## anonymous one year ago can someone solve this ? |x+1|^2 + 2 |x+2| ≥ 2

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1. Michele_Laino

we can note this: $\Large {\left| {x + 1} \right|^2} = {\left( {x + 1} \right)^2}$

2. Empty

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

welp

4. Michele_Laino

so, we can rewrite your inequality, as below: $\Large {\left( {x + 1} \right)^2} + 2\left| {x + 2} \right| \geqslant 2$

5. anonymous

so @Michele_Laino (x+1)^2 = x^2+2x+1 ?

6. Michele_Laino

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