How do you find #sqrt(1521)# ?

1 Answer
Apr 15, 2017

Answer:

#39#

Explanation:

This link here lists several methods to compute square roots: https://en.wikipedia.org/wiki/Methods_of_computing_square_roots

The digit-by-digit method is an extremely useful method to find exact answers but is extremely slow.

I will use an estimation method here. Don't worry if you do not understand everything in the following paragraph.

Newton's method states that the zeros of a function #P# can be approximated as #x_1=x_0-(P(x_0))/(P'(x_0))#, where #x_0# is our initial guess, #x_1# is our next guess, and #P'# is the derivative of #P# with respect to #x#. Finding the square root of a number #A# is essentially finding a number #k# such that #k^2-A=0#. Substituting these in, we get #x_1=x_0-(x_0^2-A)/(2x_0)=(x_0^2+A)/(2x_0)#.

This means that, if we have a guess #x_0#, we can get a better guess by calculating #(x_0^2+A)/(2x_0)#, where #A# is the number which square root we are trying to find. We can do this again with our new guess to get more accurate answers.

Take note of three things. First, our new guesses will always be slightly larger than the actual answer (except when the guess is exactly the answer, then the new guess would be equal to the previous guess). We can round our final answer down to account for this. Second, some guesses may not work (for example, if our first guess is #0#, we would be dividing by #0#). Third, the closer the initial guess is to the actual number, the better the new guess would be.

Let's first try finding the square root of #121#. #121# is really close to #100#, which square root is #10#. So guess #x_0=10# first. Then, compute #(x_0^2+A)/(2x_0)=(10^2+121)/(2*10)=(100+121)/(2*10)=221/20=11 1/20#. This is pretty close (slightly larger) to the actual answer #11#. You can repeat the whole process again to get more accurate answers.

Now, find the square root of #1521#. Suppose that we guess the number #35#. Then, #(35^2+1521)/(35*2)=2746/70=39 8/35#. Notice that #35# is really far away from #39 8/35#. This means that our new answer is likely not really accurate. But it would be really tedious to use #39 8/35# in our calculations. Let's round down to #39# (remember that all guesses from this method is slightly larger than the actual number). Then, #(39^2+1521)/(39*2)=3042/78=39#. Since our previous guess is equal to this guess, it means that #39# is the exact answer.