Solving an equation involving surds

[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.

 

[chwala]

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

sorry for not seeing then, ...$u=2$

 

[kumusta]

@chawala
Thanks for the explanation.