The solution for x2 + 2x + 8 ≤ 0 is a) all real numbers b) the empty set c) x= 2 or x =4 d) x = -2 or x = 4

Step-by-step explanation:

Please use " ^ " for exponentiation:  x^2 + 2x + 8 ≤ 0.

Let's solve this by completing the square:

x^2 + 2x + 8 ≤ 0  =>  x^2 + 2x + 1^2 - 1^2 + 8 ≤ 0.  Continuing this rewrite:

                                    (x + 1)^2 + 7 ≤ 0

Taking the sqrt of both sides:   (x + 1)^2 = i*sqrt(7)

Then the solutions are x = -1 + i√7 and x = -1 - i√7

There's something really wrong here.  I've graphed your function, x^2 + 2x + 8, and can see from the graph that there are no real roots, but only complex roots.  Please double-check to ensure that you've copied down this problem correctly.

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step-by-step explanation:

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~