What is this special type of square root called?
554358
its nested bhai......
964858
That just means that keep it in the square root instead of simplifying it.\nEx: Keep it as square root of 2 instead of 1.414\n\nSo if you're finding the square root of 27, for example.  You can split up 27 into 9*3.  Square root of 9 is 3, but 3 is not a perfect square.  So your answer would be : 3 square root of 3.  Instead of 5.196
410829
The commonly taught method for computing square roots with pencil and paper can be found in the Wikipedia article, "Methods of computing square roots".\n\nScroll down to the heading "An exact long-division-like algorithm".
522997
ok, lets say that Sqrt() means square root \n\nSqrt(2x-5)=13   , elevate both sides of the equation to square\n (Sqrt(2x-5))^2 = 13^2  , the square root and the square power cancell each other, so\n\n(2x-5) = 169\n2x = 164, x = 82\n\nThe second one is the same, only that you have to leave the square root alone in the left side of the equation\n\nsqrt(3x-5)+7 = 3 , \nsqrt(3x-5) = -4\n\nand do the same as the first one\n\n3x-5 = 16\nx = 7\n\nthe last one is a bit harder cause there's an 'x' outside of the square root, and you do it like this\n\nsqrt(x+2) = x - 4    , elevate both sides to the square\n\n(x+2) = (x-4)^2    , and do the special product \n(x-4)^2 = x^2 -8x + 16 , then you get\n\n(x+2) = x^2 - 8x + 16 ,  this is a quadratic equation, take it to the general form\n\nx^2 -9x + 14 = 0    and factorize\n\n(x-7)(x-2) = 0  then you get\n\nx1 = 7 and x2 = 2 \n\nthat's it, if you have doubts contact me
678825
The square root of -1\nis an imaginary 1\nit is depicted generally with a character 'i'\nif u want root of -2\nthen itd be i x root(2)\nor 1.414i
460242
ur answer is [sqrt(3) - sqrt(7)].it can written as sqrt[x +/- 2*sqrt[y]] when x=a^2 + b^2 and y=a*b is equal to sqrt(a) +/- sqrt(b)
705926
let cube root of 1 be x. \n=> x^3 = 1\n=> x^3 - 1 = 0\n=> (x-1) (x^2 + x + 1) = 0\n=> x=1 or x^2+ x+1=0\n=> x=1 or x= [1+ i(3)^1/2 ] /  2  or  [1- i(3)^1/2 ] /  2 \n  where i is iota, (-1)^1/2\nHence cube root of one has three roots, one real and 2 imaginary. these imaginary roots are called w (omega) and omega square. \n1+w +w^2 =0\n1* w * w^2 = 1
612102
you DONT get the same value back !\n\nyou get TWO different values back whenever you obtain the square root of ( s^2 ).  for example the sq rt of 25 is 5 or (-5) .\n\nthus, solution to a square root always involves TWO values ( +/- ).\n\nin other words, "s"  can be positive or negative only but the solution is not unique ( +s or -s ) and unspecifiable.\n\nthanks gore .... :)
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.
1069254
I think it's log 3x2[6y]/z\n\n*square root of 6y\n*3x squared
739917
Many mathematical operations have an inverse, or opposite, operation. Subtraction is the opposite of addition, division is the inverse of multiplication, and so on. Squaring, lesson (exponents), has an inverse too, called "finding the square root." Remember, the square of a number is that number times itself. The perfect squares are the squares of the whole numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 … \nThe square root of a number, n, written \nsquare root of n\nis the number that gives n when multiplied by itself. For example,\nsquare root of 100 =10\nbecause 10 x 10 = 100\nFinding square roots of of numbers that aren't perfect squares without a calculator.\n1. Estimate - first, get as close as you can by finding two perfect square roots your number is between.\n2. Divide - divide your number by one of those square roots.\n3. Average - take the average of the result of step 2 and the root.\n4. Use the result of step 3 to repeat steps 2 and 3 until you have a number that is accurate enough for you.\nExample: Calculate the square root of 10 to 2 decimal places.\n1. Find the two perfect square numbers it lies between.\nSolution:\n3 squared = 9 and 42 = 16, so the square root of 10  lies between 3 and 4.\n2. Divide 10 by 3. 10/3 = 3.33 (you can round off your answer)\n3. Average 3.33 and 3. (3.33 + 3)/2 = 3.1667\nRepeat step 2: 10/3.1667 = 3.1579\nRepeat step 3: Average 3.1579 and 3.1667. (3.1579 + 3.1667)/2 = 3.1623\nTry the answer --> Is 3.1623 squared equal to 10? 3.1623 x 3.1623 = 10.0001\nIf this is accurate enough for you, you can stop! Otherwise, you can repeat steps 2 and 3.\nNote: There are a number of ways to calculate square roots without a calculator. This is only one of them.
1031543
There is no exact square root but it is\n\n7414.85009963114
633045
The answer that others have offered (that the square root of infinity is infinity) is not really helpful, I think.  The problem is that infinity is not a number, but a direction, like "east," or to use a more mathematical term, like "less." It makes no sense to ask for the square root of infinity (or how big is "infinity plus one," or anything else that that treats infinity as a specific number). \n\nJust to confuse things a little, however, there is such a thing as "degrees of infinity."  ;-)
50988
The square root of 144 is 12 and -12. The principal square root is 12.
668644
"triangular numbers" they're called
458634
The equation becomes x^(-4)=32+22; or x^(-4)=54; x^4=1/54\n.: x^2=±1/√54; so that x=±1/∜54 are two real roots - that is regular numbers. There are two more roots, x=(±1/∜54).√(-1), which are special numbers called imaginary numbers.\n\nQuadratic eqn is an equ of the form ax^2+bx+c=0. where a, b, c are known, x is unknown. if you put x^(-2)=u, the eqn becomes u^2=54 which is a quadratic eqn. solve it and put back the value of u. What you get can be written as another pair of quadratic eqn. Solve this to get the value of x. The end.
683937
Looks like someone is trying to get there math homework done.  Tsk tsk.  Square roots are easy - there is a button to do them on the calculator you should probably blow the dust off of.\n\nDid I say that?  Thanks for my 2 points and have fun cheating your way though high school math!\n\nAnd since it doesn't look like you're listening to your teacher, here's a math web site to get you started on the road to learning!