Solving an equation involving surds

[chwala]

Homework Statement

Find $x$,
if $x^{0.5}+x^{1/3}=12$

Relevant Equations

surds

$x^{1/6}=12x^{1/3}-1$

 

[chwala]

is there a way of solving this algebraically or one has to use numerical methods?Newtons finite methods or something like that. i know that $x=64$.

 

[fresh_42]

Your second equation is wrong.
Anyway. What do you get if you substitute $u=x^\frac{1}{6}\,?$

 

[chwala]

fresh how are you...yeah the equation is supposed to be;

$x^{1/6}$=$12x^{-1/3}-1$. i thought of that substitution,
i thought of,
$x^{({1/2})({1/3})}=12x^{-1/3}-1$

 

[fresh_42]

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 $u=x^{1/6}$?

 

[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 $u=x^{1/6}$?

i divided each term by $x^{1/3}$...
i wanted to simplify in a way, and have some common term...

 

[chwala]

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

 

[WWGD]

There is a general solution formula for cubics.

 

[chwala]

There is a general solution formula for cubics.

i am getting $u= [{-3±i√(15)}]/2$

 

[WWGD]

Sorry, I am on my phone, no paper, hard to check. Can you try Wolfram? But notice you must end up with 3 roots up to multiplicity. Where is the third?

 

[chwala]

Sorry, I am on my phone, no paper, hard to check. Can you try Wolfram? But notice you must end up with 3 roots up to multiplicity. Where is the third?

sorry, i was not keen, $u=2$ is another solution,
now, $x^{1/6}=2 →x= 2^6 = 64$

Bingo! WWGD and Fresh!

 

[fresh_42]

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

Yes, and $12=12\cdot 1 = 6 \cdot 2 = 4 \cdot 3$. Which one do we get for $u$?

 

[chwala]

so the general idea or thinking in similar kind of equations is to think of the lcm of the exponents and come up with another variable...

 

[fresh_42]

so the general idea or thinking in similar kind of equations is to think of the lcm of the exponents and come up with another variable...

Yes, but it depends on the case. I would say: "The general idea is to get rid of what disturbs!" And since $1/2$ and $1/3$ disturbs in this case, we need $1/6$ to get rid of both at the same time.

 

[chwala]

just going on with my thoughts on similar kind of problems, consider,

$x^{1/10} + x^{1/5}=1056$
the lcm is $10$
therefore, letting $u^{10}= x$ we shall have,
$u+u^2=1056$
is my thinking correct? i earlier indicated that establishing the lcm may be key in solving such kind of problems.
regards,

 

[Mark44]

just going on with my thoughts on similar kind of problems, consider,

$x^{1/10} + x^{1/5}=1056$
the lcm is $10$
therefore, letting $u^{10}= x$ we shall have,
$u+u^2=1056$
is my thinking correct? i earlier indicated that establishing the lcm may be key in solving such kind of problems.
regards,

Yes, that's the right idea. The equation you ended with above can be factored without too much difficulty. Keep in mind, though, that you need to verify that any solutions you get must satisfy the original equation, so it's possible that some solutions of the equation in u don't work in the original equation.

 

[chwala]

Yes, that's the right idea. The equation you ended with above can be factored without too much difficulty. Keep in mind, though, that you need to verify that any solutions you get must satisfy the original equation, so it's possible that some solutions of the equation in u don't work in the original equation.

now in this equation, i get $u=32$ and $u=-33$ how do we solve for $x$
i am getting $x= 32^{0.1}$ which does not satisfy the original equation, maybe i am tired ..i can't seem to see

aaaaahhhhhhh seen it

if $x=32^{0.1}$ then substituting in the original we will have;

$32+ [(1.125899907⊗10^{15})]^{0.2}=1056$ Bingo

 

[chwala]

I may need to further explore this kind of questions, and see how lcm works in simplifying them, let's pick another question,
$x^{1/3}+x^{1/9}=512$

 

[Mark44]

now in this equation, i get $u=32$ and $u=-33$ how do we solve for $x$
i am getting $x= 32^{0.1}$ which does not satisfy the original equation, maybe i am tired ..i can't seem to see

aaaaahhhhhhh seen it

if $x=32^{0.1}$ then substituting in the original we will have;

$32+ [(1.125899907⊗10^{15})]^{0.2}=1056$ Bingo

Or more simply, without resorting to tenth root approximations,
$(32^{10})^{1/10} + (32^{10})^{1/5} = 32^{10/10} + 32^{10/5} = 32 + 32^2 = 1056$

 

[Mark44]

I may need to further explore this kind of questions, and see how lcm works in simplifying them, let's pick another question,
$x^{1/3}+x^{1/9}=512$

This one is harder, since the likely substitution will turn it into a cubic equation. There is a formula for solving cubic equations, but it is much harder to apply than the quadratic formula.

