- #1

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^{4}√(x

^{14}).

I said x

^{4}√x

^{10}.

I realize I have 3 x

^{4}ths and x

^{2}but I'm not sure if I can pull out more xs.

What are the rules for this? Ideas, insight?

- Thread starter Apollinaria
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- #1

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I said x

I realize I have 3 x

What are the rules for this? Ideas, insight?

- #2

symbolipoint

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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.^{4}√(x^{14}).

I said x^{4}√x^{10}.

I realize I have 3 x^{4}ths and x^{2}but I'm not sure if I can pull out more xs.

What are the rules for this? Ideas, insight?

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

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

- #3

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

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

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

- #4

I like Serena

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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:

$$\sqrt[4]{x^{14}} = (x^{14})^{\frac 1 4} = x^{14 \cdot \frac 1 4} = x^{3 + \frac 1 2} = x^3 \cdot x^{\frac 1 2} = x^3 \sqrt x$$

- #5

Khashishi

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That doesn't work if x is negative. If you are considering all the complex roots, then [itex]\sqrt[4]{x^2}[/itex] has 4 roots and [itex]\sqrt{x}[/itex] has 2.Can you simplify [itex]\sqrt[4]{x^2}[/itex] still further to [itex]\sqrt{x}[/itex], or does that fall foul of something like principal roots?

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

- #6

I like Serena

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

With the assumption that x is a non-negative real, it can be safely simplified.That doesn't work if x is negative. If you are considering all the complex roots, then [itex]\sqrt[4]{x^2}[/itex] has 4 roots and [itex]\sqrt{x}[/itex] has 2.

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

If x can be a negative real, we have that √(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

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