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Christos
 2 years ago
Solve the inequality x^2+2x>=0
Christos
 2 years ago
Solve the inequality x^2+2x>=0

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

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

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

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

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

DrKingSchultz
 2 years 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
 2 years 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
 2 years 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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