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Is a polynomial of degree 3 with irrational roots possible?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
It is easy to give an example of a polynomial of degree 3 with integer coefficients having:
(a) three distinct rational roots,
(b) one rational root and two irrational roots.
But for a while I am trying to construct one that all its roots are irrational but I can't. It seems that it is not possible at all?
Also, can a polynomial with integer coefficients of degree 3 have two rational roots and one irrational root?
algebra-precalculus elementary-number-theory
asked 46 mins ago
Edi
850930
 |Â
show 2 more comments
up vote
2
down vote
favorite
It is easy to give an example of a polynomial of degree 3 with integer coefficients having:
(a) three distinct rational roots,
(b) one rational root and two irrational roots.
But for a while I am trying to construct one that all its roots are irrational but I can't. It seems that it is not possible at all?
Also, can a polynomial with integer coefficients of degree 3 have two rational roots and one irrational root?
algebra-precalculus elementary-number-theory
asked 46 mins ago
Edi
850930
3
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
7
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
3
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It is easy to give an example of a polynomial of degree 3 with integer coefficients having:
(a) three distinct rational roots,
(b) one rational root and two irrational roots.
But for a while I am trying to construct one that all its roots are irrational but I can't. It seems that it is not possible at all?
Also, can a polynomial with integer coefficients of degree 3 have two rational roots and one irrational root?
algebra-precalculus elementary-number-theory
asked 46 mins ago
Edi
850930
It is easy to give an example of a polynomial of degree 3 with integer coefficients having:
(a) three distinct rational roots,
(b) one rational root and two irrational roots.
But for a while I am trying to construct one that all its roots are irrational but I can't. It seems that it is not possible at all?
Also, can a polynomial with integer coefficients of degree 3 have two rational roots and one irrational root?
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
asked 46 mins ago
Edi
850930
asked 46 mins ago
Edi
850930
edited 41 mins ago
asked 46 mins ago
Edi
850930
asked 46 mins ago
Edi
850930
asked 46 mins ago
Edi
850930
850930
3
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
7
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
3
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago
 |Â
show 2 more comments
3
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
7
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
3
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago
3
3
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
7
7
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
3
3
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago
 |Â
show 2 more comments
4 Answers
4
active
oldest
votes
up vote
3
down vote
By Vieta the sum of the roots must be rational, hence this excludes a single irrational.
All other cases are possible.
$$beginalign0&:x(x^2-1)=0,
\2&:x(x^2-2)=0,
\3&:8x^3-6x-1=0.endalign$$
The last one was built from
$$cos3theta=4cos^3theta-3costheta=cosfracpi3$$
so that the roots are
$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$
answered 22 mins ago
Yves Daoust
116k667211
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
add a comment |Â
up vote
2
down vote
Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)cdot q(x)$$
then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)
However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.
answered 37 mins ago
5xum
84.3k386151
1
two rationals one irrational, is my question
– Edi
35 mins ago
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
add a comment |Â
up vote
1
down vote
Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.
answered 44 mins ago
José Carlos Santos
126k17102189
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
add a comment |Â
up vote
1
down vote
To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.
Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+sqrtb$ is $a-sqrtb$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.
For the case with all 3 irrational roots, see here.
answered 37 mins ago
tarit goswami
1,378219
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
By Vieta the sum of the roots must be rational, hence this excludes a single irrational.
All other cases are possible.
$$beginalign0&:x(x^2-1)=0,
\2&:x(x^2-2)=0,
\3&:8x^3-6x-1=0.endalign$$
The last one was built from
$$cos3theta=4cos^3theta-3costheta=cosfracpi3$$
so that the roots are
$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$
answered 22 mins ago
Yves Daoust
116k667211
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
add a comment |Â
up vote
3
down vote
By Vieta the sum of the roots must be rational, hence this excludes a single irrational.
All other cases are possible.
$$beginalign0&:x(x^2-1)=0,
\2&:x(x^2-2)=0,
\3&:8x^3-6x-1=0.endalign$$
The last one was built from
$$cos3theta=4cos^3theta-3costheta=cosfracpi3$$
so that the roots are
$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$
answered 22 mins ago
Yves Daoust
116k667211
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
By Vieta the sum of the roots must be rational, hence this excludes a single irrational.
All other cases are possible.
$$beginalign0&:x(x^2-1)=0,
\2&:x(x^2-2)=0,
\3&:8x^3-6x-1=0.endalign$$
The last one was built from
$$cos3theta=4cos^3theta-3costheta=cosfracpi3$$
so that the roots are
$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$
answered 22 mins ago
Yves Daoust
116k667211
By Vieta the sum of the roots must be rational, hence this excludes a single irrational.
