the square root of 3 + the square root of 27?  how do you come to an answer of  4square root3?
483565
the factors of 27 are 3-3-3.  when you have two of the same number in a factor tree, you can take them outside the root.  So now you have 3squareroot3.  squareroot3 + 3squareroot3 = 4 squareroot 3.  Make sense?
370534
37 is between 36 and 49 on the number line.  Therefore, the square root of 37 is between the square root of 36 and the square root of 49, or between 6 and 7.\n\n27 is between 25 and 36, so its square root is between 5 and 6.\n\n130 is between 121 and 144, so its square root is between 11 and 12.\n\n170 is between 169 and 196, so its square root is between 13 and 14.\n\nYou need to know the perfect squares by heart.  Usually you won't need to know beyond 15^2 or 20^2.  Then, given a number, figure out what perfect square is closest to it below the number, and closest to it above the number.  The square root of the number will lie between the square roots of those two numbers.
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).
530142
use the quadratic equation:\nx=-b +/- the square root of b^2 - 4ac all over 2a\na=1,b=7,c=10 so plug it in\nx=-7 +/- the square root of 49-40 all over 2\n-7 +3 = -4 divided by 2 x=-2\n-7-3=-10 divided by 2 = -5\nso x = -2 or x=-5
942253
the answer is -3x square root of 2
517819
start pulling out common factors\n\n2 (x^3 + 27)\n\n2(x^3 + 3^3)
500945
A number, to the power of 3. (i.e. 27=3-cubed, or 3X3X3)
902935
27 = 3^3\n\n27^(2/3) = (3^3)^(2/3)\n= 3^(3*2/3) [ because (a^b)^c = a^(b*c) ]\n= 3^(2)\n= 9
793181
sqrt(x+4)=3\nx+4=9\nx=9-4\nx=5\n\ns
768314
take 3 as a common factor\n=3(x^2-4x-8)\n=3((x-2)^2-12) putting in form of square\n\nthis has got a minimum
408683
When we square a number we multiply it by itself\n\nThe square root of a number is the number that when multiplied by itself makes that number.\n\nFor example:\n\nThe square root of 9 is 3 (3 * 3 = 9)\n\nThere are different ways of finding the square rot of a number (by hand - using a calculator is just too easy lol) and the source below shows you some of them.
404426
sqrt(81) = 9\n\nso take 9 x 3 = 27\n\nthen 27 plus 17 = 44\n\nfinaly 44 - 12 = 32\n\nso your final answer is 32.
593847
The answer is zero isn't it?  The square root of (7 squared) is 7.  If you minus 14 you get -7. Add another 7, brings you up to zero.
1157777
sqrt() means square root of:\n\n18)\n\nsqrt (3yx)/x\n\nI got that by multiplying both top and bottom by sqrt(x)\n\n\n19)\n11*sqrt(3)/3\ndid the same thing\n\n20) sqrt (6) divided by 3\nsame thing
142254
-3 = -45/sqrx^2 - 1  :: multiply both sides by sqrt(x^2-1)\n\n-3*sqrt(x^2-1) = -45   :: divide both sides by -3\n\nsqrt(x^2-1) = 15 :: square both sides\n\n(sqrt(x^2-1))^2 = 15^2\n\nx^2 - 1 = 15^2  :: add 1 to both sides\nx^2 = 15^2 + 1\n\nx^2 = 226
47031
Your equation is:\n\nf(x) = x - 3Sqrt(x) + 4 = 5\n\n==> x - 3Sqrt(x) - 1 = 0\n\nTo deal with the Square root...\n\nLet z = Sqrt(x)\n==> z^2 = x\n\nSubstitute this into your equation:\n\n==> z^2 - 3z - 1 = 0\n\nNow...solve this like you would any quadratic. I'm not going to do al that for you....I think you can do that, but I will give you the results.....ok?\n\nYou'll end up with z = [3 +- Sqrt(13)]/2\n\nSince z = Sqrt(x)\n\nSqrt(x) = [3 +- Sqrt(13)]/2    Square both sides.\n\n==> x = {[3 +- Sqrt(13)]^2}/4\n\nHope this helps.
923612
1) 24 is the product of 4*6 , we can take the square root of 4 which is 2,but we can't simplify the 6 any more\nso square root of 24 is 2square root of 3.\n2)a^3 = a^2 * a^1  ,so we can take the square root of a^2 which is = a\nso square root of a^3 =a square root of a.\n3) b^6 = b^2 * b^2 * b^2 , now we can take the square root of each b^2  , so the answer for this part will be b^3,but we have to put this b^3 inside an absolute value ( check the algebra II textbook ).\n4) c will stay inside the square root.\n\nso the final answer is : 2aIb^3I square root of 6ac.\n                               or: 2a absolute value of b^3 times square root \n                                       of 6ac  .\n\nI hope this would help