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3 years ago

The solution for x2 + 2x + 8 ≤ 0 is

Mathematics

2 answers:

Marina CMI [18]3 years ago

5 0

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.

Valentin [98]3 years ago

3 0

Answer:

B. The Empty Set

Step-by-step explanation:

Hope this helps!!! Have a great day!!!! : )

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Answer:

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Step-by-step explanation:

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3 years ago

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saveliy_v [14]

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4 years ago

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9966 [12]

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other odd integer = x +2

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