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## what is the square root of 58

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.

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 i2 is evenly dividable into N.  Each time you find that it is, you re-assign N/i2 to N, and continue the process, starting from the same i you just found.  If i2 is not evenly divisible into N, then you move on to the next prime number.  Once you get to the point where i2 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        -> i2 > 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.

For any number x, the square root of x = x^(1/2).

Take the derivative of x^(1/2)= (1/2)x^(-1/2)dx.

58 is between 7^2=49 and 8^2=64. It is closer to 64 than 49 so we will use 64 as the value of x.

64-58=6. This will be the value of dx, but with a negative sign.

(1/2)x^(-1/2)dx= (1/2)((64)^(-1/2))(-6)= -3/8.

8-(3/8)= 7 5/8.

(7 5/8)^2=58.14. (Pretty close to the calculator value of 7.62)

All the arithmetic can be done without a calculator. The actual value was

Obtained with a calculator.

You can use a calculator or use a handy pencil-paper method:

7 . 6   1   5   7 ...

Sqrt 58.00 00 00 00 00...

49 = 72

9 00

8 76 = (20*7+6)*6

24 00

15 21 = (20*76+1)*1

8 79 00

7 61 25 = (20*761+5)*5

1 17 75 00

1 06 61 49= (20*7615+7)*7

11 13 51 00

...

Roman, I haven't seen this method before... can you explain it in words because I am having a terrible time trying to follow the math with no idea how you are getting to the next step...

This is based on the fact that (10a+b)2 - (10a)2 = 20ab+b2 = (20a+b)b. The right hand side of this equation is easier to compute which is why it is used.

For example, (20*7+6)*6 = 762-702, and (20*76+1)*1=7612-7602

Here, 76 is the largest integer fitting into v5800 and 761 is the largest integer fitting into v580000, etc.

The algorithm builds the answer up digit by digit so that it's square exhausts the number whose square root is taken. For every next digit, the largest one for which the answer so far doesn't exceed the true square-root is chosen.

It is easy to verify the identity (10a+b)2 - (10a)2 = (20a+b)b, which is what is used starting at the second step onward. RHS is easier than LHS to calculate with.

For example (20*7+6)*6 = 762-702 and (20*76+1)*1 = 7612-7602 etc.

Notice also the 762 is the largest square not exceeding 5800 and 7612 is the largest square not exceeding 580000, etc.

Now I know why people use slide rules!