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

[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 $-2x$.

 

[Sat-P]

Put x= 1

 

[FactChecker]

Put x= 1

Ha! Good catch! But I can't help but wonder what mathematics principles this problem is teaching?

 

[RChristenk]

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.

 

[FactChecker]

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.

 

[Sat-P]

 

[Sat-P]

Did you put the value and checked it??

There must be another way. There is another solution at about x= 60.9118882395753. That can not be found by "prudent observation".

 

[FactChecker]

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.

 

[Sat-P]

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

 

[Borek]

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.

 

[RChristenk]

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.

 

[Sat-P]

Indeed!

 

[Sat-P]

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.

 

[RChristenk]

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.

Thanks for your help.

 

[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.