apply the method of bisectIon to compute sqr3

(1) Consider the positive root of f(x) = x2 - 3 = 0, we start the iteration from the range 1 < x < 2 since f(1) < 0 and f(2) > 0 as follows:

圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Jun08/Crazynumeric1.jpg

So, corr. to 4 sig. fig., square root of 3 is 1.732.
(2) To give a 12 sig. fig. value, it should be 11 decimal places and therefore, the criteria should be, when assuming that n steps of bisection is required:
(1/2)n < 0.5 x 10-11
(1/2)n-1 < 10-11
- (n - 1) log 2 < -11
(n - 1) log 2 > 11
n - 1 > 11/log 2 = 36.5
n > 37.5
n = 38.
So 38 steps of bisection is needed.

係唔係要由
x^3+x^2+x^1+C = 0
變成(x+a)(x+b)(x+c)=0??