詳解：(1)  \(x = \sqrt 3 \; \Rightarrow {x^2} = 3\quad  \Rightarrow {x^2} - 3 = 0\)

(2) 牛頓求根法之遞迴公式為

\({x_{n + 1}} = {x_n} - \frac{{f({x_n})}}{{f'({x_n})}}   \Rightarrow {x_{n + 1}} = {x_n} - \frac{{x_n^2 - 3}}{{2{x_n}}}  \Rightarrow {x_{n + 1}} = \frac{{x_n^2 + 3}}{{2{x_n}}}\)

(3)依序將數字代入遞迴公式

\({x_1} = \frac{{{2^2} + 3}}{4} = \frac{7}{4} \buildrel\textstyle.\over= 1.75\)

\({x_2} = \frac{{{{1.75}^2} + 3}}{{2 \cdot \left( {1.75} \right)}} = \frac{{6.0625}}{{3.5}} \buildrel\textstyle.\over= 1.732\)