Simplify Square Roots?
354088
sqrt(75a^2) can be written as\n\nsqrt(15) * sqrt(5) * sqrt(a^2)\n\nSo cancel out the sqrt(5) in the numerator and denominator, take the square root of a^2, and you have\n\na * sqrt(15)\n\nThis cannot be simplified further.
593752
log ( sqrt (x+ 1) *x^5)
879678
The first thing to do is to find the factors of the number you're taking the root of.\nFor example, teh factors of 24 include 2*12, 3*8, and 4*6.\nSince four is a perfect square (2^2), we'll use that set of factors.\nSo:\nsqrt(24)= sqrt(4*6)= sqrt(4)*sqrt(6)= 2sqrt(6)\n\nSame with sqrt(18):\n18=9*2\n9=3^2\nsqrt(18)= sqrt(9)*sqrt(2)= 3sqrt(2)\n\nsqrt(3) +sqrt(3) can only be reduced by realizing that it's\n2*sqrt(3), which could be the sqrt(12). THis is because you can take the two under the radical by squaring it. so  you get sqrt (2^2+3)= sqrt(12), but technically 2sqrt(3) is more reduced.\n\nsqrt(27)=3 sqrt(3)\nsqrt(12)= 2 sqrt(3)\nso 3sqrt(3) + 2sqrt(3) = 5sqrt(3)\nin this case, you're treating the sqrt(3) sort of like it was a variable. think of 3x +2x = 5x
946285
square root of 2x+3=6\n<=> 2x+3 = 36\n<=> x = 33/2
942253
the answer is -3x square root of 2
424740
Unfortunately there is not. There is however an algorithm that will allow you to work out square roots by hand.
787487
roots: x= (-12 +/- sqrt(144-4k^2) ) / 2k\n\nset them equal, assume k is not zero and cancel 2k:\n  -12 + sqrt(144-4k^2) ) = -12 - sqrt(144-4k^2) )\nso it mus be that \n  sqrt(144-4k^2) ) =0\nor \n  144=4k^2\n  k=6
772527
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
256804
5x^2  -10 x +5\ncan be factorised into (x-1)(5x-5)\nsimplified to 5(x-1)(x-1)\n\nthus square root of given sum becomes square root of \n5(x-1)^2\n\nwhich is (x-1)*5^1/2
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
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.
1123203
You can multiply and divide them easier.\n\nfifth root of 2x times 2x^20\n           -or-\n2x^(1/5) times 2x^20
803501
1.  add 1 toboth sides, then multiply by 3\n\n(x+4)^2=18  now take the square root of both sides\n\nx+4= +/- 3sqrt2\nx=-4 +/- 3sqrt 2\n\nyou also could use the quadratic formula\n\n\n2.  multiply num and denom by 3 - 2 sqrt3\n\n12-8sqrt3 +2sqrt3 -2sqrt6  all over -3\n\n\n3.  mult num and denom by 2 - sqrt 12 and get \n\n12-6sqrt12 all over -8\n\n12-12sqrt2  all over -8\n\n3-3sqrt2 all over -2
941657
Nah, sorry, but something in your thinking is incorrect.  You seem to be saying that if you add the square root of 3 four times, you want to get 9.  That's not right.  If you take sqrt(3)^4, you WILL get 9.   Taking the fourth root of a number is not like dividing a pie into four pieces.
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)
1185125
√3 (√3+2) is of the form a(b+c)\na(b+c) = (ab+ac)\nso we simplify to \n√3√3+√3*2\n√a√a = √a*a = √a² = a\nso it's 3+2√3.\nThe rules for roots are exactly the same as the rules for exponents, since a (square) root is merely a exponent of (1/2).  A cube root is an exponent of (1/3).\nⁿ√a root is the same as a^(1/n).
694690
determine the square of 6 - sqrt[(x-2)^2 + y^2]\n\n-->\n\n36 + [(x-2)^2 + y^2] - 12*sqrt[(x-2)^2 + y^2]\n= 40 + x^2 + 2x + y^2 - 12*sqrt[(x-2)^2 + y^2]