Finding simplest radical form of a 4th root?

[Apollinaria]

I haven't taken math in years and am having trouble understanding how to find simplest radical form of a ⁴√(x¹⁴).

I said x⁴√x¹⁰.

I realize I have 3 x⁴ths and x² but I'm not sure if I can pull out more xs.

What are the rules for this? Ideas, insight?

 

[symbolipoint]

I haven't taken math in years and am having trouble understanding how to find simplest radical form of a ⁴√(x¹⁴).

I said x⁴√x¹⁰.

I realize I have 3 x⁴ths and x² but I'm not sure if I can pull out more xs.

What are the rules for this? Ideas, insight?

For each occurance of x^4 as a factor in the term of the radicand, you have x to move to outside of the radical function.

$\sqrt[4]{x^{14}}$=$\sqrt[4]{x^4\cdot x^4\cdot x^4\cdot x^2}$

=$x^3\sqrt[4]{x^2}$

 

[sjb-2812]

For each occurance of x^4 as a factor in the term of the radicand, you have x to move to outside of the radical function.

$\sqrt[4]{x^{14}}$=$\sqrt[4]{x^4\cdot x^4\cdot x^4\cdot x^2}$

=$x^3\sqrt[4]{x^2}$

Can you simplify $\sqrt[4]{x^2}$ still further to $\sqrt{x}$, or does that fall foul of something like principal roots?

 

[I like Serena]

Hi Apollinaria!

There are a few rules for dealing with radical form, powers from powers, and sums of powers.
Here's how it works in your case:
$$\sqrt[4]{x^{14}} = (x^{14})^{\frac 1 4} = x^{14 \cdot \frac 1 4} = x^{3 + \frac 1 2} = x^3 \cdot x^{\frac 1 2} = x^3 \sqrt x$$

 

[Khashishi]

Can you simplify $\sqrt[4]{x^2}$ still further to $\sqrt{x}$, or does that fall foul of something like principal roots?

That doesn't work if x is negative. If you are considering all the complex roots, then $\sqrt[4]{x^2}$ has 4 roots and $\sqrt{x}$ has 2.

But, I think it works if you group together the solutions like $\sqrt[4]{x^2}$ = $\sqrt{x}$ or $\sqrt{-x}$.

 

[I like Serena]

Can you simplify $\sqrt[4]{x^2}$ still further to $\sqrt{x}$, or does that fall foul of something like principal roots?

That doesn't work if x is negative. If you are considering all the complex roots, then $\sqrt[4]{x^2}$ has 4 roots and $\sqrt{x}$ has 2.

But, I think it works if you group together the solutions like $\sqrt[4]{x^2}$ = $\sqrt{x}$ or $\sqrt{-x}$.

With the assumption that x is a non-negative real, it can be safely simplified.

If x can be a negative real, we have that √(x²) = |x| and ⁴√(x²)=√|x|.
However, in general we need to be very careful with negative real numbers and fractional powers.
They are generally not well-defined.
See for instance: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents
(The last couple of lines of the section.)

If x can be a complex number, it becomes even worse:
See for instance: http://en.wikipedia.org/wiki/Exponentiation#Failure_of_power_and_logarithm_identities