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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Observe that if you want \(x \times (x + 2)\) nonnegative, both \(x\) and \(x + 2\) are nonnegative.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0First time I am solving a 2nd degree inequality :S

Christos
 one year ago
Best ResponseYou've already chosen the best response.0Could someone help me continue? :)

DrKingSchultz
 one year ago
Best ResponseYou've already chosen the best response.0I can offer an alternative. Just sketch the graph. That never fails.

Christos
 one year ago
Best ResponseYou've already chosen the best response.0I need to do it without a graph

nitz
 one year ago
Best ResponseYou've already chosen the best response.0\[x \le 2 \] and \[x \ge0\]

DrKingSchultz
 one year ago
Best ResponseYou've already chosen the best response.0Okay, then first find the xaxis 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

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0Two 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:

Mertsj
 one year ago
Best ResponseYou've already chosen the best response.0And so we see that the product is positive if x <2 OR x>0 and that is the solution.
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