Can an answer with a fractional exponent be considered as simplest form?

[Mackattack]

If it has to be a radical, do I need to expand the square? Also, do I have to distribute the 3 in the denominator or leave it as it is?

 

[Harry_the_cat]

Because the brackets contain the same expression, it can be simplified further to:
\(\displaystyle \frac{2}{3 \sqrt[3]{2x^3-4x+1}}\)

 

[Mackattack]

Because the brackets contain the same expression, it can be simplified further to:
\(\displaystyle \frac{2}{3 \sqrt[3]{2x^3-4x+1}}\)

That was actually my answer before I rationalized it. Does that mean that there's no need for rationalization here?

 

[Harry_the_cat]

Good point. To me, this expression is "simpler". I suppose it depends on exactly what "simplified" means.

 

[JeffM]

I believe rationalization relates primarily to numerical expressions and reflects the greater simplicity of manual calculation when dividing by an integer.

 

[Mackattack]

Good point. To me, this expression is "simpler". I suppose it depends on exactly what "simplified" means.

The answer before rationalization indeed looks simpler. Hahahaha! Thank you.

 

[Cubist]

View attachment 25658
If it has to be a radical, do I need to expand the square? Also, do I have to distribute the 3 in the denominator or leave it as it is?

IF you have studied complex numbers, then note the following. Else, just ignore this post for now!

[math] \frac{2\left( 2x^3-4x+1\right)^{(2/3)}}{3 \left( 2x^3-4x+1\right) } \ne \frac{2\sqrt[3]{\left( 2x^3-4x+1\right)^{2}}}{3 \left( 2x^3-4x+1\right) } \,\text{ if }\, 2x^3-4x+1<0 [/math]
...try plugging in x=1. The left hand expression will be complex, and the right hand expression will be a real number.

 

[Mackattack]

IF you have studied complex numbers, then note the following. Else, just ignore this post for now!

[math] \frac{2\left( 2x^3-4x+1\right)^{(2/3)}}{3 \left( 2x^3-4x+1\right) } \ne \frac{2\sqrt[3]{\left( 2x^3-4x+1\right)^{2}}}{3 \left( 2x^3-4x+1\right) } \,\text{ if }\, 2x^3-4x+1<0 [/math]
...try plugging in x=1. The left hand expression will be complex, and the right hand expression will be a real number.

I've studied complex numbers before, but it wasn't in depth. How is it that they're not equal if the fractional exponent can technically be changed to a radical form?

 

[lookagain]

If it has to be a radical, do I need to expand the square? Also, do I have to distribute the 3 in the denominator or leave it as it is?

Mackattack, I would write it more simplified as \(\displaystyle \ \dfrac{2}{3}\bigg(2x^3 - 4x + 1\bigg)^{\tfrac{-1}{3}}\).


If your original problem had the fractional exponents and stated to simplify, then you should not
be using the radical form for the final answer. There would have to be some extra clarification
with that "simplify" elsewise.

 

[Cubist]

I've studied complex numbers before, but it wasn't in depth. How is it that they're not equal if the fractional exponent can technically be changed to a radical form?

This is like applying the "power of power" rule in reverse. There are conditions when this can, and can't, be applied (see this post and the one after)

Try these expressions in a calculator that works with complex numbers, or perhaps Matlab/ Octave if you have it (this is a simplified version of what happens when x=1 in your expression)...
(-1)^(2/3) = ( (-1)^(1/3) )^2 ≈ -0.5 + 0.866i
(-1)^(2/3) ≠ ( (-1)^2 )^(1/3) = 1

There are some big threads about principal root VS all roots on this forum! Most complex calculators will return the principal root, even if a real root exists (a real root always exists with cube roots). Just something to be aware of!

 

[Mackattack]

Mackattack, I would write it more simplified as \(\displaystyle \ \dfrac{2}{3}\bigg(2x^3 - 4x + 1\bigg)^{\tfrac{-1}{3}}\).


If your original problem had the fractional exponents and stated to simplify, then you should not
be using the radical form for the final answer. There would have to be some extra clarification
with that "simplify" elsewise.

You make a point as well. The problem uses a fractional exponent rather than a radical sign. There were actually no indications to simplify; it's usually assumed because my teacher doesn't seem to check the answers that were not simplified in his exams.