Mixing
$||x-1|-|x+2||=p$ find p for which the equation has one solution
$begingroup$
Consider the equation $||x-1|-|x+2||=p$
Find the value of $p$ for which the above equation has one solution.
algebra-precalculus absolute-value
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
add a comment |
$begingroup$
Consider the equation $||x-1|-|x+2||=p$
Find the value of $p$ for which the above equation has one solution.
algebra-precalculus absolute-value
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07
add a comment |
$begingroup$
Consider the equation $||x-1|-|x+2||=p$
Find the value of $p$ for which the above equation has one solution.
algebra-precalculus absolute-value
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
Consider the equation $||x-1|-|x+2||=p$
Find the value of $p$ for which the above equation has one solution.
algebra-precalculus absolute-value
algebra-precalculus absolute-value
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
edited Jan 30 at 15:52
YuiTo Cheng
2,1963937
2,1963937
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
asked Jan 30 at 15:02
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
1,6191520
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07
add a comment |
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
Notice the range of $||x-1|-|x+2||=[0,3]$, so $pin [0,3]$
for $xleqslant-2$ or $x geqslant1$ , $p=3$
if $pneq0$, there are $2$ distinct solutions for $x$ (why?)
so $p=0$
Edit:
The graph of $||x-1|-|x+2||$
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
$endgroup$
add a comment |
$begingroup$
You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
$endgroup$
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%2f3093636%2fx-1-x2-p-find-p-for-which-the-equation-has-one-solution%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
Notice the range of $||x-1|-|x+2||=[0,3]$, so $pin [0,3]$
for $xleqslant-2$ or $x geqslant1$ , $p=3$
if $pneq0$, there are $2$ distinct solutions for $x$ (why?)
so $p=0$
Edit:
The graph of $||x-1|-|x+2||$
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
$endgroup$
add a comment |
$begingroup$
Hint:
Notice the range of $||x-1|-|x+2||=[0,3]$, so $pin [0,3]$
for $xleqslant-2$ or $x geqslant1$ , $p=3$
if $pneq0$, there are $2$ distinct solutions for $x$ (why?)
so $p=0$
Edit:
The graph of $||x-1|-|x+2||$
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
$endgroup$
add a comment |
$begingroup$
Hint:
Notice the range of $||x-1|-|x+2||=[0,3]$, so $pin [0,3]$
for $xleqslant-2$ or $x geqslant1$ , $p=3$
if $pneq0$, there are $2$ distinct solutions for $x$ (why?)
so $p=0$
Edit:
The graph of $||x-1|-|x+2||$
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
$endgroup$
Hint:
Notice the range of $||x-1|-|x+2||=[0,3]$, so $pin [0,3]$
for $xleqslant-2$ or $x geqslant1$ , $p=3$
if $pneq0$, there are $2$ distinct solutions for $x$ (why?)
so $p=0$
Edit:
The graph of $||x-1|-|x+2||$
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
edited Jan 30 at 15:50
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
answered Jan 30 at 15:16
YuiTo ChengYuiTo Cheng
2,1963937
2,1963937
add a comment |
add a comment |
$begingroup$
You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
$endgroup$
add a comment |
$begingroup$
You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
$endgroup$
add a comment |
$begingroup$
You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
$endgroup$
You need $|x-1|=|x+2|$ and you need that this has only one solution, which is the case for $x=-0.5$. So $p=0$is indeed the answer.
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
answered Jan 30 at 15:19
AndreasAndreas
8,4161137
8,4161137
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.
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%2f3093636%2fx-1-x2-p-find-p-for-which-the-equation-has-one-solution%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
$begingroup$
My intuition tells me p=0
$endgroup$
– YuiTo Cheng
Jan 30 at 15:07