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Christos
Solve the inequality x^2+2x>=0
Observe that if you want \(x \times (x + 2)\) nonnegative, both \(x\) and \(x + 2\) are nonnegative.
First time I am solving a 2nd degree inequality :S
Could someone help me continue? :)
I can offer an alternative. Just sketch the graph. That never fails.
I need to do it without a graph
\[x \le -2 \] and \[x \ge0\]
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
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:
And so we see that the product is positive if x <-2 OR x>0 and that is the solution.