Square root (sqrt) in JavaScript using a binary search

sqrt.js

/*

sqrt(8)

(0 + 8) / 2 = 4
4 * 4 = 16
(0 + 4) / 2 = 2
2 * 2 = 4
(2 + 4) / 2 = 3
3 * 3 = 9
(2 + 3) / 2 = 2.5
2.5 * 2.5 = 6.25
...

*/

function _sqrt(number, low_bound, high_bound) {
  let guess = (low_bound + high_bound) / 2;
  let aproximation = guess * guess;

  if (Math.abs(number - aproximation) < 0.01)
    return guess;
  
  if (aproximation > number)
    high_bound = guess;
  else
    low_bound = guess;

  return _sqrt(number, low_bound, high_bound);
}

function sqrt(number) {
  if (number < 0)
    return -1;
  
  return _sqrt(number, 0, number);
}

console.log(sqrt(8));
// 2.828125

if (Math.abs(number - aproximation) < 0.01) - this will not give you accurate result
use tolerance = 1e-8 or tolerance = 1e-14 for accurate result
if (Math.abs(number - aproximation) < tolerance)