# How to Find Nth Roots by Hand

Here is that funny long division-like method for finding square and cube roots generalized to nth roots. These are all really extensions of the Binomial Theorem.

## Steps

- 1
**Partition your number.**Separate the number you want to find the nth root of into n-digit intervals before and after the decimal.*If there are fewer than n digits before the decimal, then that is the first interval. And if there are no digits or fewer than n digits after the decimal, fill in the spaces with zeroes.* - 2
**Find an initial estimate.**Find a number (a) raised to the nth power closest to the first n digits (or the fewer than n digits before the decimal) as a base-ten number without going over. This is the first and only digit of your estimate so far. - 3
**Modify the difference.**Subtract your estimate to the nth power (a^{n}) from those first n digits and bring down the next n digits next to that difference to form a new number, a modified difference. (Or multiply the difference by 10^{n}and add the next n digits as a base-ten number.) - 4
**Find the second digit of your estimate.**Find a number b such that (_{n}C_{1}a^{n - 1}(10^{n-1}) +_{n}C_{2}a^{n - 2}b (10^{n - 2}) ) + . . . +_{n}C_{n - 1}a b^{n - 2}(10 ) +_{n}C_{n}b^{n - 1}(100 ) )b is less than or equal to the modified difference above (10^{n}(d ) + d_{1}d_{2}. . . d_{n}). This becomes the second digit of your estimate so far.- The combinations notation
_{n}C_{r}represents n! divided by the product of (n - r)! and r!, where n! = n(n - 1)(n - 2)(n - 3) . . . (3)(2)(1). The notation_{n}C_{r}is sometimes expressed as n over r within tall parentheses without a division bar, and it can be calculated simply as the first r factors of n! divided by r!, which is often written as_{n}P_{r}divided by r!

- The combinations notation
- 5
**Find your new modified difference.**Subtract the two quantities in the last step above (10^{n}(d ) + d_{1}d_{2}. . . d_{n}minus_{n}C_{1}a^{n - 1}(10^{n-1}) +_{n}C_{2}a^{n - 2}b (10^{n - 2}) ) + . . . +_{n}C_{n - 1}a b^{n - 2}(10 ) +_{n}C_{n}b^{n - 1}(100 ) )b) to form your new modified difference by bringing down the next set of n digits next to that result. (Or multiply the difference by 10^{n}and add the next n digits as a base-ten number.) - 6
**Find the third digit of your estimate.**Find a new number c and use your estimate so far, a (which is now 2 digits), such that (_{n}C_{1}a^{n - 1}(10^{n - 1}) +_{n}C_{2}a^{n - 2}c (10^{n - 2}) + . . . +_{n}C_{n - 1}a c^{n - 2}(10 ) +_{n}C_{n}c^{n - 1}(100 ) ) c is less than or equal to the new modified difference in above (10^{n}(d ) + d_{1}d_{2}. . . d_{n}). This becomes the third digit of your estimate so far. - 7
**Repeat.**Keep repeating the last two steps above to find more digits of your estimate.- This is basically a rolling binomial expansion minus the lead term, where the two terms involved are the prior estimate multiplied by 10 and the next digit to improve the estimate.

## Article Info

Categories: Algebra