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anonymous

  • one year ago

can someone solve this ? |x+1|^2 + 2 |x+2| ≥ 2

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  1. Michele_Laino
    • one year ago
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    we can note this: \[\Large {\left| {x + 1} \right|^2} = {\left( {x + 1} \right)^2}\]

  2. Empty
    • one year ago
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    There's really no difference between \(|x+1|^2\) and \((x+1)^2\) since they both evaluate to the same thing, so you can safely throw these absolute value signs away

  3. Empty
    • one year ago
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    welp

  4. Michele_Laino
    • one year ago
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    so, we can rewrite your inequality, as below: \[\Large {\left( {x + 1} \right)^2} + 2\left| {x + 2} \right| \geqslant 2\]

  5. anonymous
    • one year ago
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    so @Michele_Laino (x+1)^2 = x^2+2x+1 ?

  6. Michele_Laino
    • one year ago
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    next, we have to consider the subsequent two cases: \[\large \left\{ \begin{gathered} {\left( {x + 1} \right)^2} + 2\left( {x + 2} \right) \geqslant 2 \hfill \\ x + 2 \geqslant 0 \hfill \\ \end{gathered} \right.,\quad \left\{ \begin{gathered} {\left( {x + 1} \right)^2} - 2\left( {x + 2} \right) \geqslant 2 \hfill \\ x + 2 < 0 \hfill \\ \end{gathered} \right.\] please solve both systems above

  7. madhu.mukherjee.946
    • one year ago
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    did you solve it

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