How do you find the square root of a number by hand?
536258
It is not possible to find out sqare root of number by hand.I think Terror2287 is wrong.
597652
The square root of 1 is 1.\nsqrt(6.6) = 2.569\nsqrt(15) = 3.873
105888
square root(pi) = 1.77245385
352163
The circumference divided by pi then get the square root of that.
998294
square root of 13 is the answer
801012
inverse square root of a number say"x" is\nx^[-(1/2)]
709225
Finding the square root\n\nClick on the URL below for additional information concerning finding the roots without a caculator\n\nwww.nist.gov/dads/HTML/squareRoot.html \n\nwww.homeschoolmath.net/teaching/square-root-algorithm.php \n\n- - - - - - -s-
634605
Mathematicians have designated a special number 'i' which is equal to the square root of minus 1. Then, it follows that i2 = -1. To determine the square root of a negative number (-16 for example), take the square root of the absolute value of the number (square root of 16 = 4) and then multiply it by 'i'. So, the square root of -16 is 4i.\nAs a double check, we can square 4i (4*4 = 16 and i*i =-1), producing -16.\nAll negative square roots are called "imaginary numbers" (now you know where that letter 'i' comes from).
1113001
take squre root of both sides\nx-1=+ or - sqrt(7)\n\nthen \nthe first root \nx1=1-sqrt(7)\n\nthe second root\nx2=1+sqrt(7)\n\nBy factoring method:\n\ntake 7 to left\n(x-1)^2-(sqrt(7))^2=0\nfactor like a^2-b^2=(a-b)(a+b)\n\n[x-1-sqrt(7)] * [x-1+sqrt(7)]=0\n0*0=0\nx=1+sqrt(7)\nor\nx=1-sqrt(7)
171301
The square root of 3 is an irrational number, however, we can approximate it using a bit of calculus.\n\nWe know the sqrt of 4 = 2.  4 is pretty close to 3, so we can estimate the sqrt of 3 to be,\nsqrt(4) - (1/(2*sqrt(4)))*(1) = 1.75\n\nBy plugging sqrt(3) into a calculator, we get about 1.732060808, which is pretty close to our estimate.  The close the numebr you choose to 3, the closer your estimate will be to the "true" value.
946285
square root of 2x+3=6\n<=> 2x+3 = 36\n<=> x = 33/2
522755
Download XSET from the link given below and use its /MATH option to get any power of any number. Check the documentation and you should be using something like the following in your batch script to get the square root of 34:\n\nXSET /MATH var = "34^0.5"\n\nCheck the exact syntax from the file xset.txt in the file mentioned below. You can also read more about it from the second link below.
909546
The square root of a number is just the number which when multiplied by itself gives the first number. So 2 is the square root of 4 because 2 * 2 = 4.\n\nStart with the number you want to find the square root of. Let's use 12. There are three steps:\n\n   1. Guess\n   2. Divide\n   3. Average.\n\n      ... and then just keep repeating steps 2 and 3. \n\nFirst, start by guessing a square root value. It helps if your guess is a good one but it will work even if it is a terrible guess. We will guess that 2 is the square root of 12.\n\nIn step two, we divide 12 by our guess of 2 and we get 6.\n\nIn step three, we average 6 and 2: (6+2)/2 = 4\n\nNow we repeat step two with the new guess of 4. So 12/4 = 3\n\nNow average 4 and 3: (4+3)/2 = 3.5\n\nRepeat step two: 12/3.5 = 3.43\n\nAverage: (3.5 + 3.43)/2 = 3.465\n\nWe could keep going forever, getting a better and better approximation but let's stop here to see how we are doing.\n\n    3.465 * 3.465 = 12.006225\n\nThat is quite close to 12.
98145
the square root of 2 is 1.4142
918825
Yes. When 'c' is given, find the value of c^2.  Then, find out the possible numbers, whose sum = that number. Find out the root of all these numbers, till you arrive a perfect square root( if possible ).\n\nEg: when given, c = 10, find c^2. Here, it is 100. \n   therefore, a few possibilities are : 50 + 50 ( none are perfect squares)\n                                                         25 + 75 ( only 1 of them is a perfect square)\n                                                         36 + 64 ( both are perfect squares)\n\n Therefore, the answers being looked for are 6 and 8.
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!
879678
The first thing to do is to find the factors of the number you're taking the root of.\nFor example, teh factors of 24 include 2*12, 3*8, and 4*6.\nSince four is a perfect square (2^2), we'll use that set of factors.\nSo:\nsqrt(24)= sqrt(4*6)= sqrt(4)*sqrt(6)= 2sqrt(6)\n\nSame with sqrt(18):\n18=9*2\n9=3^2\nsqrt(18)= sqrt(9)*sqrt(2)= 3sqrt(2)\n\nsqrt(3) +sqrt(3) can only be reduced by realizing that it's\n2*sqrt(3), which could be the sqrt(12). THis is because you can take the two under the radical by squaring it. so  you get sqrt (2^2+3)= sqrt(12), but technically 2sqrt(3) is more reduced.\n\nsqrt(27)=3 sqrt(3)\nsqrt(12)= 2 sqrt(3)\nso 3sqrt(3) + 2sqrt(3) = 5sqrt(3)\nin this case, you're treating the sqrt(3) sort of like it was a variable. think of 3x +2x = 5x