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.

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation:

50- 250 is correct i did that one too

step-by-step explanation: