What are the cube roots of -1?
915401
this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
801012
inverse square root of a number say"x" is\nx^[-(1/2)]
977047
The cube root of x is x^(1/3), so the cube root of a^2 is a^(2/3)
207106
x^2-3=-2\nx^2=-2+3\nx^2= 1\nx=1\n\n1^2+-3+2=0
26092
The Babylonian method is easy and fast:\n\nn[i+1] = (n + (r/n))/2\nwhere n is the approximate root, and r is the number you wish to find the sqaure root of. Check sources for more info.
798797
As you point out, 8 has exactly 3 distinct cube roots.  Only one of these is a real number.  Namely, 2.\n\n( If  you are not comfortable with complex exponentials, I've added a more algebraic method is below )\nExamine the complex exponential a*e^(i*x), where i is the imaginary unit ( a square root of -1 ), and a is a positive real number.  If this is a cube root of 8, then:\n\n( a*e^(i*x) )^3 = a^3 * e^( i * 3x ) = 8 = 8 * e^( i*0 )\n\nFrom this we see that a^3 = 8, so a = 2.\nMoreover, e^( i * 3x ) = e^( i*0 )\nAs e^(z) is 2*pi*i periodic, it must be that\n3x = 2*pi*k for some integer k \n\n = > x = 0, 2*pi/3, 4*pi/3  ( and then repeats )\n\nEuler's Identity says that e^( i * x ) = cos(x) + i * sin(x)\nso e^(i*0) = 1 ( as we already knew )\ne^( i*2*pi/3 ) = cos( 2*pi/3) + i * sin( 2*pi/3)\n                    = (-1/2)  + i * ( sqrt(3) / 2 )\ne^( i*4*pi/3)  = cos( 4*pi/3) + i * sin( 4*pi/3)\n                    = (-1/2)  + i * ( - sqrt(3) / 2 )\n\nMultiplying by the value of a we found above,  the 3 cube roots of 8 are:\n\n2, -1 + i * sqrt(3), and -1 - i*sqrt(3).\n\n\nAn alternate method for cube roots is the following:\n\nThe cube roots of 8 are the solutions of the equation x^3 = 8\nRewriting we find a difference of two cubes which we can factor.\n\nx^3 - 8 = 0\n(x-2)(x^2 + 2x + 4) = 0\nso either\nx - 2 = 0     or    x^2 + 2x + 4 = 0\nx = 2          or    x^2 + 2x + 1 = -3\n                         (x + 1)^2 = -3\n                          x + 1 = +/- i * sqrt(3)\n                          x = -1 +/- i * sqrt(3)\n\nThis method does not work with finding all roots, however.  It works best with cube roots.
1175698
cube root of (-125)=-5 \n\n3*sqrt(-125)=\n3*sqrt(25*5*-1)=\n3*5*i*sqrt(5)=\n15i*sqrt(5)\n\nnot sure which you were asking. For future reference, if you are talking cube or higher roots, use the n^(1/y) notation. A lot less confusing. sqrt is just for square roots.
117623
first, accept the fact that (i)^2 = -1. It's way of doing of mathematicians to solve for complex numbers.\n\nHence, sq root of -1 is i. \n\nif you're learning elementary indices, then most probably this is a trick question by the teacher.
599219
you have to solve the ones in the brackets first. so in this instance, 4(-1) = -4.\n\nand then, you solve -4 - 3 = -7
801068
-4(-1)^2 - 3(-1)\n-4(1) - -3\n-4 + 3\n= - 1
1130519
Seperate the constants - you get:\n\n2x = -3\n\nDivide both sides by 2\n\n\nx = -3/2\n\ntake the cube root of -3/2\n\nIts an irrational about -1.2599
832515
(1) any of the following may be the root of the expression,+1,-1,+2,-2,+3,-3,+6 or -6.\nNow putting the value of x=-1,see ifthe value of the expression becomes zero or not?If it becomes zero,then x=1 is one factor of the expression,if it does not become zero try with =1,then -2.+2 and so on.\nLuckily for us,with x=-1,the value of the expression becomes zero\nTherfore,x+1 is one factor\nx^4++x^3-7x^2-13x-6\n=x^3(x+1)-7x(x+1)-6(x+1)\n=(x+1)(x^3-7x-6)\n=(x+1){x^2(x+1)-x(x+1)-6(x+1)}\n=(x+1)(x+1)(x^2-x-6)\n=(x+1)^2(x-3)(x+2)\n(2)Will you please check up the second one once more and if there is any correction send it to me through Yahoo answers.
966844
f(0) = -i\nf(1) = 1² - i = 1-i\nf(2) = 2² - i = 4-i\n\nI take it you are dealing with complex numbers?  i is the square root of -1.\n\nNRT
322093
Hello.  First you have to determine if you are going to find 'real' answers.  use b^2 - 4*a*c, where b=-3, a=1, c=5 (from ax^2 + bx + c).  \n\nNow, b^2 - 4*a*c = 3^2 - (4*1*5) = 9-20=-11, hence, you will have imaginary roots in your answer.\n\nNote that sqrt(-1) = i, then,\n\nx^2 - 3x + 5 = 0\n(x - [3/2])^2 + 5 - (3/2)^2 = 0 ;(Completing the roots)\n(x - [3/2])^2 = -11/4\n(x - [3/2])^2 = (i^2) * 11/4 ; (note that i^2 = -1)\n\nx - 3/2 = +/- i * 0.5 * sqrt(11)\nx = (3/2) +/- i * 0.5 * sqrt(11)\n\n------------\nAlways remember to check if b^2 - 4*a*c > 0 or < 0 then you will know if there is imaginary i in the answer..
820030
i is define as sqrt(-1)\ntherfore  i*i=sqrt(-1)*sqrt(-1)= -1
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).
849035
1/2\n\nIf z^5 = 1, then z^5 - 1 = 0\nor, after factoring,\n(z - 1)(z^4 + z^3 + z^2 + z + 1) = 0\n\nIf a and b are roots of this equation, and neither a nor b is =1, then a and b both satisfy\nz^4 + z^3 + z^2 + z + 1 = 0\n\nso a^4 + a^3 + a^2 + a + 1 = 0\nand b^4 + b^3 + b^2 + b + 1 = 0\n\nso 2+a+a^2+a^3+a^4 = a^4 + a^3 + a^2 + a + 1 + 1 = 0+1 = 1\nand 3+b+b^2+b^3+b^4 = b^4 + b^3 + b^2 + b + 1 + 2 = 0+2 = 2\ntherefore\n(2+a+a^2+a^3+a^4) / (3+b+b^2+b^3+b^4) = 1/2