Solve ##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##

• RChristenk
RChristenk
Homework Statement
Solve ##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##
Relevant Equations
Algebra solving skills
##x^{\frac{-1}{2}}-2x^{\frac{1}{2}}+x^{\frac{2}{3}}=0##

##\Leftrightarrow x^{\frac{-1}{2}}(1-2x+x^{\frac{7}{6}})=0##

Then I'm at a loss as to what to do next because ##x^{\frac{7}{6}}## can't be factored here in a way to get me ##-2x##.

Put x= 1

FactChecker
Sat-P said:
Put x= 1
Ha! Good catch! But I can't help but wonder what mathematics principles this problem is teaching?

The answer is 1, but it would be a strange problem indeed if it cannot be factored and is to be solved by prudent observation.

RChristenk said:
The answer is 1, but it would be a strange problem indeed if it cannot be factored and is to be solved by prudent observation.
There must be another way. There is another solution at about x= 60.9118882395753. That can not be found by "prudent observation".

Sat-P
Indeed if you really need an actual method, I guess there is none except using graph and this has to be done online too. So an observation is necessary.

Did you put the value and checked it??
FactChecker said:
There must be another way. There is another solution at about x= 60.9118882395753. That can not be found by "prudent observation".

Sat-P said:
Did you put the value and checked it??
Yes. I programmed it in GeoGebra. When I zoomed in enough, it told me what the root was. It's not exact. I checked it then with a calculator and the value was very close to 0. There clearly is a crossing near that point.

FactChecker said:
Yes. I programmed it in GeoGebra. When I zoomed in enough, it told me what the root was. It's not exact. I checked it then with a calculator and the value was very close to 0. There clearly is a crossing near that point.
I also checked it right now, and it is somewhere around 7.5*10^-6

FactChecker
Numerical solutions are never exact, but WA plots the function in a way that is obvious there is another root. You can also check that the function changes sign between 60 and 61, so (assuming the function is continuous) there must be a root there.

FactChecker
Sat-P I followed what you did.

##x^{-\frac{1}{2}}(1-2x+x^{\frac{7}{6}})=0##

Set ##x^{\frac{1}{6}}=t, x=t^6##

##t^{-3}(1-2t^6+t^7)=0 \tag{1}##

##t^{-3}(1-t^6-t^6+t^6 \cdot t)=0 \tag{2}##

##t^{-3}(-t^6+1+t^6 \cdot t - t^6)=0 \tag{3}##

##t^{-3}(-(t^6-1)+t^6(t-1))=0 \tag{4}##

##t^{-3}(-(t-1)(t^5+t^4+t^3+t^2+t+1)+t^6(t-1))=0 \tag{5}##

##t^{-3}((t-1)(t^6-t^5-t^4-t^3-t^2-t-1))=0 \tag{6}##

This doesn't look much different than ##x^{-\frac{1}{2}}(1-2x+x^{\frac{7}{6}})=0##, but of course I could take notice of the factor ##(t-1)## and realize ##t=1## is a solution. The other solution seems not possible by paper and must be found by computer. I'm just asking to make sure I'm not missing anything here thanks.

renormalize
Indeed!

RChristenk said:
Sat-P I followed what you did.

##x^{-\frac{1}{2}}(1-2x+x^{\frac{7}{6}})=0##

Set ##x^{\frac{1}{6}}=t, x=t^6##

##t^{-3}(1-2t^6+t^7)=0 \tag{1}##

##t^{-3}(1-t^6-t^6+t^6 \cdot t)=0 \tag{2}##

##t^{-3}(-t^6+1+t^6 \cdot t - t^6)=0 \tag{3}##

##t^{-3}(-(t^6-1)+t^6(t-1))=0 \tag{4}##

##t^{-3}(-(t-1)(t^5+t^4+t^3+t^2+t+1)+t^6(t-1))=0 \tag{5}##

##t^{-3}((t-1)(t^6-t^5-t^4-t^3-t^2-t-1))=0 \tag{6}##

But at this juncture the equation still relies on observing you can plug ##t=1## and not necessarily because it has been reduced to simplest terms.
You can conclude from this, since t^-3 can't be equal to zero since this will give an indeterminate form so t-1 must be equal to 0. But what you said about observing the question to find it's solution/s is imminent before you do some hardcore methods and waste half your time.

Sat-P said:
You can conclude from this, since t^-3 can't be equal to zero since this will give an indeterminate form so t-1 must be equal to 0. But what you said about observing the question to find it's solution/s is imminent before you do some hardcore methods and waste half your time.

Sat-P
Don't mention, and try the question after breaking it don't be confined to one or two methods, think wide. Practice lots of question but the quality matters. I would suggest that practice relevant questions only for now which can help you conquer the exam.

FactChecker
FactChecker said:
There must be another way. There is another solution at about x= 60.9118882395753. That can not be found by "prudent observation".
I think that your graphical approach is quite prudent.

FactChecker
SammyS said:
I think that your graphical approach is quite prudent.
Sir, but you won't know about such a thing in the middle of examination, like the graph will again intercept x axis between 60 and 61.

Sat-P said:
Sir, but you won't know about such a thing in the middle of examination, like the graph will again intercept x axis between 60 and 61.

Your approach of "put x=1" won't do that too.

This just doesn't look like kind of a problem that can be solved using basic algebra.

Sat-P said:
Sir, this is indeed basic algebra for us. Maybe they don't teach this in high school

It can be easily deducted from the question.

Sat-P said:
It can be easily deducted from the question.

You are missing the point: what you found is just a partial answer. It doesn't qualify as a complete solution. The other root seems to be in the realm of 7th degree equations, which is way beyond the basic algebra.

There is always a chance of finding the other root with clever manipulation of the equation, but so far we haven't seen it.

FactChecker
Borek said:
You are missing the point: what you found is just a partial answer. It doesn't qualify as a complete solution. The other root seems to be in the realm of 7th degree equations, which is way beyond the basic algebra.

There is always a chance of finding the other root with clever manipulation of the equation, but so far we haven't seen it.
What you are saying is indeed true and if we follow the general trend there might or might not be 7 solutions to that equation which can't be solved using basic algebra.

Algebra solving skills are useless at this problem. Numerical analysis can help.

FactChecker
Sat-P said:
What you are saying is indeed true and if we follow the general trend there might or might not be 7 solutions to that equation which can't be solved using basic algebra.
It is not a very satisfying example problem but your answer of x=1 is probably all that they expected. The people who wrote this example might not even realize that there is another solution that is difficult to solve for.
IMO, you should probably not spend more time on this one problem.

Gavran

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