How do you square a #?
469007
A way of visualising this is to think of soldiers on parade. If there are 7 soldiers in a row and 7 rows of soldiers (making a square formation on the parade ground) how many soldiers are there in the square formation?\n\nClearly it is the number of soldiers per row x the number of rows (as it would be for a rectangle) but in the special case of a square formation, these two numbers are the same, so multiply 7 by itself
146354
x square-12x-85\nI think?
875144
you have to take the square root of each side, then just simplyfy\n\nx = 1.2472191289247
243669
divde the area by pi, then take the square root.  Multiply that reult by 2 and then by pi and your done.
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
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.
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 !!!
134165
On top..\n4 plus or minus the square root of 8\non bottom..\nall over 4\n=(4 +or- sqrt8)/4\nIf you can find any factor of the number inside the radical sign which is a perfect square, then you may rewrite that number into the product of that perfect square and the 'other". Let's say 45. 9 is a factor of 45 and a perfect square, so 45 = 9 * 5.\nIn your given, 8 = 4 * 2. (4 is a perfect square) You may substitute this to the original equation:\n=(4 +or- sqrt(4*2))/4\nyou may then get the square root of that perfect square and place it out of the radical sign, leaving the 'other' number. Since 2 is the sqrt of 4, then\n=(4 +or- 2 (sqrt 2))/4\nThe numerator has a common monomial factor of 2 (One technique is to find if all terms are even). you can factor this out and use the cancellation method.\n=2(2 +or- sqrt 2)/4\nCancelling 2 from both numerator and denominator,\n=(2 +or- sqrt 2)/2\nYou have a +or- sign, so there are to answers:\n=(2 + sqrt 2)/2 or\n=(2 - sqrt 2)/2\nThese answers are accepted, but if you want, you can simplify sqrt 2 into 1.4142. So you will get:\n=1.7071 or\n=0.2929\nHope this helps ;-)
352166
8\ntake away o from top and you're left with o on bottom - half - nothing\nor i guess 10 could work, too (or any double digit ending in 0) take away the first number and you're left with 0
540085
(a-2)(a+2)  \n(difference of squares)
737774
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!
59409
Because I like to waste time I'll answer this. If somebody tells you to simplify the square root of 8 you'd need to take two numbers out and on of them has to be a perfect square or a number that has an exact square root with no remaining numbers. So. Since 8 is not a perfect square you have to break it up. Take 4 and 2 out of it because 4 and 2 multiply to get 8. So take the perfect square of 4 and bring it outside the house (to the left) and leave the 2 (already a perfect square in the house). So. You would get 2 (radical sign) 2. Another example would be 56. You'd get 2 (radical sign) 14 because if you square 2, to get 4 and multiply it by 14 you get 56. You can't do 64 because it already has a perfect square: 8, because 8 times 8 is 64 and you don't need to break it down. If you still dont get it just talk to a teacher.
1110740
I -1 digit\nLOVE - 4 digits\nYOU - 3 digits
255771
The first ten square numbers are - \n1, 4, 9, 16, 25, 36, 49, 64, 81, 100 ...\nEach is the result of multiplying a number by itself - \n1 x 1, 2 x 2, 3 x 3, 4 x 4, 5 x 5 ... \n- which can also be written - \n1, 2, 3, 4, 5 ... \n- where the small  means 'squared'.\n\nThe square of 7 is 49, and working backwards, we say the square root of 49 is 7.\n\nAll square numbers end in either a 0, 1, 4, 5, 6 or a 9. So if someone were to ask you if 225,736,192 was a square number, you could say 'no' right away because it ends in a 2.\n\nIf you subtract a square number from the next square number you get the odd numbers 3, 5, 7, 9 ...
316466
Another answerer gave an example, but I thought it\nmight help to explain it to.\n\nFor any large (or small) number, move the decimal\npoint to just after the first non-zero digit (so\nthat you end up with a number between 1 and 10.\nThen specify the number of digits that the decimal\nwas moved using the "times 10 to the something power"\nnotation.\n\nFor example, 123 would be "1.23 X 10^2"\nOr .00023 would be "2.3 X 10^-4" \n\n(where "^" means "to the power of")
978895
Can you use a square root symbol?  If so, sqrt(9) + 9 - 9
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)