Seemingly Easy??

[bandofhorses11]

Hi, I've been trying to solve this problem algebraically for a while, but I can't seem to figure it out. How would you solve this step by step? Thanks!

x^3 + x = 40

 

[chrisr]

The answer lies between 3 and 4.
Have you thought of the Newton-Raphson method to locate the root of x[sup:34ir76rh]3[/sup:34ir76rh]+x-40=0 ?

 

[bandofhorses11]

Thanks, how would you use that? I know approx what the answer would be... but I guess my question would specifically be how to solve a problem of that nature. I mean, the problem could be x^3 + x = 2, or = 49, or = 100009.

 

[chrisr]

You know the techniques for solving quadratics, I guess,
and the cubics that are factoriseable.

There are other ways of solving cubics also.
The Newton-Raphson is a homing device.
It seeks out the solution using an iterative process.
It's fun to see how it works.
It boils down to picking a value between 3 and 4 in this case
(could be wider of the mark too, of course).

Then you apply the formula and you get a "next guess" which will be much closer to the answer.
If you do it again using the "next guess", you get much closer again.

The formula is x[sub:91j7msad]next[/sub:91j7msad]=x-{f(x)/f[sup:91j7msad]'[/sup:91j7msad](x)}

or x[sub:91j7msad]n+1[/sub:91j7msad]=x[sub:91j7msad]n[/sub:91j7msad]-{f(x[sub:91j7msad]n[/sub:91j7msad])/(f[sup:91j7msad]'[/sup:91j7msad](x[sub:91j7msad]n[/sub:91j7msad]))}.

In this case, f(x) is the graph of x[sup:91j7msad]3[/sup:91j7msad]+x-40 and we are discovering the x-axis crossing point
where it is zero.

So, differentiate f(x) to get 3x[sup:91j7msad]2[/sup:91j7msad]+1.
Use x=initial guess,
place in the formula for x, f(x) and f[sup:91j7msad]'[/sup:91j7msad](x) and the result of that is your "next approximation"
to the solution of x[sup:91j7msad]3[/sup:91j7msad]+x-40=0.

You zoom in on the answer quickly.