- #1
chwala
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- Homework Statement
- Find ##x##,
if ##x^{0.5}+x^{1/3}=12##
- Relevant Equations
- surds
##x^{1/6}=12x^{1/3}-1##
fresh_42 said: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}##?
WWGD said:There is a general solution formula for cubics.
sorry, i was not keen, ##u=2## is another solution,WWGD said: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?
Yes, and ##12=12\cdot 1 = 6 \cdot 2 = 4 \cdot 3##. Which one do we get for ##u##?chwala said:if ##u = x^{1/6}##
then, we shall have, ##u^3+u^2=12##
##u^2(u+1)=12##
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 said: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, 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 said: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 said: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.
Or more simply, without resorting to tenth root approximations,chwala said: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
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.chwala said: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##
What do you mean? You have ##u^{10}=x=32^{10}## which results inchwala said: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
fresh_42 said: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?
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...fresh_42 said: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 said: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.chwala said: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...
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 said: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 said: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.
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 said: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...
maybe in future it might be possible, which groups are you reffering to ...group theory?fresh_42 said: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.
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.chwala said:maybe in future it might be possible, which groups are you reffering to ...group theory?
No, it is ##u^4+u^2+u=u(u^3+u+1)=22## with ##u=2##.chwala said: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}##
Surds are numbers that cannot be expressed as a ratio of two integers and involve an irrational root, such as √2 or √3.
To solve an equation involving surds, you need to isolate the surd term on one side of the equation and square both sides to eliminate the surd. Then, solve the resulting equation for the variable.
Yes, surds can be simplified by finding the largest perfect square that is a factor of the surd and taking its square root. For example, √32 can be simplified to 4√2.
Some common mistakes when solving equations involving surds include forgetting to square both sides to eliminate the surd, incorrectly simplifying the surd, and not checking for extraneous solutions.
Yes, equations involving surds can have multiple solutions. This is because when we square both sides to eliminate the surd, we may introduce extraneous solutions that do not satisfy the original equation.