طرق حساب الجذر التربيعي

إذا وجد عددان بحيث ${\displaystyle a=x^{2}+y^{2}}$

${\displaystyle {\sqrt {a}}={\sqrt {x^{2}+y^{2}}}=x+{\frac {y}{{\frac {2x}{y}}+{\frac {1}{{\frac {2x}{y}}+{\frac {1}{{\frac {2x}{y}}+{\frac {1}{{\frac {2x}{y}}+...}}}}}}}}}$

لحساب ${\displaystyle {\sqrt {S}}}$, حيث S = 125348,

${\displaystyle x_{0}=6\cdot 10^{2}=600.000.\,}$

${\displaystyle x_{1}={\frac {1}{2}}\left(x_{0}+{\frac {S}{x_{0}}}\right)={\frac {1}{2}}\left(600.000+{\frac {125348}{600.000}}\right)=404.457.}$

${\displaystyle x_{2}={\frac {1}{2}}\left(x_{1}+{\frac {S}{x_{1}}}\right)={\frac {1}{2}}\left(404.457+{\frac {125348}{404.457}}\right)=357.187.}$

${\displaystyle x_{3}={\frac {1}{2}}\left(x_{2}+{\frac {S}{x_{2}}}\right)={\frac {1}{2}}\left(357.187+{\frac {125348}{357.187}}\right)=354.059.}$

${\displaystyle x_{4}={\frac {1}{2}}\left(x_{3}+{\frac {S}{x_{3}}}\right)={\frac {1}{2}}\left(354.059+{\frac {125348}{354.059}}\right)=354.045.}$

${\displaystyle x_{5}={\frac {1}{2}}\left(x_{4}+{\frac {S}{x_{4}}}\right)={\frac {1}{2}}\left(354.045+{\frac {125348}{354.045}}\right)=354.045.}$

هكذا, ${\displaystyle {\sqrt {125348}}\approx 354.045\,.}$

لحساب ${\displaystyle {\sqrt {S}}}$, حيث S = 27,

${\displaystyle x_{0}={\sqrt {25}}=5.\,}$

${\displaystyle x_{1}={\frac {1}{2}}\left(x_{0}+{\frac {S}{x_{0}}}\right)={\frac {1}{2}}\left(5+{\frac {27}{5}}\right)=5.2.}$

${\displaystyle x_{2}={\frac {1}{2}}\left(x_{1}+{\frac {S}{x_{1}}}\right)={\frac {1}{2}}\left(5.2+{\frac {27}{5.2}}\right)=5.196.}$

${\displaystyle x_{3}={\frac {1}{2}}\left(x_{2}+{\frac {S}{x_{2}}}\right)={\frac {1}{2}}\left(5.196+{\frac {27}{5.196}}\right)=5.196.}$

هكذا, ${\displaystyle {\sqrt {27}}\approx 5.196\,.}$