How do you approximate #sqrt(17.02)#?

Question

Christopher Agresta · Accepted Answer

4.1255302699 to 9 Decimal Places

# 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

Aurora Alvarado · Answer

#sqrt(17.02)~~color(green)4.125508#

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

Ezra Adams · Answer

#sqrt(17.02)~~4.125531# (after 20 iterations using the Newton Method)

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

Ezekiel Alexander · Answer

undefined One way to approximate √17.02 is to recognize that it falls between √16 (which is 4) and √25 (which is 5). Since 17.02 is closer to 16, we can start with 4.1. Squaring 4.1 gives 16.81, which is close to 17.02. So, a reasonable approximation for √17.02 is approximately 4.1.