Help with absolute value inequalitie












0














I am having trouble finding what the answer to this inequalitie is:
$|x+4|x + |x+2| > -2$



first i did this:



$|x+4|=left{ begin{align}
x+4 & text{ , if }xgeq -4 \
-(x+4) & text{ , if }x <-4
end{align}
right}$



$|x+2|=left{ begin{align}
x+2 & text{ , if }xgeq -2 \
-(x+2) & text{ , if }x <-2
end{align}
right}$



Then i got this:



$1. x<-4: $



$ -(x+4)x + (-(x+2)) > -2$



$2. -4 $le$x<-2:$



$ (x+4)x + (-(x+2)) > -2$



$3. x$geq$-2: $



$ (x+4)x+x+2 > -2$



After i calculated top 3 inequalities i got this answers:



$1. x$^2$ + 5x < 0$



$2. x$^2$ + 3x > 0$



$3. x$^2$+5x+4 > 0$



I think i calculated everything correctly, but now i do not know how to find the answer to this absolute value inequalitie.



Thank you for your help.






inequality absolute-value







share|cite|improve this question





























    0














    I am having trouble finding what the answer to this inequalitie is:
    $|x+4|x + |x+2| > -2$



    first i did this:



    $|x+4|=left{ begin{align}
    x+4 & text{ , if }xgeq -4 \
    -(x+4) & text{ , if }x <-4
    end{align}
    right}$



    $|x+2|=left{ begin{align}
    x+2 & text{ , if }xgeq -2 \
    -(x+2) & text{ , if }x <-2
    end{align}
    right}$



    Then i got this:



    $1. x<-4: $



    $ -(x+4)x + (-(x+2)) > -2$



    $2. -4 $le$x<-2:$



    $ (x+4)x + (-(x+2)) > -2$



    $3. x$geq$-2: $



    $ (x+4)x+x+2 > -2$



    After i calculated top 3 inequalities i got this answers:



    $1. x$^2$ + 5x < 0$



    $2. x$^2$ + 3x > 0$



    $3. x$^2$+5x+4 > 0$



    I think i calculated everything correctly, but now i do not know how to find the answer to this absolute value inequalitie.



    Thank you for your help.






    inequality absolute-value







    share|cite|improve this question



























      0












      0








      0







      I am having trouble finding what the answer to this inequalitie is:
      $|x+4|x + |x+2| > -2$



      first i did this:



      $|x+4|=left{ begin{align}
      x+4 & text{ , if }xgeq -4 \
      -(x+4) & text{ , if }x <-4
      end{align}
      right}$



      $|x+2|=left{ begin{align}
      x+2 & text{ , if }xgeq -2 \
      -(x+2) & text{ , if }x <-2
      end{align}
      right}$



      Then i got this:



      $1. x<-4: $



      $ -(x+4)x + (-(x+2)) > -2$



      $2. -4 $le$x<-2:$



      $ (x+4)x + (-(x+2)) > -2$



      $3. x$geq$-2: $



      $ (x+4)x+x+2 > -2$



      After i calculated top 3 inequalities i got this answers:



      $1. x$^2$ + 5x < 0$



      $2. x$^2$ + 3x > 0$



      $3. x$^2$+5x+4 > 0$



      I think i calculated everything correctly, but now i do not know how to find the answer to this absolute value inequalitie.



      Thank you for your help.






      inequality absolute-value







      share|cite|improve this question















      I am having trouble finding what the answer to this inequalitie is:
      $|x+4|x + |x+2| > -2$



      first i did this:



      $|x+4|=left{ begin{align}
      x+4 & text{ , if }xgeq -4 \
      -(x+4) & text{ , if }x <-4
      end{align}
      right}$



      $|x+2|=left{ begin{align}
      x+2 & text{ , if }xgeq -2 \
      -(x+2) & text{ , if }x <-2
      end{align}
      right}$



      Then i got this:



      $1. x<-4: $



      $ -(x+4)x + (-(x+2)) > -2$



      $2. -4 $le$x<-2:$



      $ (x+4)x + (-(x+2)) > -2$



      $3. x$geq$-2: $



      $ (x+4)x+x+2 > -2$



      After i calculated top 3 inequalities i got this answers:



      $1. x$^2$ + 5x < 0$



      $2. x$^2$ + 3x > 0$



      $3. x$^2$+5x+4 > 0$



      I think i calculated everything correctly, but now i do not know how to find the answer to this absolute value inequalitie.



      Thank you for your help.






      inequality absolute-value




      inequality absolute-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 20 '18 at 9:50









      Akash Roy

      1




      1










      asked Nov 20 '18 at 9:42









      krneki

      6




      6






















          1 Answer
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          3














          Guide:



          For the first case, we need $x<-4$ and $x^2+5x<0$



          that is $x<-4$ and $x(x+5) <0$ which is equivalent to



          $x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.



          Do the same thing for the other cases as well.



          Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:



          enter image description here






          share|cite|improve this answer





















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            1 Answer
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            active

            oldest

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            3














            Guide:



            For the first case, we need $x<-4$ and $x^2+5x<0$



            that is $x<-4$ and $x(x+5) <0$ which is equivalent to



            $x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.



            Do the same thing for the other cases as well.



            Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:



            enter image description here






            share|cite|improve this answer


























              3














              Guide:



              For the first case, we need $x<-4$ and $x^2+5x<0$



              that is $x<-4$ and $x(x+5) <0$ which is equivalent to



              $x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.



              Do the same thing for the other cases as well.



              Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:



              enter image description here






              share|cite|improve this answer
























                3












                3








                3






                Guide:



                For the first case, we need $x<-4$ and $x^2+5x<0$



                that is $x<-4$ and $x(x+5) <0$ which is equivalent to



                $x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.



                Do the same thing for the other cases as well.



                Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:



                enter image description here






                share|cite|improve this answer












                Guide:



                For the first case, we need $x<-4$ and $x^2+5x<0$



                that is $x<-4$ and $x(x+5) <0$ which is equivalent to



                $x<-4$ and $-5<x<0$. which can be summarized as $-5<x<-4$.



                Do the same thing for the other cases as well.



                Remark: $x(x+5)<0$ is equivalent to $-5<x<0$ can be seen from its graph:



                enter image description here







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 20 '18 at 9:49









                Siong Thye Goh

                99.4k1464117




                99.4k1464117






























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