Is there a trick for finding square numbers?
455350
Kind of, if you look at the numbers below:\n\n1,4,9,16,25,36,49,64,...\n\nYou can see that they are increasing by the numbers:\n\n3,5,7,9,11,13,15,...\n\nSo you can find the next square number if you have two preceding ones
972961
Sites are there. But I would recommend to sole in this form\n\nthere is no x term.\n\nso collect x^2 and constants on 2 sides\n\n4x^2 = 9-189 = -180\nsimplify \n\nx^2 = -45\n\ntake square root of both sidex\n\nx = +/-3 sqrt(5) i\n\nIf yo still require a site you can visit
87810
Perfect squares between 1 and 49 are:\n1*1=1,\n2*2=4,\n3*3=9,\n4*4=16,\n5*5=25,\n6*6=36,\n7*7=49,\ntherefore \n\n49 = 49\n\nOnly one term is needed
886898
Find next closest square to 80 that is also a factor: 16\n\nSince 4*4 is 16, you can bring the 4 outside the radical and leave the product of the other factors underneath the radical: 4rt(5)
716502
CLUE 1: The number squared < 1200\nTake square roots: the number < 34.64...\nIt is an integer, so the number ≤ 34.\n\nCLUE 2: The number cubed > 29000\nTake cube roots: the number > 30.72...\nIt is an integer, so the number ≥ 31.\n\nCLUE 3: The fourth power of the number < 930000\nTake fourth root: the number < 31.05...\nIt is an integer, so the number ≤ 31.\n\nFrom CLUE 2 and CLUE 3 the number must be 31 (which is consistent with CLUE 1).
832512
either square root or erauqs
671808
4 sides and corners in a square
621145
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
416099
The simplest, but most tedious, way to calculate the decimal digits of a square root is to figure out each digit in turn, as follows. \nFirst of all, find how many digits to the left of the decimal point your square root will have. \n\nA number with one digit to the left of the decimal point will be somewhere from 1 up to, but not including, 10; therefore, its square will be somewhere from 1 up to, but not including, 100, and hence will have either one or two digits to the left of the decimal point. \n\nA number with two digits to the left of the decimal point will be somewhere from 10 up to, but not including, 100; therefore, its square will be somewhere from 100 up to, but not including, 10000, and hence will have either three or four digits to the left of the decimal point. \n\nAnd so on. \n\nTherefore, if the number you're trying to find the square root of has one or two digits to the left of the decimal point, its square root will have one; if it has three or four, its square root will have two; and so on. \n\nNow, having found how many digits the square root has, just start from the left and figure out each digit in turn. \n\nFor example, to find the square root of 1234 to 3 decimal places: \n\nThe square root has two digits to the left of the decimal point, so we're looking for a number of the form AB.CDE. . . (and we want to find A through E). \n\nNow, 10^2=100, 20^2 = 400, 30^2=900, all too small, but 40^2=1600, too big. This tells us the first digit A is 3. \n\nNext, 31^2 = 961, 32^2 = 1024, 33^2 = 1089, 34^2=1156, 35^2 = 1225, all too small; 36^2 = 1296, too big, so the second digit B is 5. \n\nThirdly, 35.1^2 = 1232.01, too small, but 35.2^2 = 1239.04, too big, so the third digit C is 1. \n\nAnd so on; continuing in this way, you see that . \n\nThis method is horribly slow and inefficient but that's no problem in an era of calculators. All the necessary ideas are there. \n\nYou can apply lots of little shortcuts to this procedure to drastically reduce the number of computations you need to do. Then you come up with a procedure similar to the long-division algorithm. Its complicated to remember, though; I don't remember it offhand. It is seldom taught today because there is little need for it. \n\nTo actually calculate a square root to a large number of digits quickly, one uses a different method. For example, saying that x^2 = a is the same as saying x = (1/2)(x + a/x) and you can use this as the basis of an iteration. Start with one value of x, for example, x=1. Then get a second value y = (1/2)(x + a/x). Then get a third value z = (1/2)(y + a/y). And so on. It turns out that these numbers x, y, z, . . . converge quite quickly to the desired square root, especially if you start with a number that is already close. \n\nDoing this calculation in the above example, for a=1234, and starting with the number a=40 which is obviously just a little bit bigger than the desired square root, you get the numbers 40, 35.425, 35.129578. . . , 35.128336. . . , 35.128336. . . and you see that just four calculations are enough to get the square root accurate to six decimal places.
701990
Okay, this is corny, but say you were arranging a number of something on a table, and you wanted to make a square. You would take the square root of however many you have, and then you would know how many to put in the first row, and you could just line up each column after that to match. Or, if you wanted to stack them in a cube, you would take the cube root.
593727
I can remember upto 4 to 5 ways of doing it and not more than that.  btw you are talking about halving a square (diagram)! Right ?
654434
Yes, there is. The square root of negative 1 is the number i, which is an "Imaginary Number." \n\nDespite the name, imaginary numbers are real quantities that are used in equations to get real answers. However, they are a somewhat confusing concept, mathematically. \n\nOne way to visualize them is to consider that multiplication by an imaginary number rotates a function orthogonally on a number axis. The answers given by performing this procedure once may not seem to be applicable to real situatons, but if you multiply by an imaginary number again, in rotates the function back onto the initial axis, yielding real number results. \n\nI am including links to a number of good sites on imaginary numbers. Although it can be difficult to understand the implications of imaginary numbers beyond the fact that their squares are negative, if you spend some time on these sites, I'm sure you'll come to understand them.
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.
771462
The answer is +/- the square root of 119, over 8.  Divide both sides of the equation by 8, and you get X squared = 119/64.  Take the square root of this and you have the answer.  64 is a perfect square, but 119 isn't.  If it had been 121 on top, your answer would be +/- 11/8.
844185
square-ish in shape\ncome in different sizes\nmany varieties
751528
Yes.  For example, take the two from the top left corner, and the two from the bottom right corner, and make a new square touching the top right corner.\n\n\nAAAA BBBBB\nC.........D..........E\nC.........D..........E\nC.........D..........E\nFFFFFF GGGG\nH.........I............J\nH.........I............J\nH.........I............J\nKKKKK LLLLL\n\nbecomes\n . . . . . . . . . . . . . AAAA\n . . . . . . . . . . . . .C . . . . J\n . . . . . . . . . . . . .C . . . . J\n . . . . . . . . . . . . .C . . . . J\n. . . . . . .BBBBB LLLLL\n. .........D..........E\n. .........D..........E\n. .........D..........E\nFFFFFF GGGG\nH.........I\nH.........I\nH.........I\nKKKKK
896906
The strategy is to reduce the number of square roots by repeatedly squaring and rearranging:\n\nSquare\n\n     11 - 2 √11√10 + 10 < 6 - 2 √6√5 + 5\n\nRearrange\n\n     5 <  √11√10 - √6√5\n\nSquare\n\n     25 < 110 - 2 √11√10 √6√5 + 30\n\nRearrange\n\n     2 √11√10 √6√5 < 115\n\nSimplify\n\n     4 √3√11 < 23\n\nSquare\n\n     4² × 3 × 11 < 23²\n\n     528 < 529\n\nwhich is true.\n\nNote that all steps are reversible because both sides of the "<" are positive. For example, √11 - √10 > 0. Else you get into trouble "proving" that 1 < -2 from 1² < (-2)² !!!