Answer
$$\sqrt 5 \approx 2.236067978$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2} - 5,{\text{ so }}f'\left( x \right) = 2x \cr
& {\text{Using the Newton's Method }}{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}{\text{ }} \cr
& {\text{Then evaluating }}f\left( {{x_n}} \right){\text{ and }}f'\left( {{x_n}} \right) \cr
& {x_{n + 1}} = {x_n} - \frac{{x_n^2 - 5}}{{2{x_n}}} \cr
& {\text{Taking }}{x_1} = 1 \cr
& {x_1} = 1 \cr
& {x_{1 + 1}} = {x_2} = 1 - \frac{{{{\left( 1 \right)}^2} - 5}}{{2\left( 1 \right)}} = 3 \cr
& {x_{2 + 1}} = {x_3} = 3 - \frac{{{{\left( 3 \right)}^2} - 5}}{{2\left( 3 \right)}} \approx 2.333333333 \cr
& {x_{3 + 1}} = {x_4} = 2.333333333 - \frac{{{{\left( {2.333333333} \right)}^2} - 5}}{{2\left( {2.333333333} \right)}} \approx 2.238095238 \cr
& {x_{4 + 1}} = {x_5} = 2.238095238 - \frac{{{{\left( {2.238095238} \right)}^2} - 5}}{{2\left( {2.238095238} \right)}} \approx 2.236068896 \cr
& {x_{5 + 1}} = {x_6} = 2.236068896 - \frac{{{{\left( {2.236068896} \right)}^2} - 5}}{{2\left( {2.236068896} \right)}} \approx 2.236067978 \cr
& {x_{6 + 1}} = {x_7} = 2.236067978 - \frac{{{{\left( {2.236067978} \right)}^2} - 5}}{{2\left( {2.236067978} \right)}} \approx 2.236067978 \cr
& {\text{We obtain that }}{x_6} \approx {x_7} \cr
& {\text{Thus}}{\text{,}} \cr
& \sqrt 5 \approx 2.236067978 \cr} $$
