what are the solutions to the equation: x squared +2x+2=0?

Answer 1

since u cannot factor u have to use the general formula

x=[-2(+-)sqrt(2^2-4(1)(2)) ]/ 2

this is

x= [-2(+-)Sqrt(-4) ] /2

since Sqrt-4 is not a real number there is no solution ifu know already imaginary numbers solutions are

x1=(-1-i) and x2=(-1+i)

Answer 2

The most straightforward way to answer this question is to realize that:

x squared + 2x + 2 =0

is the same as

(x + 1) quantity squared = 0

Therefore, in order to make the quantity square to 0, x = -1

Answer 3

x² +2x + 2 = 0
(x+1)² + 1 = 0
(x+1)² = - 1
therefore
x = - 1 ± i

Answer 4

Since x^2+2x+1 is a perfect square, we can solve this as follows: (x+1)^2 + 1 = 0, so (x+1)^2=-1
Then x= -1 +/- i, where i is the sqrt(-1).

Answer 5

-b±sqrt(b^2-4ac) throughout 2a; this is the quadratic equation. = -2±sqrt(2^2-4*a million*2) throughout 2*a million = -2±sqrt(-4) throughout 2 = -2±2i throughout 2 = -a million±i So x=-a million±i, the place i is the imaginary selection that equals the sq. root of a million.

Answer 6

x^2 + 2x + 2 = 0 n.b.: i^2 = -1 => i = +-sqrt(-1)

x = (-2 + - sqrt(4 -4(1)(2)))/2

= (-2 + - sqrt(-8))/2


sqrt(-8) = sqrt(4 x -2) = sqrt 4 x sqrt(-2) = 2 sqrt(-2)
= 2(sqrt(2 x (-1))
= 2 (sqrt(2) x sqrt(-1))
= 2(sqrt(2) x i
= 2i(sqrt(2))

We obtain two complex roots:

x1 = (-2 + 2i(sqrt(2))/2 => 2(-1 + i)sqrt(2)/2 => (-1+ i)sqrt(2)

x2 = (-2 - 2i(sqrt(2))/2 => 2(-1 - i)sqrt(2)/2 => (-1 - i)sqrt(2)

Answer 7

I don't know, use completing the square, i just can't think well enough to help you right now, you probably think i am really retarded but i am not.

Answer 8

x^2+2x+2=0
x=(-2±√(4-8))/2
x=-1±i
x=-1+i
x=-1-i

Answer 9

have to do quadratic formula... x=(-b+-sqrt(b^2-4ac))/(2a)
(-2+-sqrt(-4))/2
x= -1+2i or -1-2i

Answer 10

huh?