - 6, which leads to a cubic polynomial x^3 - 18x^2 + 107x - 216. Participants express uncertainty about the solution process and the nature of cubic equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss methods such as cubing both sides and applying the Rational Roots Test. Some express confusion about the factorability of the resulting polynomial and consider numerical methods for finding roots. Others mention the possibility of using long division to simplify the polynomial after finding one root.

Discussion Status

The conversation is ongoing, with various methods being proposed and explored. Some participants have shared their attempts and reasoning, while others seek clarification on the steps involved in solving the cubic equation. There is no explicit consensus on a single approach, but several productive lines of inquiry are being pursued.

Contextual Notes

Participants note that the problem is part of classwork and express a desire to learn the solution process step by step. There is mention of the complexity of solving higher-degree polynomials and the limitations of existing methods.

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cuberoot(x) = x-6

I know the answer is 8, but I don't know how to solve it... how embarrassing.
Taking Ap chem and not knowing how to find this out. I am just in Algebra 2.
:cry:
 
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Haxx0rm4ster said:
cuberoot(x) = x-6

I know the answer is 8, but I don't know how to solve it... how embarrassing.
Taking Ap chem and not knowing how to find this out. I am just in Algebra 2.
:cry:

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
 
[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 into a substitute teacher... but no one knew how to do it.

I just want to learn step by step.
 
Last edited:
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.
 
Haxx0rm4ster said:
[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 into a substitute teacher... but no one knew how to do it.

I just want to learn step by step.

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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