# Finding simplest radical form of a 4th root?

by Apollinaria
 P: 82 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?
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P: 2,692
 Quote by 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?
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.

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

=$x^3\sqrt[4]{x^2}$
P: 418
 Quote by symbolipoint 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?

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P: 6,189

## Finding simplest radical form of a 4th root?

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:
$$\sqrt[4]{x^{14}} = (x^{14})^{\frac 1 4} = x^{14 \cdot \frac 1 4} = x^{3 + \frac 1 2} = x^3 \cdot x^{\frac 1 2} = x^3 \sqrt x$$
P: 833
 Quote by sjb-2812 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 $\sqrt{-x}$.
HW Helper
P: 6,189
 Quote by sjb-2812 Can you simplify $\sqrt[4]{x^2}$ still further to $\sqrt{x}$, or does that fall foul of something like principal roots?
 Quote by Khashishi 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/Exponen...onal_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/Exponen...thm_identities

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