The solution for x2 + 2x + 8 ≤ 0 is

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.

B. The Empty Set

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Hope this helps   : )

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$35.59 in pounds

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