Can you tell me the answer and how to find out between what two consecutive interfers is each square root?
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.
203508
Two.\n\nYou can rewrite it as x^2 = 15 or x = sqrt(15)\n\nx solves to (rounded) 3.87 or -3.87
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
386376
My computer gives me this 4 real results, there are two more Imaginary results.\n\n1) -1\n2) 1.8698\n3) -0.260\n4) -2.688\n\nImaginary roots are\n\n5) 1.039 + j1.40\n6) 1.039 - j1.40
596727
Figure out what other number times itself equals your number.  That's its square root.
889816
1/rt3 *1/rt2-1/rt3\n[1/rt3](1/rt2-1)\n[1/rt3](1-rt2/rt2)\nrationalising\n[(1-rt2)*rt6]/6
1144250
2) new-old divided by old\n3) find the square root first\n4) i think its 184... not sure\n5) Distribute 3 to 4x and 4 to get 12x+12 and the 2 tho 5x and 9 to get: 12x+12+10x+18-12 Then simplify (I'm not gonna do all ur work 4 u!)
888483
First question:\nTake the sq. root of both sides:  x+2=+/-(sqrt5)\nThen subtract 2:  \n                               x = -2 +/-(sqrt 5)\n\nSecond prob:  subtract 5 first:\n\n(a+2)^2 = -3\nTake the square root of both sides:  (a+2) = +/-(sqrt(-3)\n    \nsqrt (-3) = i*sqrt 3\n\na+2 = +/-(i*sqrt 3)\n\nThen subtract 2: final answer:  -2 +/-(i*sqrt3)
540401
(3 + i)(5 + i)(7 + i)(8 + i) = 650 + 650i\n\nwhere i is the square root of -1.
713860
1.Xc=1/2pifc\nc=1/2pifXc\n\n2.ib=ibq+rt(2is)\nib(1-q)=rt(2is)\nsquaring\n[ib(1-q)]^2=2is\ns=[ib(1-q)]^2/2i\n\n3.Q=Xl/R\nXl=Q/R\n\n4.Lt=L1+L2+2Lm\n2Lm=Lt-L1-L2\nLm=[Lt-L1-L2]/2\n\n5.Es=E1+E2+E3\nE2=[Es-E1-E3]\n\n6.R1=i2R^2/i2\ni2R^2=i2R1/i2\nR=rt[i2R1/i2]\n\n7.R1/Rx=R2/R3\nRx=R1R3/R2\n\n8.y=mx+b\nb=y-mx\n\n9.i2=i1R1/i1
773471
√2 . √6 \n---- x ----- =\n. 3 .... 22\n\n√12\n------- =\n66\n\n\n√(2² x 3)\n------------- =\n66\n\n2√3\n------ =\n66\n\n√3\n-----\n33
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 .... :)
607013
Another Method.\n\n-1 = -1\n\n-1/1 = 1/-1\n\nSquare root both sides,\n\nSQR(-1/1) = SQR (1/-1)\n\nExpanding,\n\nSQR(-1)/SQR(1) = SQR(1)/SQR(-1)\n\nSQR(-1)/1 = 1/SQR(-1)\n\nCross Multiply,\n\nSQR(-1)*SQR(-1) = 1\n\n(SQR(-1))^2 = 1\n\n-1 = 1\n\nDivide by 2 on both sides,\n\n-1/2 = 1/2\n\nAdd 3/2 to both sides,\n\n-1/2 + 3/2 = 1/2 + 3/2\n\n1 = 2 (ans)
606423
Here's how to work it out:\n\n(Sorry, because I can't use special symbols, I'm using _/ for the square root symbol and <= for the less than or equal to symbol.)\n\n_/3x + 6 + 4 <= 7\n_/3x + 10 <= 7\n3x + 10 <= 7 x 7\n3x + 10 <= 49\n3x + 10 - 10 <= 49 - 10\n3x <= 39\n3x/3 <= 39/3\nx <= 13
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
868180
the roots are (2a-1)x(2a-1)
539132
both roots are complex numbers