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Finding simplest radical form of a 4th root?

  1. Jan 15, 2013 #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?
  2. jcsd
  3. Jan 15, 2013 #2


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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]

  4. Jan 16, 2013 #3
    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?
  5. Jan 16, 2013 #4

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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:
    $$\sqrt[4]{x^{14}} = (x^{14})^{\frac 1 4} = x^{14 \cdot \frac 1 4} = x^{3 + \frac 1 2} = x^3 \cdot x^{\frac 1 2} = x^3 \sqrt x$$
  6. Jan 16, 2013 #5
    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].
  7. Jan 16, 2013 #6

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    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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