Solving an equation involving surds

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Homework Statement:

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

Homework Equations:

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

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

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

Gold Member
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...

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

WWGD
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2019 Award
There is a general solution formula for cubics.

Gold Member
There is a general solution formula for cubics.
i am getting $u= [{-3±i√(15)}]/2$

WWGD
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2019 Award
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?

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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
Mentor
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 $u$?

Gold Member
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
Mentor
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.

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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
Mentor
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.

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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

Last edited:
Gold Member
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
Mentor
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
Mentor
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
Mentor
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?

Gold Member
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
Mentor
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.

Gold Member
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....

Gold Member
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...