Solving an equation involving surds

[chwala]

I also came up with this problem,
$x^{1/3}+x^{1/6}+x^{1/12}=22$
i know that $x=4096$
now using $u^{12}=x$
we shall have $u^4+u^2+u^{12}=22$ of course, here the degree of polynomial is greater than 4.
it follows that,
$u^2(u^{10}+u^2+1)=22$
$u^2[u^2(u^8+1)+1]=22$

most of these equations are ending up in the above form, now its a matter of exploring 'numerical methods' that may solve this. my thinkin am crazy at times
if i let $(u^8+1) =m$ then we shall have,
$u^2[u^2m+1]=22$
dividing both $u^2$
$u^2m+1=22u^{-2}$

 

[fresh_42]

maybe in future it might be possible, which groups are you reffering to ...group theory?

Yes. The alternating groups $A_n$ are not solvable for $n>4$, they are simple. This is a provable fact which is responsible that it is impossible to find formulas for $n>4$. It is a consequence of Galois' theory.

 

[fresh_42]

I also came up with this problem,
$x^{1/3}+x^{1/6}+x^{1/12}=22$
i know that $x=4096$
now using $u^{12}=x$
we shall have $u^4+u^2+u^{12}=22$ of course, here the degree of polynomial is greater than 4.

No, it is $u^4+u^2+u=u(u^3+u+1)=22$ with $u=2$.

it follows that,
$u^2(u^{10}+u^2+1)=22$
$u^2[u^2(u^8+1)+1]=22$

most of these equations are ending up in the above form, now its a matter of exploring 'numerical methods' that may solve this. my thinkin am crazy at times
if i let $(u^8+1) =m$ then we shall have,
$u^2[u^2m+1]=22$
dividing both $u^2$
$u^2m+1=22u^{-2}$

 

[chwala]

lol, i made some mathematical error, i appreciate your comments...i will still look at this once i am free and see what else come up. Thanks Fresh...

 

[chwala]

I thought of a problem,
find $x$
if $x^{1/3}+x^{1/18}=66$
i know that $x = 262144$
it follows that, if i let $u^{18}=x$
then we shall have,
$u^6+u=66$ so here we have a polynomial of degree 6,...

 

[fresh_42]

Such small examples with high powers are easy. $n\longmapsto n^6$ quickly exceeds $66$, so $u$ can only be $1$ or $2$. If we seek especially integer solutions, the fastest algorithm is certainly to check all possibilities.

Edit:

If we have an integer polynomial $p(x)=x^n+a_1x^{n-1}+\ldots +a_{n-1}x+a_n$ and a number $z$ is a zero, that means $p(z)=0$, then we can write $p(x)=(x-z)\cdot q(x)$ with a polynomial $q(x)$ of degree $n-1$. This means especially that $z$ divides the constant term $a_n\, : \,z|a_n\,.$

So in your example of $p(u)=u^6+u-66$ we can only have divisors of $66$ as possible integer solutions. So regardless that $4$ and $5$ are already too big, they do not divide $66$ either.

 

[chwala]

My interest in on how we can come up with a solution from this step
$u^6+u=66$
$(u^2)^3+u=66$ or $(u^3)^2+u=66$
if let say $u^2=m$, then it follows that,
$m^3+m^{1/2}-66=0$
Note
i already know that $u=2$
i am just interested in how we can find the solution systematically...

$(u^3)^2+u=66$
$m^2+ m^{1/3}=66$
$m^2+m^{1/3}-66=0$
or
$u^6=66-u$

 

[mfb]

Well, try the divisors of 66.

If there is an integer solution it is nice to find, but if there is not... try to find an exact solution to e.g. $u^6+u=65$.

 

[chwala]

so are we implying that we have no formula in the world to solve polynomials of degree more than 4. In particular $u^6+u=66?$

 

[SammyS]

so are we implying that we have no formula in the world to solve polynomials of degree more than 4. In particular $u^6+u=66?$

As fresh_42 said. There is no general formula to find the roots of any polynomial of degree greater than 4. Furthermore, this fact has been proven .

That does not mean that we can never find the roots, or even 1 root, of some particular degree 6 polynomial .

 

[chwala]

As fresh_42 said. There is no general formula to find the roots of any polynomial of degree greater than 4. Furthermore, this fact has been proven .

That does not mean that we can never find the roots, or even 1 root, of some particular degree 6 polynomial .

thanks... so finding these roots is by trial and error method and by use of graphs? and its long since i refreshed on numerical methods, are there no numerical methods that may give an approximate solution if not exact?

 

[fresh_42]

thanks... so finding these roots is by trial and error method and by use of graphs? and its long since i refreshed on numerical methods, are there no numerical methods that may give an approximate solution if not exact?

