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.
inequality absolute-value
edited Nov 20 at 9:50
Akash Roy
51214
asked Nov 20 at 9:42
krneki
6
New contributor
krneki is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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.
inequality absolute-value
edited Nov 20 at 9:50
Akash Roy
51214
asked Nov 20 at 9:42
krneki
6
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 |
up vote
0
down vote
favorite
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.
inequality absolute-value
edited Nov 20 at 9:50
Akash Roy
51214
asked Nov 20 at 9:42
krneki
6
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.
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
edited Nov 20 at 9:50
Akash Roy
51214
asked Nov 20 at 9:42
krneki
6
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.
edited Nov 20 at 9:50
Akash Roy
51214
asked Nov 20 at 9:42
krneki
6
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.
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
51214
asked Nov 20 at 9:42
krneki
6
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.
asked Nov 20 at 9:42
krneki
6
asked Nov 20 at 9:42
krneki
6
6
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.
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.
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 |
add a comment |
1 Answer
1
active
oldest
votes
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:
answered Nov 20 at 9:49
Siong Thye Goh
93.6k1462114
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
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:
answered Nov 20 at 9:49
Siong Thye Goh
93.6k1462114
add a comment |
up vote
3
down vote
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:
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:
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
93.6k1462114
add a comment |
add a comment |
krneki is a new contributor. Be nice, and check out our Code of Conduct.
krneki is a new contributor. Be nice, and check out our Code of Conduct.
krneki is a new contributor. Be nice, and check out our Code of Conduct.
krneki is a new contributor. Be nice, and check out our Code of Conduct.
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