How do you approximate #sqrt(17.02)#?

Possibly NOT the intended method since this was asked under "Using Newton's Method"
We could take #4# as a really gross estimate of #sqrt(17.02)# since #4^2=16# is closer to #17.02# than #5^2=25#

We know that #4# really is too small though,
so lets suppose our answer is #4+epsilon# (with #abs(epsilon) < 1#)
#color(white)("XXX")(4+epsilon)^2=17.02#

#color(white)("XXX")16+8epsilon+epsilon^2=17.02#

#color(white)("XXX")8epsilon =1.02-epsilon^2#

#color(white)("XXX")epsilon = 1.02/8-epsilon^2/8#

Since #abs(epsilon) < 1# then #epsilon^2# won't be very big and #epsilon^2/8# will be even smaller.
So as an initial estimate for #epsilon# we could try using
#color(white)("XXX")epsilon' = 1.01/8=0.12625#
in place of #epsilon# on the right side of this equation.

#color(white)("XXX")epsilon ~~(1.02-(0.12625)^2)/8#

(I used a calculator from this point on)

#color(white)("XXX")epsilon ~~0.125508#

#4+epsilon = 4.125508#

Testing this value as the root of #17.02#
#color(white)("XXX")4.125508^2~~17.01981#
which (to my mind) is close enough

If you wanted the answer using Newton's Method; here it is
as the product of some iterative expressions in a spreadsheet:

# sqrt(17.02) # is a solution to the equation
#x^2=17.02 => x^2-17.02 =0#

Let # f(x)=x^2-17.02 # then # f'(x)=2x # and we can use Newton's method using the iterative formula;

# x_n = x_(n-1) - f(x_(n-1)) / (f'(x_(n-1))) #

If we start with # x_0=4 #, then we can tabulate the results as follows (in this case using Excel);

So we see that very rapidly the method converges to a solution 4.1255302699 to 9 Decimal Places