i^2=-1 is our common beleif but i^2=i*i =>root 1=+1or-1 then why not i^2=+1?
820030
i is define as sqrt(-1)\ntherfore  i*i=sqrt(-1)*sqrt(-1)= -1
758984
i^(-1) = 1/i = -i\n\nsummation from i = 1 to n i^(-1) = -ni\n\nput if you mistyped and meant which is likely\n\ni^(-n) it is geometic series\n\n= -(i+i^2+i^3+..... )\n= -(1-i^(n+1))/(1-i)
19513
This seems to be a popular question. There is no real number that is the square root of -1.  However, it turns out that it is useful to develop as set of numbers that allows the square root of -1.    This is called "i" as is known as an imaginary number.  The set of numbers that includes i is called the set of complex numbers.
89444
The answer is "i" because there is no number, that when mutliplied by itself, will give you -1.  Think about it: the only possible combination that would make sense is 1 X 1, which equals 1...not -1.
211940
This is your fifth question in math...Do you find the subject difficult?...Ok...If your question is correct...\n\n3x^2 + 13 - 10 = 0\n\n3x^2 + 3 = 0\n\n3x^2 = -3\n\nx^2 = -3/3\n\nx^2 = -1\n\nx^2 = i^2 ===>i is a symbol in complex numbers\n\nwhere i^2 = -1. \n\nx^2 - i^2 = 0\n\n(x + i)(x - i) = 0\n\nTherefore:\n\nx + i = 0 ......... x - i = 0\n\nx = -i ............. x = i ANSWERS or\nx = - sq. root of -1..... x = sq. root of -1 ANSWERS\n\nThe solution is imaginary and NOT real.
860947
sqrt ( -1 * -1 ) doesn't equal sqrt(-1) * sqrt(-1).\n\nanother example:\n\nsqrt ( 5 * 5 ) = sqrt ( 25) = 5\n\nbut it doesn't equal sqrt(5) * sqrt(5) because this is an irrational answer..\n\nhope this helps
25136
1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1)=i*i = i^2 = -1\nSo 1 = -1.  Add 1 to both sides and 2 = 0.  Divide both sides by 2 and 1 = 0.
719816
Round roots??? V(-1) = i
841095
Who said sqrt(-1)² isn't -1?\n\nsqrt(-1) = i\nsqrt(-1)² = i²\nsqrt(-1)² = -1
752580
It just takes foiling-\n\n(5*2)+(5*5i)+(-3i*2) +(-3i*5i)\n(10)+(25i)+(-6i)+(-15*i^2)\ni^2=-1\nso...\n25+19i\n\nTo those who have said that you can't simplify it past 10 + 19i - 15i^2, that isn't correct considering that since i is the square root of -1, i^2 is -1. That's why you can simplify -15i^2 to 15, thus making it 10+19i+15, or 25+19i
152568
2i. The character "i" denotes the square root of -1, which is not a real number. So, the sq. root of -4 is 2i.
1060909
Let z=x+iy\n\nThen |Re(z)|=|x|\n|Im(z)|=|y|\n\nAnd |z|=sqrt(x^2+y^2), by definition.\n\nSo now you seek to prove:\n|x|+|y|<= sqrt(2)sqrt(x^2+y^2)=sqrt(2x^2+2y^2)\n\nSince |x|>0 and |y|>0, we can square both sides:\n\nx^2 +2|xy|+y^2<=2x^2+2y^2\n\nThis is true if and only if\n\nx^2-2|xy|+y^2>=0, which is true iff\n\n(|x|-|y|)^2>=0. This however is necessarily true, since x and y are real numbers, and the square of a real number is always positive. Since this statement is true, the original statement \n\n|Re(z)|+|IM(z)| =< sqrt 2 * |z|\n\nis also necessarily true.
142343
As the responses above show, the answer is -1.  However, this implies two remarkable results.\n\nResult #1 is that e^i*n*pi is equal to -1 where n is any odd number.  The reason for this is as follows:\nFrom Eulers identity (mentioned above) we know that \ne^i*x  = cos(x) + i*sin(x)  \nso, as a result\ne^i*n*pi = cos(n*pi) + i*sin(n*pi)\nbut note that when n is an odd number\ncos(n*pi) is always -1    and\nsin(n*pi) is always 0.\nso the sum is always -1.\n\nResult #2 is that we can rewrite e^i*pi = -1 as\ne^i*pi + 1 = 0\n\nThis gives us e, i, pi, 1 and 0, five of the most fundamental numbers of mathematics, in one tidy pakage!\n\nBeautiful indeed!
761784
Well, by inspection -2 is one root...\n\nP(-2) = (-2)^3 - 4(-2)^2 - 7(-2) + 10\nP(-2) = -8 - 16 + 14 + 10\nP(-2) = -24 + 24\nP(-2) = 0\n\nNow take your equation and divide by (x - (-2)).  In other words (x + 2):\n\n............ 1 -6 5\n1 2 ) 1 -4 -7 10\n....... 1 2\n....... ---------\n........... -6 -7\n........... -6 -12\n........... ----------\n............... 5 10\n............... 5 10\n............... ------\n.................... 0\n\nSo the result is x^2 - 6x + 5\n\nThis can be factored to (x - 5)(x - 1)\n\nThis results in (x + 2)(x - 5)(x - 1)\n\nAnd your roots are:\n-2, 5, 1\n\n(I don't know why I didn't notice 1 as a root... I usually check that first...)
728155
Lots of great answers which deserve the 10 points, but a short-cut would be to use \nproduct of roots = -40\n\nThe product of (3 + i) and (3 - i) is 10, so the other root must be -4.\n\nOf course, the advantage of the other methods is that they check up that the given information is correct, whereas this method accepts the information that 3 + i is a root (so that 3 - i must be, too).
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..
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)