If  i  is a root of X^3 + i = 0, what are the other two roots? (i means imaginary)?
371573
If i is a root, you may express it into x = i, a root of x^3 + i = 0.\nsince x = i, x - i = 0\nTherefore, x^3 + i has a root x - i. to get the other 2 factors, divide.\n(x^3 + i)/(x - i) = 0\nDivide it using long division, and you'll get\nx^2 + ix - 1 = 0\nActually, \nx^3 + i factors as (x - i)(x^2 + ix - 1) \ntherefore, \nx^2 + ix - 1 = 0\nuse the quadratic formula:\na = 1\nb = i\nc = -1\nx = [-b +or- sqrt(b^2 - 4ac)]/2a\nx = [-i +or- sqrt(i^2 - 4(1)(-1))]/2(1)\nx = [-i +or- sqrt(-1 + 4)/2\nx = (-i +or- sqrt -3)/2\nx = (-i +or- i sqrt3)/2\nTherefore, the other roots are\nx = (-i + i sqrt3)/2\nx = (-i - i sqrt3)/2\n\n^_^
291373
I have no idea what you're trying to ask!  This looks a little like a math question.  Does the "i" in your question stand for an imaginary number?
650764
Remember that i^2 is -1 so\n-3i+2(-1)-2+i=-4-2i
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
241669
Y=x^3(x+2)(3x-1)\n we have three possibility for x\nx^3=0 than X=0\nx+2=0 than X=-2\n3x-1=0 than x=1/3\n\nso the answer is a
122155
(x + 1)(x + 3)(x - 2) = 0\n(x^2 + 4x + 3)(x - 2) = 0\n(x^2)(x - 2) + 4x(x - 2) + 3(x - 2) = 0\nx^3 - 2x^2 + 4x^2 - 8x + 3x - 6 = 0\nx^3 + 2x^2 - 5x - 6 = 0 is the answer ;-)
539132
both roots are complex numbers
977047
The cube root of x is x^(1/3), so the cube root of a^2 is a^(2/3)
191203
(x-7i)(x+7i)=x^2 + (-i)^2 * 49 = x^2 + 49\n(which is your polinom having the roots -7i and +7i)
540401
(3 + i)(5 + i)(7 + i)(8 + i) = 650 + 650i\n\nwhere i is the square root of -1.
1164880
If x-2 and x+1 is a factor, then P(2) = 0 and P(-1) = 0\nP(2) = 2^4 + a*2^3 - 2^2 +b*2 - 12\n= 16 + 8a - 4 + 2b - 12\n= 8a + 2b = 0\nP(-1) = 1 - a - 1 - b - 12\n= -a - b - 12 = 0\nYou now have a system of equations.\n8a + 2b = 0\n-a - b - 12 = 0\nSolve by substitution. a = 4 and b = -16\nNow use synthetic division or long division to get the other roots.\nx = 2 and x = -1 are given.  The other roots are -3 and -2.\n\n2) Not sure\n\n3) Use sythetic division or long division.\nYou will get that f(x) divided by (x+2) is 2x^2 - 2ax + 3a.\nIf -2 is the only real root, then the other roots must be imaginary.\nIf the roots of a quadratic equation are imaginary, then the discriminant is less than 0.\ndiscriminant = b^2 - 4ac, where a = 2, b = -2a and c = 3a\nb^2 - 4ac = (-2a)^2 - 4(2)(3a)\n= 4a^2 - 24a <0\nFactor:\n4a(a-6) < 0\nIf a<0 or a>6 then 4a(a-6) is positive.\nIf 0<a<6, then 4a(a-6) is negative.\nSo 0<a<6.\n\n4) Not sure, but I'll think about it.
768805
Factor out the 3:\n3(x^2+3x+2)\nUnfoil:\n3(x+1)(x+2)\nTake the roots:\nx=-1 or -2
649881
4) n is even.\nStarting with an even number (0) and ending with an even number, means you have an odd number off roots.
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
527455
If those are the zeros, you just do it like this for the first one:\n\ny=(x-5)(x-3i)(x+3i)\n\n\nIf your teacher wants it in general form, you have to multiply out all the terms. \n\ny=(x-5)(x^2+9)\n\ny=x^3+4x^2-45\n\n\nThe second set doesn't make sense because you have an i without a partner. In polynomial equations, the complex factors always come as conjugate pairs, like the -3i and +3i.\n\nAnyway, I hope this helps. Basically, you set up a bunch of factors that are x-each root.\n\nI just changed it because I realized that the 1. meant problem number 1.
299363
Well, the square root is 27i, that makes it easy, cause 3X3X3 = 27, so 3 to the 6th is 729, and i X i is -1, so the sixth root of -729 is 3i.
83855
y = a*x^2 + b*x + c\n\nThat is, y = constant times x squared plus constant times x plus a third constant. The solution is Eqn 5 on the link below.\n\nIf the stuff under the radical in the solution is negative then you will have an imaginary number on your hands.