Root of number - this practice passes all test cases, using binary search, error range is defined as 0.0001 instead of 0.001, round up value is used.

RootOfNumber_Practice.cs

using System;

class Solution
{
    public const double RANGE = 0.0001; 
  
    public static double Root(double x, int n) // x = 7, n= 3
    {
      if(x < 0 || n <= 0) //false
      {
        return -1; 
      }
      
      return binarySearch(x, n, 0, x); // 0.0001, 0, 7
    }
  
    private static double binarySearch(double x, int n, double start, double end)
    {
      // base case
      var isSame = Math.Abs(end - start) < RANGE; 
      if(isSame)
      {
        return toThreeDecimal(start); // edge 
      }
      
      var middle = start + ( end - start)/ 2; // middle = 7.0/2 = 3.5 
      
      var middleValue = Math.Pow(middle, n); // 3.5 ^3 = 27 > 7 
      
      var diff =  x - middleValue; // 7 - 27
      
      var found = Math.Abs(diff) < RANGE; // false 
      var less  = diff  > RANGE; // middleValue < x , false 7 < 3.5^3 
      
      if(found)
      {
        return toThreeDecimal(middle); // base case 
      }
      
      if(less)
      {
        start = middle + RANGE; 
        
      }
      else 
      {
        end = middle - RANGE; // 3.5 - 0.0001 = 3.4999
      }
      
      return binarySearch(x, n, start, end);       
    }
  
    private static double toThreeDecimal(double value) // 0.1234 -> 0.123, 0.125 -> 0.126 
    {
      int round1000 = (int)(value * 1000 + 0.5); 
      return round1000/ 1000.0; // round up 
    }

    static void Main(string[] args)
    {
        Console.WriteLine(Root(7, 3)); 
    }
}
// x = 7, n = 3 
// 1.913
// binary search tree 0 - 7 -> middle = 3.5, 3.5^3 > 7 , right half 
// Math.abs (middle^3 - 7) < 0.0001 -> 
// API powerFunction(x, n)  -> x^n -> Math.power
// x >= 0, 0^3 = 0 
// n > 0