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Christos

  • 2 years ago

Solve the inequality x^2+2x>=0

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  1. ParthKohli
    • 2 years ago
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    \[x(x + 2) \ge 0\]

  2. ParthKohli
    • 2 years ago
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    Observe that if you want \(x \times (x + 2)\) nonnegative, both \(x\) and \(x + 2\) are nonnegative.

  3. Christos
    • 2 years ago
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    yes

  4. Christos
    • 2 years ago
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    hm

  5. Christos
    • 2 years ago
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    First time I am solving a 2nd degree inequality :S

  6. Christos
    • 2 years ago
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    Could someone help me continue? :)

  7. DrKingSchultz
    • 2 years ago
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    I can offer an alternative. Just sketch the graph. That never fails.

  8. Christos
    • 2 years ago
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    I need to do it without a graph

  9. nitz
    • 2 years ago
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    \[x \le -2 \] and \[x \ge0\]

  10. Christos
    • 2 years ago
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    How and why?

  11. DrKingSchultz
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    |dw:1362224143800:dw|

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

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