There is, and probably more than one. An easy one is by Newton's method:
https://en.wikipedia.org/wiki/Newton's_method

Btw., this algorithm can be used to calculate (square) roots manually.

 

[chwala]

There is, and probably more than one. An easy one is by Newton's method:
https://en.wikipedia.org/wiki/Newton's_method

Btw., this algorithm can be used to calculate (square) roots manually.

yeah, i remember studying that ,plus other methods like runge kutta method ... Anyway, are you able to try solve my problem using the approximate method(s), you've just mentioned?

 

[mfb]

As fresh_42 said. There is no general formula to find the roots of any polynomial of degree greater than 4. Furthermore, this fact has been proven .

There are formulas that rely on infinite fractions or similar things. There is no general formula that would produce an algebraic solution (using only roots and things like that).

 

[fresh_42]

yeah, i remember studying that ,plus other methods like runge kutta method ... Anyway, are you able to try solve my problem using the approximate method(s), you've just mentioned?

Sure, but why? Choose $x_0=1$ and create the sequence in the article for $f(x)=x^6+x-66$. What didn't you understand?

 

[chwala]

Sure, but why? Choose $x_0=1$ and create the sequence in the article for $f(x)=x^6+x-66$. What didn't you understand?

ok mate i hear you...The Newton- Raphson method cannot be used on this problem...
my perspective is let
$f(x)= x^6+x-66$
put say $x=o$
$f(0)=-66$
$f(1)=-64$
$f(-1)=-66$ and $f^{''}(x)=30x^4$ there is no convergence as the iterations are going to increase(diverge)... mythoughts

 

[mfb]

Did you actually calculate it or why did you dismiss it?

$f'(x)=6x^5+1$
f(0)=-66, f'(0)=1, so our next point is 66.
f(66)=82653950016, f'(66)=7513995457, the ratio is 11, so our next point is 66-11=55.
The following points are approximately 45.8, 38.2, 26.5, ... you get 2.0027 after 22 steps and 2.0000090 after 23 steps.

0 is very poor choice for the starting point but it still converges to one of the solutions.

 

[chwala]

Sure, but why? Choose $x_0=1$ and create the sequence in the article for $f(x)=x^6+x-66$. What didn't you understand?

Did you actually calculate it or why did you dismiss it?

$f'(x)=6x^5+1$
f(0)=-66, f'(0)=1, so our next point is 66.
f(66)=82653950016, f'(66)=7513995457, the ratio is 11, so our next point is 66-11=55.
The following points are approximately 45.8, 38.2, 26.5, ... you get 2.0027 after 22 steps and 2.0000090 after 23 steps.

0 is very poor choice for the starting point but it still converges to one of the solutions.

i will take time during the december holidays and refresh on this...they're not difficult. I just need to create time and do some reading. Bingo

 

[chwala]

You have $x^{1/2}+x^{1/3}=12$ which is not the same as $x^{1/6}=12x^{1/3}-1$.

Again, what do you get from $x^{1/2}+x^{1/3}=12$ if $u=x^{1/6}$?

just trying to go through my previous working...what of using the approach of,
$x^{1/2} + x^{1/3} = 12$
then,
$x^{1/6}+ 1 = 12x^{-1/3}$
on squaring both sides, we get;
$x^{1/3} + 2x^{1/6} + 1= \frac{144}{x^{2/3}}$
$x + 2x^{5/6} + x^{2/3}= 144$
$x^{6/6}+ 2x^{5/6} + x^{4/6}= 144$
letting $x^{1/6}=p$
it follows that,
$p^6+2p^5+p^4=144$
on solving numerically, we get
$p=2$ bingo...

 

[Mark44]

if $u = x^{1/6}$
then, we shall have, $u^3+u^2=12$
$u^2(u+1)=12$

The factorization on the left side doesn't do you any good. Factorization is useful if you have a product of factors on one side, and zero on the other.

 

[fresh_42]

just trying to go through my previous working...what of using the approach of,
$x^{1/2} + x^{1/3} = 12$
then,
$x^{1/6}+ 1 = 12x^{-1/3}$
on squaring both sides, we get;
$x^{1/3} + 2x^{1/6} + 1= \frac{144}{x^{2/3}}$
$x + 2x^{5/6} + x^{2/3}= 144$
$x^{6/6}+ 2x^{5/6} + x^{4/6}= 144$
letting $x^{1/6}=p$
it follows that,
$p^6+2p^5+p^4=144$
on solving numerically, we get
$p=2$ bingo...

This is too complicated and I'm not sure whether it is right.

You set $p=x^{1/6}$ anywhere in the calculation. I see no reason why you don't do it from the start.
$x^{1/2}=p^3\, , \,x^{1/3}=p^2$

 

[chwala]

This is too complicated and I'm not sure whether it is right.

