Sum and Product of Root
$begingroup$
Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?
algebra-precalculus elementary-number-theory
edited Jan 18 at 2:10
clathratus
4,478336
asked Jan 18 at 2:09
sumisumi
72
$endgroup$
add a comment |
$begingroup$
Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?
algebra-precalculus elementary-number-theory
edited Jan 18 at 2:10
clathratus
4,478336
asked Jan 18 at 2:09
sumisumi
72
$endgroup$
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
3
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16
add a comment |
$begingroup$
Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?
algebra-precalculus elementary-number-theory
edited Jan 18 at 2:10
clathratus
4,478336
asked Jan 18 at 2:09
sumisumi
72
$endgroup$
Suppose the roots of the polynomial $x^2+mx+n$ are positive prime integers (not necessarily distinct). Given that $m<20$ how many possible values of $n$ are there?
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
edited Jan 18 at 2:10
clathratus
4,478336
asked Jan 18 at 2:09
sumisumi
72
edited Jan 18 at 2:10
clathratus
4,478336
asked Jan 18 at 2:09
sumisumi
72
edited Jan 18 at 2:10
clathratus
4,478336
edited Jan 18 at 2:10
clathratus
4,478336
edited Jan 18 at 2:10
clathratus
4,478336
4,478336
asked Jan 18 at 2:09
sumisumi
72
asked Jan 18 at 2:09
sumisumi
72
asked Jan 18 at 2:09
sumisumi
72
72
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
3
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16
add a comment |
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
3
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
3
3
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming $m < 20$ means $|m| < 20$,
Say the roots are $p_1$ and $p_2$ (needn't be distinct)
then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$
$m = - (p_1 + p_2)$ and $n = p_1p_2$
So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$
number of $n = 18$
answered Jan 18 at 3:11
NeilNeil
113
$endgroup$
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
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%2f3077753%2fsum-and-product-of-root%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
$begingroup$
Assuming $m < 20$ means $|m| < 20$,
Say the roots are $p_1$ and $p_2$ (needn't be distinct)
then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$
$m = - (p_1 + p_2)$ and $n = p_1p_2$
So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$
number of $n = 18$
answered Jan 18 at 3:11
NeilNeil
113
$endgroup$
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
add a comment |
$begingroup$
Assuming $m < 20$ means $|m| < 20$,
Say the roots are $p_1$ and $p_2$ (needn't be distinct)
then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$
$m = - (p_1 + p_2)$ and $n = p_1p_2$
So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$
number of $n = 18$
answered Jan 18 at 3:11
NeilNeil
113
$endgroup$
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
add a comment |
$begingroup$
Assuming $m < 20$ means $|m| < 20$,
Say the roots are $p_1$ and $p_2$ (needn't be distinct)
then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$
$m = - (p_1 + p_2)$ and $n = p_1p_2$
So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$
number of $n = 18$
answered Jan 18 at 3:11
NeilNeil
113
$endgroup$
Assuming $m < 20$ means $|m| < 20$,
Say the roots are $p_1$ and $p_2$ (needn't be distinct)
then, the quadratic becomes $x^2 - (p_1 + p_2)x + p_1p_2$
$m = - (p_1 + p_2)$ and $n = p_1p_2$
So, $p_1$ and $p_2$ can take all combinations (repeat) from$ ({2, 3, 5, 7, 11, 13, 17} | p_1 + p_2 < 20)$
number of $n = 18$
answered Jan 18 at 3:11
NeilNeil
113
edited Jan 18 at 3:52
answered Jan 18 at 3:11
NeilNeil
113
answered Jan 18 at 3:11
NeilNeil
113
answered Jan 18 at 3:11
NeilNeil
113
113
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
add a comment |
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
You should have $17$ instead of $19$, and unless I'm missing something, there are only $18$ valid combinations.
$endgroup$
– Daniel Mathias
Jan 18 at 3:23
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
$begingroup$
@DanielMathias Yeah, I've repeated some. I corrected it :)
$endgroup$
– Neil
Jan 18 at 3:54
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%2f3077753%2fsum-and-product-of-root%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
$begingroup$
What have you tried? Have you, say, found some possible values for $n$?
$endgroup$
– lulu
Jan 18 at 2:12
3
$begingroup$
Also, did you mean to write $|m|<20$? After all, if the roots are both positive then $m$ is negative, so, as written, there are infinitely many possible values of $n$.
$endgroup$
– lulu
Jan 18 at 2:16