## Christos Group Title Solve the inequality x^2+2x>=0 one year ago one year ago

1. ParthKohli Group Title

$x(x + 2) \ge 0$

2. ParthKohli Group Title

Observe that if you want $$x \times (x + 2)$$ nonnegative, both $$x$$ and $$x + 2$$ are nonnegative.

3. Christos Group Title

yes

4. Christos Group Title

hm

5. Christos Group Title

First time I am solving a 2nd degree inequality :S

6. Christos Group Title

Could someone help me continue? :)

7. DrKingSchultz Group Title

I can offer an alternative. Just sketch the graph. That never fails.

8. Christos Group Title

I need to do it without a graph

9. nitz Group Title

$x \le -2$ and $x \ge0$

10. Christos Group Title

How and why?

11. DrKingSchultz Group Title

Okay, then first find the x-axis intercepts by treating the inequality sign as an equals sign. $x ^{2}+2x=0$ $x(x+2)=0$ $x=-2,0$ Sub in values for $x<-2$ e.g when x=-3, y=3 Then sub in values for x>0 e.g when x=1, y=3 Since $x \in (-\infty,-2)$and $x \in (0,\infty)$yield positive answers Therefore $x ^{2}+2x>=0$ for x=<-2 and x>=0

12. Mertsj Group Title

Two factors: x and x+2 If the product is positive, x and x+2 must both be positive OR x and x+2 must both be negative. Let us make a picture of the factors and their signs at various points on the number line:

13. Mertsj Group Title

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14. Mertsj Group Title

And so we see that the product is positive if x <-2 OR x>0 and that is the solution.