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Finding simplest radical form of a 4th root? 
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#1
Jan1513, 08:42 PM

P: 82

I haven't taken math in years and am having trouble understanding how to find simplest radical form of a ^{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? 


#2
Jan1513, 09:37 PM

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[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
Jan1613, 06:06 AM

P: 427




#4
Jan1613, 06:29 AM

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P: 6,188

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


#5
Jan1613, 10:26 AM

P: 887

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
Jan1613, 10:46 AM

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P: 6,188

If x can be a negative real, we have that √(x^{2}) = x and ^{4}√(x^{2})=√x. However, in general we need to be very careful with negative real numbers and fractional powers. They are generally not welldefined. 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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