Finding simplest radical form of a 4th root?

  • #1
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?
 

Answers and Replies

  • #2
symbolipoint
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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.

[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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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]
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?
 
  • #4
I like Serena
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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:
$$\sqrt[4]{x^{14}} = (x^{14})^{\frac 1 4} = x^{14 \cdot \frac 1 4} = x^{3 + \frac 1 2} = x^3 \cdot x^{\frac 1 2} = x^3 \sqrt x$$
 
  • #5
Khashishi
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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?
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].
 
  • #6
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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?
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].
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
 

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