Simplyfing radicals: 4rt(64)

[Polyphile]

Hey guys,

I'm having a bit of a problem here; or maybe it's my math book, I don't know. Let me see if I can explain it.

In a logarithmic equation problem (I'm not asking about the logarithmic equation, just a specific part of what x turns out to be), there's a part where I have x^4 = 64. So, I take 4rt(64) and I simplify it. I eventually get 2*4rt(4), which is equal to about 2.83. However, my math book simplifies to 2sqrt(2), which also is equal to about 2.83. However, I can't figure out how they were able to simplify to a square root from a 4th root! Every resource I can find online simplifies to my answer (2*4rt(4)), but obviously my math book's (Saxon Algebra II) answer is also correct, because it is equal to my answer.

So, I know neither of these answers are "wrong", per se, since they represent the same number. However, I was wondering if anyone could explain to me how they got 2*sqrt(2), and if there is a process to simplify event roots (4th roots, 6th roots, whatever) into square roots, and whether this represents the most simplified form.

I hope I've made sense. Thank you so much!

 

[pka]

\(\displaystyle \sqrt[4]{{64}} = \left( {2^6 } \right)^{\frac{1}{4}} = 2^{\frac{3}{2}} = 2\sqrt 2\)

 

[Polyphile]

Thank you, sir, you are awesome!

I have a final question though; is such a simplification as \(\displaystyle 2\sqrt2\) preferred, or is \(\displaystyle 2\sqrt[4]4\) preferred?

 

[Deleted member 4993]

Thank you, sir, you are awesome!

I have a final question though; is such a simplification as \(\displaystyle 2\sqrt2\) preferred.........Yes,

or is \(\displaystyle 2\sqrt[4]4\) preferred...........No

?

 

[pka]

As a rule of thumb, it is best to reduce fractions: \(\displaystyle \sqrt[4]{4} = \left( {2^2 } \right)^{\frac{1}{4}} = 2^{\frac{1}{2}}\)

 

[Polyphile]

I understand. Thank you very much!