# What is the square root of 26?

What is the square root of 26?

          You can only have an approximation: 5. 09901951. . .                                     <h4 class=answerHeader>Explanation:</h4>                                      The square root of a number <mathjax>%x%</mathjax> is a number <mathjax>%y%</mathjax> such that <mathjax>%y^2=x%</mathjax>.  So, were looking for a number <mathjax>%y%</mathjax> such that <mathjax>%y^2=26%</mathjax>.  Since <mathjax>%5^2=25%</mathjax> and <mathjax>%6^2=36%</mathjax>, the square root of <mathjax>%26%</mathjax> is between <mathjax>%5%</mathjax> and <mathjax>%6%</mathjax>.  Since there is no algorithm to compute it exactly, you can only have an approximation.  A possible way is the following: we know that <mathjax>%sqrt(26)%</mathjax> is between <mathjax>%5%</mathjax> and <mathjax>%6%</mathjax>.  So, since <mathjax>%5^2=25%</mathjax> and <mathjax>%5. 1^2=26. 01%</mathjax>, <mathjax>%sqrt(26)%</mathjax> must be between <mathjax>%5%</mathjax> and <mathjax>%5. 1%</mathjax>. Iterating this process gives you all the decimal digits you need.                       <hr>              %sqrt(26)% does not simplify, but you can calculate an approximation efficiently using Newton Raphson method as:%sqrt(26) ~~ 54100801 / 10610040 ~~ 5. 099019513592786%                                    <h4 class=answerHeader>Explanation:</h4>                                      <mathjax>%26 = 2 * 13%</mathjax> has no square factors, so <mathjax>%sqrt(26)%</mathjax> cannot be simplified. If you want to calculate an approximation by hand, then I would recommend a form of Newton Raphson method, starting with first approximation <mathjax>%a_0 = 5%</mathjax>. To iterate you can use the formula:<mathjax>%a_(i+1) = (a_i^2 + n)/(2a_i)%</mathjax>where <mathjax>%n = 26%</mathjax> is the number you are approximating the square root of. Personally, I like to deal with these approximations as rational approximations in the form <mathjax>%p_i/q_i = a_i%</mathjax> where <mathjax>%p_i%</mathjax> and <mathjax>%q_i%</mathjax> are integers as follows:<mathjax>%n = 26%</mathjax><br></br><mathjax>%p_0 = 5%</mathjax><br></br><mathjax>%q_0 = 1%</mathjax>Iterate using:<mathjax>%p_(i+1) = p_i^2 + n q_i^2%</mathjax><br></br><mathjax>%q_(i+1) = 2 p_i q_i%</mathjax>So:<mathjax>%p_1 = 5^2 + 26*1^2 = 25+26 = 51%</mathjax><br></br><mathjax>%q_1 = 2*5*1 = 10%</mathjax><mathjax>%p_2 = 51^2 + 26*10^2 = 2601 + 2600 = 5201%</mathjax><br></br><mathjax>%q_2 = 2*51*10 = 1020%</mathjax><mathjax>%p_3 = 5201^2 + 26*1020^2 = 27050401 + 27050400 = 54100801%</mathjax><br></br><mathjax>%q_3 = 2*5201*1020 = 10610040%</mathjax>Stop when you think you have enough significant digits (typically about the number of significant digits of <mathjax>%p_i%</mathjax> + the number of significant digits of <mathjax>%q_i%</mathjax>). <mathjax>%sqrt(26) ~~ 54100801 / 10610040 ~~ 5. 099019513592786%</mathjax>Actually <mathjax>%sqrt(26) ~~ 5. 099019513592785%</mathjax>