math problem please help

Wyatt,

Your question is a bit ambiguous.... In the answers below, George has used calculus to iteratively estimate and then drive closer to the answer (this is the fundamental procedure of Newton's method which Robert also demonstrated in the comment-answer). Roman has provided a somewhat complicated way to calculate square roots by hand. Both are valid and correct methods, and both provided approximations to the calculator decimal answer of 7.615773105863909... (The true answer written in decimal form never ends and never repeats, by the way, which means it is an irrational number.)

However, there is another answer that you may be expecting, or if not, you are likely to use in the future. This answer preserves the exact true value of the answer, but does not usually eliminate the radical (it only does this if the number is a perfect square). This is called "simplifying the radical" and is performed like this:

To simplify the square root of a number N, start from i=2, and decide whether i^{2} is evenly dividable into N. Each time you find that it is, you re-assign N/i^{2} to N, and continue the process, starting from the same i you just found.
If i^{2} is not evenly divisible into N, then you move on to the next *
prime *number. Once you get to the point where i^{2} is greater than N, or if N=1, or if N is prime, then you stop. Then, all of the i's you took out get multiplied together, and that gets multiplied by the remaining square root of N.

Here's how to do this for N = 96 (not the number you are asking about):

`N=96 i=2 96/4 = 24 -> Take out a 2`

N=24 i=2 24/4 = 6 -> Take out a 2 (again)

N=6 i=2 6/4 -> Doesn't divide evenly

N=6 i=3 9 > 6 -> STOP

Now you multiply all the i's you took out, and multiply that by the square root of what was left. So the sqrt(96) = 2 * 2 * sqrt(6) = 4 * sqrt(6)

Here's a more complicated example - sqrt(19800):

`N=19800 i=2 19800/4 = 4950 -> Take out 2`

N= 4950 i=2 4950/4 -> Doesn't divide evenly

N= 4950 i=3 4950/9 = 550 -> Take out 3

N= 550 i=3 550/9 -> Doesn't divide evenly

N= 550 i=5 550/25 = 22 -> Take out 5

N= 22 i=5 22/25 -> i^{2} > N - STOP

Therefore, sqrt(19800) = 2 · 3 · 5 · sqrt(22) = 30·sqrt(22)

Your question, sqrt(58) is much simpler, because it is the product of two primes, 2 and 29, and so nothing can be "taken out", and thus cannot be reduced.

## Comments

Wyatt,

You can use Newton's method.

f(x) = x^2 - 58

f'(x) = 2x

x(0) = 7

x(i+1) = x(i) - f(x(i))/f'(x(i))

x(1) = 7.642857

x(2) = 7.615821

...

x(10) = 7.615773106 <==Very close to v(58)

Newton's method has a benefit of quadratic convergence (the new error is about the square of the old error with each step). Thus in the long run, the number of digits of accuracy doubles with each step. Such fast convergence only occurs if the root has multplicity 1, so that f(r)=0 but f'(r) ? 0, just like in this case.

The comment system replaced my "not equal to" sign with a question mark.