You set $p=x^{1/6}$ anywhere in the calculation. I see no reason why you don't do it from the start.
$x^{1/2}=p^3\, , \,x^{1/3}=p^2$

It should be correct, even by using factor theorem by letting $f(p) = 0$, should realize the value of $p$ and consequently find $x$ as required.

 

[chwala]

This is too complicated and I'm not sure whether it is right.

You set $p=x^{1/6}$ anywhere in the calculation. I see no reason why you don't do it from the start.
$x^{1/2}=p^3\, , \,x^{1/3}=p^2$

I am just trying to use a different approach, that's the beauty of Math...of course your approach is straightforward...

 

[Office_Shredder]

If you are going to use a numerical method to find solutions, there's no reason to turn it into a polynomial in the first place, other than maybe avoiding awkwardness where the method tries to take you to a negative number.

 

[chwala]

I think there is a wrong step somewhere, I'll check this tomorrow...

i just checked. Working is 100% correct.

 

[fresh_42]

I am just trying to use a different approach, that's the beauty of Math...of course your approach is straightforward...

You made $p^6+2p^5+p^4=144$ out of the equation. But this is $p^3+p^2=12$ squared. You can as well get $p^3+p^2=12$ directly from the first line. To guess $p=2$ is even easier from the unsquared equation. Now, with $p=2$ you can calculate $f(p):=(p^3+p^2-12)\, : \,(p-2)$ and get the other solutions from $f(p)=0.$

 

[chwala]

You made $p^6+2p^5+p^4=144$ out of the equation. But this is $p^3+p^2=12$ squared. You can as well get $p^3+p^2=12$ directly from the first line. To guess $p=2$ is even easier from the unsquared equation. Now, with $p=2$ you can calculate $f(p):=(p^3+p^2-12)\, : \,(p-2)$ and get the other solutions from $f(p)=0.$

That's true Fresh, the question is, 'if a student was to use this approach', would he be penalised for his approach? also your approach may be similar to mine, just a different substitution if you get what i mean..you used $u$= $x^{1/6}$ which realizes a polynomial of degree $3$...

 

[fresh_42]

That's true Fresh, the question is, 'if a student was to use this approach', would he be penalised for his approach?

Depends on the penalty. An equation $x^3=a$ has three solutions, $x^6=a^2$ has six! If you take all these six solutions and check which ones do actually satisfy the given equation and come up with the remaining three, then it is a deviation, but still correct.

Always remember that $f(x)=c \Longrightarrow \ldots \Longrightarrow c \in \{c_1,\ldots,c_n\}$ is only the necessary part. It does not say that $f(c_k)=0$. Sufficiency has to be considered. And squaring is a one-way process. E.g. $x+2=3$ results in $x=1$ and $1+2=3$ is so obvious, that we do not have to check it. However, $(x+2)^2=x^2+4x+4=3^2=9$ yields $x\in \{-5,1\}$ but $-5+2\neq 3.$

 

[kumusta]

I find this thread confusing. First, in post #1:

Homework Statement:: Find $x$,
if $x^{0.5}+x^{1/3}=12$
Relevant Equations:: surds
$x^{1/6}=12x^{1/3}-1$

The $~x^{1/2}+x^{1/3}=12~$ became $~x^{1/6}=12x^{1/3}-1~$.
Then, in post #4,

...yeah the equation is supposed to be;
$x^{1/6}$=$12x^{-1/3}-1$. i thought of that substitution,
i thought of,
$x^{({1/2})({1/3})}=12x^{-1/3}-1$

the $~x^{1/6}=12x^{1/3}-1~$ became $x^{1/6}=12x^{-1/3}-1$
I do not know which of the two is the actual equation of interest.

 

[chwala]

I find this thread confusing. First, in post #1:

The $~x^{1/2}+x^{1/3}=12~$ became $~x^{1/6}=12x^{1/3}-1~$.
Then, in post #4,

the $~x^{1/6}=12x^{1/3}-1~$ became $x^{1/6}=12x^{-1/3}-1$
I do not know which of the two is the actual equation of interest.

The question of interest is on post $1$ the question was discussed and a solution found by post$12$ (check post $12$), The author posed a question asking "what $u$ was' given the terms $4⋅3=12$... from here it was clear that our $u=2$
The other posts following this solution was a general discussion on similar kind of problems...

Now looking at post $51$;
I was attempting to solve the problem in post$1$, from a different perspective, you can follow the discussions after my post $51$ onwards. Do note that, consequently the numerical value of my $p$ in post $51$ is equivalent to the numerical value of $u$ in post $12$. I hope its clear now.