Help with absolute value inequalitie
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
add a comment |
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
add a comment |
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
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 '18 at 9:50
Akash Roy
1
1
asked Nov 20 '18 at 9:42
krneki
6
6
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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:
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006122%2fhelp-with-absolute-value-inequalitie%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:
add a comment |
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:
add a comment |
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:
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 '18 at 9:49
Siong Thye Goh
99.4k1464117
99.4k1464117
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006122%2fhelp-with-absolute-value-inequalitie%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown