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Mathematics 16 Online
OpenStudy (anonymous):

Simplify:

OpenStudy (anonymous):

\[2 \sqrt[4]{10}\]

OpenStudy (anonymous):

@DSS @Sheraz12345

OpenStudy (sheraz12345):

Clearly No idea.. :/ Sorry

OpenStudy (anonymous):

10^(1/4)=1.78 2(10^(1/4)) = 3.56 Other than that, I think it's in the simplest form

OpenStudy (sheraz12345):

Thats what i thought but thats not simplification @DSS

OpenStudy (anonymous):

The answer choices are: \[4 \sqrt[4]{5}\] \[8 \sqrt[4]{5}\] \[16 \sqrt[4]{5}\] \[32 \sqrt[4]{5}\]

OpenStudy (anonymous):

Oh sorry, the question asks to simplify \[2 \sqrt[4]{80}\]

OpenStudy (anonymous):

I was looking at another question :P

OpenStudy (agent0smith):

\[\Large 2 \sqrt[4]{10} = 2 \sqrt[4]{10}\] that's it. You can't simplify the 4th root of 10.

OpenStudy (agent0smith):

Use prime factorization on 80. Look for numbers which you can put with an exponent of 4, since you're finding the 4th root. \[\Large 80 = 2*2*2*2*2*5=2^4 *2*5\] \[\Large 2 \sqrt[4]{2^4*2*5}= 2 \sqrt[4]{2^4}* \sqrt[4]{2*5}\]Try simplifying.

OpenStudy (anonymous):

I don't understand ._. Could you explain how I get the answer, I don't understand what you wrote. @DSS Could you help?

OpenStudy (agent0smith):

Do you understand prime factorization?

OpenStudy (agent0smith):

Circles are the prime factors |dw:1389124311454:dw| 80 = 2*2*2*2*2*5

OpenStudy (anonymous):

I know how to factorize, I just don't know how to get the simplified form from that.

OpenStudy (agent0smith):

Well then tell me where you got lost. This is still the original problem, we just factored 80 \[\Large 2 \sqrt[4]{2^4*2*5}= \]

OpenStudy (anonymous):

Okay, go on..

OpenStudy (agent0smith):

Break up the 4th root \[\Large 2 \sqrt[4]{2^4*2*5}= 2 \sqrt[4]{2^4}* \sqrt[4]{2*5}\]

OpenStudy (anonymous):

Gotcha so far..

OpenStudy (agent0smith):

What's the 4th root of 2^4? hint: the nth root of x^n is x.\[\huge \sqrt[n]{x^n} = x\]

OpenStudy (agent0smith):

We can't do anything with the 2*5 so leave that as 10.\[\Large 2 \sqrt[4]{2^4}* \sqrt[4]{10}\]

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