Question

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

Answer 1

The answer to this is an empty set!

Answer 2

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.