Help with absolute value inequalitie

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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.

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

add a comment |

up vote
0
down vote

favorite

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.

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

add a comment |

up vote
0
down vote

favorite

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.

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

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

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

edited Nov 20 at 9:50

Akash Roy

51214

edited Nov 20 at 9:50

Akash Roy

51214

edited Nov 20 at 9:50

Akash Roy

51214

asked Nov 20 at 9:42

krneki

New contributor

asked Nov 20 at 9:42

krneki

asked Nov 20 at 9:42

krneki

New contributor

krneki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment |

1 Answer
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3
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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

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

add a comment |

Your Answer

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1 Answer
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1 Answer
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active

oldest

votes

up vote
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down vote

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

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

add a comment |

up vote
3
down vote

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

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

add a comment |

up vote
3
down vote

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

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

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

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

answered Nov 20 at 9:49

Siong Thye Goh

93.6k1462114

add a comment |

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