Sum and Product of Root
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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
5,1441439
asked Jan 18 at 2:09
sumisumi
654
$endgroup$
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
5,1441439
asked Jan 18 at 2:09
sumisumi
654
$endgroup$
$begingroup$
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
5,1441439
asked Jan 18 at 2:09
sumisumi
654
$endgroup$
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
5,1441439
asked Jan 18 at 2:09
sumisumi
654
edited Jan 18 at 2:10
clathratus
5,1441439
asked Jan 18 at 2:09
sumisumi
654
edited Jan 18 at 2:10
clathratus
5,1441439
edited Jan 18 at 2:10
clathratus
5,1441439
edited Jan 18 at 2:10
clathratus
5,1441439
5,1441439
asked Jan 18 at 2:09
sumisumi
654
asked Jan 18 at 2:09
sumisumi
654
asked Jan 18 at 2:09
sumisumi
654
654
$begingroup$
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 |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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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 |
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$begingroup$
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