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## Comment on

Properties of Roots## Hello Mr Hanneson, is there

thankyou

## Great question!

Great question!

There are a few roots that are easy to calculate.

For example, if want want to calculate the 4th root of 81, we can first find the square root of 81, and then find the square root of the result.

Here's why this works:

The fourth root of x = x^(1/4)

The square root of x = x^(1/2)

Now notice that [x^(1/2)]^(1/2) = x^(1/4)...voila!

We can also use a similar strategy to find the 8th root of x

To calculate the 8th root of x, first find the square root of x, and then press the square root button a second time, and then press the square root button a third time

Here's why it works:

The eighth root of x = x^(1/8)

Now notice that [[x^(1/2)]^(1/2)]^(1/2) = x^(1/8)...tada!!!

For reference, here's the link to the lesson covering the GRE: https://www.greenlighttestprep.com/module/general-gre-info-and-strategie...

Cheers,

Brent

## Hi,

in the following question answer is D

https://gre.myprepclub.com/forum/if-x-0-then-must-be-equal-to-0-07x8-0-07x4-0-7x14-0-7x8-11477.html

the result is correct but it says A instead of D.

## Good catch!

Good catch!

I've edited my response.

Cheers and thanks!

Brent

## Hey brent.

Since √x = x^1/2, can I say that √x^4 = x^(4/2) = x^2 ?

## Sorry for the delay. I'm not

Sorry for the delay. I'm not sure how I missed this question.

Yes, if x ≥ 0, we can say that (√x)^4 = √(x^4) = x^(4/2) = x^2