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Dean
 
 

How do you find the square root of a decimal

by Dean Wed Aug 15, 2007 6:25 pm

Hello,

I would like some help with the process used when finding the any numbered root of a decimal. Example, what is the fourth root of .00081

Please help,

Dean
christiancryan
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Posts: 79
Joined: Thu Jul 31, 2003 10:44 am
 

by christiancryan Thu Aug 16, 2007 11:36 am

Hi Dean,

The key is to rewrite the decimal using powers of ten. The reason is that you split up the decimal into two parts: (1) a simple number without decimals and (2) a power of ten. Then you take the appropriate root of each part, and then finally you recombine the parts.

Let's look at a modified version of your example (I think I've seen this problem before), using 0.0081, not 0.00081.

0.0081 = 81 * 10^-4 (that's "eighty-one times ten to the negative fourth power")

Notice that I didn't write 8.1 x 10^-3 (which is technically what you would do for scientific notation). The point is to have the first number be "simple," so that you can take the fourth root of it.

Now, 81 is the fourth *power* of a number: 81 = 9^2 = (3^2)^2 = 3^4. You look for clues like this. The GMAT will *never* have you take the root of a number that's weird. (Occasionally you may have to estimate a root very roughly, but you only need to use nearby simple roots: e.g., the square root of 12 is between the square root of 9 and the square root of 16 -- that is, between 3 and 4.)

So I can rewrite the problem this way:

0.0081 = 81 * 10^-4 = 3^4 * 10^-4

Now, when you take the fourth *root* of a number, you are taking it to the 1/4th power. (Note -- do NOT confuse this with taking a number to a negative power. A nth root is always taken by exponentiating to 1/n.

So

0.0081^(1/4) = (3^4 * 10^-4)^(1/4)

Now, when you take a number that's already raised to a power and raise it to another power, you multiply those powers:

0.0081^(1/4) = (3^4 * 10^-4)^(1/4) = 3^(4*1/4) * 10^(-4*1/4) = 3^1 * 10^-1 = 0.3

Definitely work this back out on your own -- "put it through your pen," so to speak.

Hope this is helpful,
Chris