Can you find the square root of a number without using a calculator?
537920
It's rough, but what I would do is find the difference between the two routes on either side of the number (Wait for the example!)  and count until you get to it, then make a fraction. \n\nEXAMPLE:\nfind square root of 2.\n\n2 is between 1 and 4.\n\nit's 1 past 4 and there are 3 numbers between the two.  So root 2 is approxamately 1 1/3.  The 1 is from the first number between one and 4, and the 3 is because there's 3 numbers between 1 and 4.  It's confusing, and it's not a good estimate, but it works if you understand it.
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
579007
square_root(2/7)=0.5345224838\n\ncosx=2cos(x/2)^2 -1 \n2cos(x/2)^2=(cosx+1)\n2cos(x/2)^2=(-3/7+1)\ncos(x/2)^2=2/7\ncos(x/2)=sqrt(2/7)
458909
You can goto http://www.math.com 's square root calculator and figure out that it is about 20.78 so since an integer is a decimal-less number the answer would be between 20 and 21.\n\nShort of using the sqrt function of a calculator if you know that 20 squared is 400 and that is close to 43 you might then try squaring 21 and get 441 then realize 432's sqrt must be between 20 and 21.
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.
999326
If you start with a number bigger than one, then the square root will always be smaller than the number, but bigger than 1.  So every time you take the square root, the number gets smaller, but never gets smaller than one.  Eventually it's so close to one that the calculator just displays it as 1.
57974
Not neccessarily. For instance, the square root of two (which is irrational) multiplied by the square root of two is two, which is very rational.\n\nGood Luck!\n\nP.S. Great Screenname. I love the phantom.
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0
218797
sqrt(157452304)= +/-12548\n\ndon't forget that 2 negatives cancel out so every number has 2 square roots (either rational or irrational) , one positive and one negative.
852187
Saw your edit, see below...\n\nSo if I understand correctly, you start with a decimal number that has a finite number of decimals, like 0.41 or 0.87293, and you want to know if you can build an irrational number that begins with those same decimals, expressing it as a function of integers only.\n\nIn your example, this can be done with 0.41, because sqrt(2) = 1.41421... (the left side of the decimal doesn't matter to you).\n\nSure then.  There are infinitely many ways to do this, but here's just one:\n\nGiven the number "0,a" having n decimals.  So for 0.41, a=41 and n=2.  Then the given number can be expressed as a/10^n.\n\nIf you take sqrt(2)=1.41421... and divide it by 10 often enough, you'll get the number 0.000...0141421... which is irrational.  All you need to do then is add this to the original number.  Just make sure the first 1 appears on the (n+1)th decimal so as not to perturb the first n decimals.  This means dividing by 10^(n+1).\n\nSo the number a/10^n + sqrt(2)/10^(n+1) satisfies the conditions.  Everything is there is integers.\n\nIn the 0.41 case, this would yield the solution 41/100 + sqrt(2)/1000 = 0.41141421356...\n\nEdit: So you don't like it because it uses a.  OK, how about this:\n\nDefine "0,x" as the square of "0,a".  By adding an extra decimal to "0,x" and taking its square root, you will get "0,a" plus an irrational component.  This can be expressed as \n\nsqrt(x/10^n+1/10^(n+1))\n\nwhere n is the number of decimals for "0,x".  In the 0.41 example, this yields the number sqrt(1681/10000 + 1/100000) = sqrt(0.16811) = 0.410012194...
1010349
Hi:\n\nI know of six ways to find a square root of a number:\n\n1) Take the logarithm of number and divided by 2 take the result and take the Antilog \n\n2) Use the School boy method of extracting a square root see this website for how :\n\nwww.nist.gov/dads/HTML/squareR...\n\n3) Use the following method ( example on the bottom of page ):\n\n(N/gn+gn)/2= gn\n\nN= number to be Square root \ngn = Guess number or Square root\n\nrepeat this for about about 5 to 10 times ( see example)\n\n4) Use trail and error by running a number value and Square it\n\n5) Graph and interpolarite it\n\n6) Look it up in a Mathematical hanbook of tables and formulas and look under Square & square roots tables.\n\n------------------------------...\n\n\n\nExample:\n\nThe square root of 27:\n\n(27/5+5)/2= gn\n\n\n(5.4+5)/2 = gn\n\n10.4/2= gn \n\n5.2 =gn \nend of run 01 \n\n(27/5.2 + 5.2)/2= gn\n\n(5 + 5.2)/2 = gn\n\n10.2/2= gn \n\n5.1= gn\n\nend of run 02\n\n((27/5.1)+ 5.1) /2= gn\n\n(5.294117647+5.1)/2 = gn\n\n10.394117647/2 = gn\n\n5.197058823 = gn\n\nEnd of run 03\n\n(27/ 5.197058823) + 5.197058823 )/2 = gn\n\n(5.195246179+ 5.197058823 )/ 2 = gn\n\n10.392305003/2 = gn\n\n5.196152501 = gn\n\nEnd of run 04 \n\n((27/5.196152501)+ 5.196152501)/2 = gn \n\n(5.196152343+5.196152501)/2 = gn\n\n10.392304844/2 = gn\n\n5.196152422= gn \n\nEnd of run 05\n\n((27/ 5.196152422) + 5.196152422 )/2 = gn\n\n(5.196152422 + 5.196152422 ) /2 = gn\n\n5.196152422 = gn \n\nEnd of run 06\n\nSquare root of 27 appoximatily = 5.196152422 ( accurate to 9 decimal places.) \n\nSource(s):\n\nwww.nist.gov/dads/HTML/squareR...\n\nwww.du.edu/~jcalvert/math/sqrt... \n\nmathforum.org/library/drmath/v...
302292
There's no solution.\n\nIf you're evil, you could argue that 102 - 78 is an answer ;)\n\nEDIT : Confirmed no solution with Telraam (http://members.chello.nl/g.de.blaauw/telraam.html). Allowing square roots gives (8*(2+.10))/.7 as a potential answer.
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)
239483
zero  you said squared, not to the zero power.  0^2=0  try it on your calculator.
766787
If this is a problem in a math class, the answer provided by abcde12345 is correct. He shows you the technique to simplify it, which is to multiply numerator and denominator by the "conjugate" of the denominator (which just means you change the sign of the square root part, so you get (3 - sqrt(5)).\n\nIf this is a problem in physics or some other class, and all you need is a value, then follow the suggestion of some others who recommend using a calculator.
1144863
Think or find the square root of the perfect number previous non-perfect and the square root of perfect number next  of non-perfect number.\nExample\nfind the square root of 23\n16<23<25\nsqrt(16)<sqrt(23)<sqrt(25\n4< sqrt(23) <5\nanswer: the square root of 23 is between 4 and 5
875144
you have to take the square root of each side, then just simplyfy\n\nx = 1.2472191289247