What is the sqaure root of 625?
960702
To take the square root of 625, the calculator would multiply by 5 to yield:\n\n3125\n\nThen it would start subtracting 100s+5 100s+15... from the leftmost digits: (s started at 0)\n\n3125\n 500-\n1500-   : 2 subtractions so first digit is 2 (so s is now 2)\nleaving 1125\n\nThen it subtracted 100s+5, 100s+15... (s = 2)\n\n1125\n 205-\n 215-\n 225-\n 235-\n 245- : 5 subtractions (so d is 5 which is appended to s yielding 25)\n        leaving 0\n\nSo the root of 625 is 25.
98145
the square root of 2 is 1.4142
1060817
Square Root:\n\nSQRT(1839472)=1356.271\n\nSQRT(75639)=275.0255\n\nSum:\n\nSQRT(1839472) + SQRT(75639)\n\n= 1631.2964 (ans)
1021313
Replace x with 6\n6^2 - 2(6) + k = 0\n36 - 12 + k = 0\n24 + k = 0\nk = -24\nThe other root is -24/6 = -4
108525
sorry, you asked it again. its 625
457868
It's possible to get 100 using even THREE eights!\nAnd this is in DECIMAL, not binary base!\n\n100 = -log_8 (log_8 (sqrt(sqrt(...sqrt(8))))),\n\nwhere log_8 is logarithm in base 8\nand the number of square roots is 300.\n\nLet's prove it.\n\nA = sqrt(sqrt(...sqrt(8))) {300 roots} =\n= ((8^(1/2))^(1/2))...^(1/2) {300 powers} =\n= 8^(1/2 * 1/2 * ... * 1/2) {300 multiples} = \n= 8^(1/8 * 1/8 * ... * 1/8) {100 multiples} = \n= 8^(1/(8^100)) =\n= 8^(8^(-100)).\n\nlog_8 (A) =\n= log_8 8^(8^(-100)) =\n= 8^(-100).\n\nlog_8 (log_8 (A)) =\n= log_8 8^(-100) =\n= -100.\n\nHence, -log_8 (log_8 (sqrt(sqrt(...sqrt(8))))) = 100.\n\nQED\n\nYou can use forth eigth, for example, to change every one 8 in that formula to sqrt(8*8).
26092
The Babylonian method is easy and fast:\n\nn[i+1] = (n + (r/n))/2\nwhere n is the approximate root, and r is the number you wish to find the sqaure root of. Check sources for more info.
807922
Here is how to do it:\n\nsqrt 500\nsqrt 100*5\n10 sqrt 5\n\nTherefore, you are correct.
399113
The square root of a number is a number that when you multiply it times itself you get that number.\n8*8 = 64 \nThus, the answer is 8.
934160
625/81\n\n(7.716049382..)
256504
because 56 is not an even root the best answer is=\n(n-squareroot of 56)(n+squareroot of 56). \nThe square root of 56 is 7.48331477355 if you need the exact number, but that's messy.
1120062
Actually, that's the approximation. The exact answer is: square root of 223423423432435. Of course I can't be sure because I've never heard of a mathematical operator called the "square route." I've heard of roots in mathematics, but not routes!
154621
If you mean 100 inside a square root symbol, the answer is 10.  However, if you mean the true square roots (as opposed the result of the square root function) the answer is 10 or -10.
608708
The basic formula for a circle is (x-h)^2 + (y-k)^2 = r^2\n\n(h,k) would be the center (where x=h and where y=k), and the radius is the positive square root of r^2\n\nSo in your example, there is no h or k, so the center would be (0,0) and the radius would be the positive square root of 25, which is 5.
1157812
There is a method called the binary search method that will find the \nsquare root of any number N.  You'll need a calculator to do this \nmethod.\n\nThe first step is to pick a value you know to be too small the be the\nsquare root of Pi.  Call this value S (for small guess).  The next\nstep is to pick a value you know to be too large to be the square root\nof Pi.  Call this value L (for large value).\n\nThen since Pi = 3.14159.... we'll say S = 1, and L = 2.\n\nS = 1\nL = 2\n\nThen find S^2 and L^2 (S and L squared):\n\nS^2 = 1\nL^2 = 4\n\nWhich of these values is farther away from Pi? S^2 is farther away \nfrom Pi then L^2, so replace S with the average of S and L.  Now we \nhave:\n\nS = 1.5\nL = 2\n\nContinue the process:\n\nS^2 = 2.25\nL^2 = 4\n\nReplace the small value again since its square is farther away.\n\nS = 1.75\nL = 2\n\nGoes to:\n\nS^2 = 3.0625\nL^2 = 4\n\nReplace the large value now:\n\nS = 1.75\nL = 1.875\n\nGoes to:\n\nS^2 = 3.0625\nL^2 = 3.515625\n\nReplace the large value again:\n\nS = 1.75\nL = 1.8125\n\nGoes to:\n\nS^2 = 3.0625\nL^2 = 3.28515625\n\nAt which point we know to replace the large value again since its \nsquare is farther away then the small value's square.\n\nSo\n\nS = 1.75\nL = 1.78125\n\nAt this point we know the square root of Pi lies between 1.75 and \n1.78125. Continuing this process we can get finer and finer bounds for \nthe value of the sqaure root of Pi.  After a while we will see that \nthe square root of Pi equals 1.77245....  Since Pi is irrational we \ncan never get an exact answer and we have to settle for an \napproximation.\n\nHope this helps.\n\n-Doctor Steven,  The Math Forum\n Check out our web site!  http://mathforum.org/dr.math/   \n    \n\nI hope this helps you a little sorry if it does not!
188735
x^256-1 is a polynomial of degree 256. It has 2 pure real roots (+1, -1)  and 254 complex roots. They are all of the form (cos(n2pi/256) + isin(n2pi/256)) n = 0, 1, ...255.\nIt has 256 factors of the form (x-a) where a is one of the 256 roots.
132069
You can find some of these at:\n\nhttp://medialab.freaknet.org/bignum/numbers.html\n\nThere is an algorithm for calculating square roots - but it takes forever once you get past 5 or 6 decimal places.