Answer:
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.