All other cases are possible.
$$beginalign0&:x(x^2-1)=0,
\2&:x(x^2-2)=0,
\3&:8x^3-6x-1=0.endalign$$
The last one was built from
$$cos3theta=4cos^3theta-3costheta=cosfracpi3$$
so that the roots are
$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$
answered 22 mins ago
Yves Daoust
116k667211
answered 22 mins ago
Yves Daoust
116k667211
answered 22 mins ago
Yves Daoust
116k667211
answered 22 mins ago
Yves Daoust
116k667211
116k667211
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
add a comment |Â
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
Your example for 2 irrational roots isn't; x(x^2 - 2) = x(x + rt(2))(x - rt(2))
– BlueHairedMeerkat
3 mins ago
add a comment |Â
up vote
2
down vote
Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)cdot q(x)$$
then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)
However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.
answered 37 mins ago
5xum
84.3k386151
1
two rationals one irrational, is my question
– Edi
35 mins ago
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
add a comment |Â
up vote
2
down vote
Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)cdot q(x)$$
then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)
However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.
answered 37 mins ago
5xum
84.3k386151
1
two rationals one irrational, is my question
– Edi
35 mins ago
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)cdot q(x)$$
then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)
However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.
answered 37 mins ago
5xum
84.3k386151
Take any second order polynomial with two irrational roots (shouldn't be hard) $q$, and take $$p(x) = (x-1)cdot q(x)$$
then, clearly, $p$ has one rational root (it's $1$) and two irrational roots (the same as $q$)
However, it is not possible for the polynomial $p$ to have two rational roots $r_1, r_2$ and one irrational one $z$. That would imply that $p=(x-r_1)(x-r_2)(x-z)$, and clearly, the expanding this polynomial shows that the coefficient at $x^2$ is $-z-r_1-r_2$. This number is rational only if $z$ is also rational.
answered 37 mins ago
5xum
84.3k386151
edited 21 mins ago
answered 37 mins ago
5xum
84.3k386151
answered 37 mins ago
5xum
84.3k386151
answered 37 mins ago
5xum
84.3k386151
84.3k386151
1
two rationals one irrational, is my question
– Edi
35 mins ago
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
add a comment |Â
1
two rationals one irrational, is my question
– Edi
35 mins ago
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
1
1
two rationals one irrational, is my question
– Edi
35 mins ago
two rationals one irrational, is my question
– Edi
35 mins ago
1
1
This case is already known by the OP.
– Yves Daoust
35 mins ago
This case is already known by the OP.
– Yves Daoust
35 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
30 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
@GEdgar You are correct, I fixed it.
– 5xum
21 mins ago
add a comment |Â
up vote
1
down vote
Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.
answered 44 mins ago
José Carlos Santos
126k17102189
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
add a comment |Â
up vote
1
down vote
Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.
answered 44 mins ago
José Carlos Santos
126k17102189
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.
answered 44 mins ago
José Carlos Santos
126k17102189
Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.
answered 44 mins ago
José Carlos Santos
126k17102189
answered 44 mins ago
José Carlos Santos
126k17102189
answered 44 mins ago
José Carlos Santos
126k17102189
answered 44 mins ago
José Carlos Santos
126k17102189
126k17102189
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
add a comment |Â
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
When there are two questions, and you say "it" is possible, we do not know which one you mean. A reason to limit questions to one only.
– GEdgar
29 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
28 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
14 mins ago
add a comment |Â
up vote
1
down vote
To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.
Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+sqrtb$ is $a-sqrtb$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.
For the case with all 3 irrational roots, see here.
answered 37 mins ago
tarit goswami
1,378219
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
add a comment |Â
up vote
1
down vote
To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.
Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+sqrtb$ is $a-sqrtb$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.
For the case with all 3 irrational roots, see here.
answered 37 mins ago
tarit goswami
1,378219
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.
Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+sqrtb$ is $a-sqrtb$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.
For the case with all 3 irrational roots, see here.
answered 37 mins ago
tarit goswami
1,378219
To the first one of your queries the answer is - No, it's not possible to construct if you want all coefficients to be rational. As, from Vieta's formula we have sum of roots of a polynomial $f(x)=ax^3+bx^2+cx+d$ equals to $-b/a$, which is rational as you want all $a,b,c,d$ to be integers.
Now,if you want one root to be irrational, you can't get the sum a rational one. As, you will always need the conjugate surd(conjugate of $a+sqrtb$ is $a-sqrtb$, which is irrational) to make the sum a rational one. For any other non-algebraic irrational number like $e$, no matter what you take, you will get the product a irrational number, but from Vieta again product of roots is $-d/a$, a rational number.
For the case with all 3 irrational roots, see here.
answered 37 mins ago
tarit goswami
1,378219
edited 14 mins ago
answered 37 mins ago
tarit goswami
1,378219
answered 37 mins ago
tarit goswami
1,378219
answered 37 mins ago
tarit goswami
1,378219
1,378219
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
add a comment |Â
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
1
1
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
24 mins ago
add a comment |Â
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3
Possible duplicate of Cubic polynomial with three (distinct) irrational roots
– lulu
43 mins ago
@ChinnapparajR, two rationals one irrational, is my question
– Edi
43 mins ago
Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
42 mins ago
7
If the polynomial has integer coefficients, then the sum of the roots is rational. Therefore, it cannot have one irrational and two rational roots.
– GEdgar
42 mins ago
3
There isn't a clear, universally accepted, convention on "irrational $implies$ real". Some people say all non-real complex numbers are irrational. See, e.g., this question Always worth specifying your intent.
– lulu
35 mins ago