 

[fresh_42]

now in this equation, i get $u=32$ and $u=-33$ how do we solve for $x$
i am getting $x= 32^{0.1}$ which does not satisfy the original equation, maybe i am tired ..i can't seem to see

aaaaahhhhhhh seen it

if $x=32^{0.1}$ then substituting in the original we will have;

$32+ [(1.125899907⊗10^{15})]^{0.2}=1056$ Bingo

What do you mean? You have $u^{10}=x=32^{10}$ which results in
$$
x^{1/10}+x^{1/5}=x^{1/10}+\left(x^{1/10}\right)^2=(32^{10})^{1/10}+\left((32^{10})^{1/10}\right)^2=32+32^2=1056
$$
For the next problem use $x^{1/3}=\left(x^{1/9}\right)^3$.
Are you sure that $512$ and the plus sign are correct?

 

[chwala]

What do you mean? You have $u^{10}=x=32^{10}$ which results in
$$
x^{1/10}+x^{1/5}=x^{1/10}+\left(x^{1/10}\right)^2=(32^{10})^{1/10}+\left((32^{10})^{1/10}\right)^2=32+32^2=1056
$$
For the next problem use $x^{1/3}=\left(x^{1/9}\right)^3$.
Are you sure that $512$ and the plus sign are correct?

i just came up with that problem, just trying to see if i can come up with a general method of solving 'such like' surd equations,
like for instance,
find $x$ given,
$x^{1/3}+x^{1/4}+x^{1/5}=20$
my approach,
let $u^{60}=x$
→$u^{20}=x^{1/3}$
→$u^{15}=x^{1/4}$
→$u^{12}=x^{1/5}$
therefore,
$u^{20}+u^{15}+u^{12}=20$
$u^{12}(u^8+u^3+1)=20$
$u^{12}[u^3(u^5+1)+1]=20$
does this make sense? can i move from here? i know that the solution will be an approximate value and that i will need to make use of a 'numerical method' in finding $x$.

 

[fresh_42]

Already your easier problem $x^{1/3}+x^{1/9}=512$ had no rational solution, so certainly no integer solution.
So the trick doesn't always work. Closed formulas for $u^n+ a_1u^{n-1} + \ldots + a_{n-1}u +a_n=0$ can only be given up to $n=4$. Hence there is a natural limit anyway. Your first two examples used the special shape of the equations.

 

[chwala]

Already your easier problem $x^{1/3}+x^{1/9}=512$ had no rational solution, so certainly no integer solution.
So the trick doesn't always work. Closed formulas for $u^n+ a_1u^{n-1} + \ldots + a_{n-1}u +a_n=0$ can only be given up to $n=4$. Hence there is a natural limit anyway. Your first two examples used the special shape of the equations.

so when $n$ is greater than $4$, you are indicating that finding a solution is not possible? i indicated that the solution may not be exact rather may take approximate values...

 

[chwala]

so when $n$ is greater than $4$, you are indicating that finding a solution is not possible? i indicated that the solution may not be exact rather may take approximate values...

I will try come up with a problem where $n$ is greater than 5 and also have the solution as an integer value...then explore on methods of solution...

 

[fresh_42]

so when $n$ is greater than $4$, you are indicating that finding a solution is not possible? i indicated that the solution may not be exact rather may take approximate values...

For $n=2$ we can write $x^2+px+q=\left(x+\dfrac{p}{2}+\sqrt{\left(\dfrac{p}{2}\right)^2-q}\right)\left(x+\dfrac{p}{2}-\sqrt{\left(\dfrac{p}{2}\right)^2-q}\right)$ and for $n=3$ and $n=4$ there are formulas, too, but much more complicated and depending on the coefficients.

For $n\geq 5$ it has been proven, that no general formula exists. Of course we can still solve special cases, e.g. $x^5-2x^3+x=0$ can easily be solved. But there is cannot be any formula fits all.

Solutions (so they exist) can always be found numerically, which are generally approximations.

 

[fresh_42]

I will try come up with a problem where $n$ is greater than 5 and also have the solution as an integer value...then explore on methods of solution...

You can always calculate $p(x)=(x-a_1)\cdot \ldots \cdot (x-a_n)$ and know beforehand which values satisfy $p(a)=0$. But not the other way around.

 

[chwala]

For $n=2$ we can write $x^2+px+q=\left(x+\dfrac{p}{2}+\sqrt{\left(\dfrac{p}{2}\right)^2-q}\right)\left(x+\dfrac{p}{2}-\sqrt{\left(\dfrac{p}{2}\right)^2-q}\right)$ and for $n=3$ and $n=4$ there are formulas, too, but much more complicated and depending on the coefficients.

For $n\geq 5$ it has been proven, that no general formula exists. Of course we can still solve special cases, e.g. $x^5-2x^3+x=0$ can easily be solved. But there is cannot be any formula fits all.

Solutions (so they exist) can always be found numerically, which are generally approximations.

i bet this would be a good area to research on, who knows maybe in the near future, mathematicians will find general ways of solving this...atleast i thought of the possibility of solving such like problems where n is greater than 4.., i am sure in the near future there might be a general way in solving this kind of problems.

 

[fresh_42]

i bet this would be a good area to research on, who knows maybe in the near future, mathematicians will find general ways of solving this...

We know for more than 100 years that it cannot be done by basic arithmetic operations and roots. It is a property of certain groups.

 

[chwala]

We know for more than 100 years that it cannot be done by basic arithmetic operations and roots. It is a property of certain groups.

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

 

[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.