SOLUTION: x^2+8x+16>=0 How do I work this out. The back of my book says that the answer is all real numbers, is that right? If it is how do I show work for that?

Question 712528: x^2+8x+16>=0 How do I work this out. The back of my book says that the answer is all real numbers, is that right? If it is how do I show work for that?
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Where does the function cross the horizontal axis? What are the zeros of the function? Does it have REAL zeros? What is the value of the discriminant?

x^2+8x+16>=0
So can this be factored? (x+4)(x+4)>=0, seems to have a zero at x=-4. That is where the quadratic will be zero. What about x<-4, and x>-4? NO PART OF THE FUNCTION IS LESS THAN ZERO. The graph opens upward, has a vertex which is a minimum and touches the horizontal axis.

So you want to show work? Remember vertex at (-4,0). Pick any x value less than -4. Is x+4 positive or negative? Negative! so (-)(-)=(+), so above the horizontal axis. Pick any x value greater than -4. Is x+4 positive or negative? Positive! so (+)(+)=(+), and above the horizontal axis. Just pick examples and show them.
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