Is a polynomial of degree 3 with irrational roots possible? [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
2
down vote

favorite













This question already has an answer here:



  • Cubic polynomial with three (distinct) irrational roots

    4 answers



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?










share|cite|improve this question















marked as duplicate by lulu, Winther, Holo, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Oct 4 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • @ChinnapparajR, two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:44







  • 2




    Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
    – lulu
    Oct 4 at 12:44










  • @Edi with rational coefficients?
    – tarit goswami
    Oct 4 at 12:44






  • 10




    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
    Oct 4 at 12:45







  • 4




    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
    Oct 4 at 12:51















up vote
2
down vote

favorite













This question already has an answer here:



  • Cubic polynomial with three (distinct) irrational roots

    4 answers



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?










share|cite|improve this question















marked as duplicate by lulu, Winther, Holo, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Oct 4 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • @ChinnapparajR, two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:44







  • 2




    Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
    – lulu
    Oct 4 at 12:44










  • @Edi with rational coefficients?
    – tarit goswami
    Oct 4 at 12:44






  • 10




    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
    Oct 4 at 12:45







  • 4




    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
    Oct 4 at 12:51













up vote
2
down vote

favorite









up vote
2
down vote

favorite












This question already has an answer here:



  • Cubic polynomial with three (distinct) irrational roots

    4 answers



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?










share|cite|improve this question
















This question already has an answer here:



  • Cubic polynomial with three (distinct) irrational roots

    4 answers



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?





This question already has an answer here:



  • Cubic polynomial with three (distinct) irrational roots

    4 answers







algebra-precalculus elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 4 at 12:45

























asked Oct 4 at 12:41









Edi

870930




870930




marked as duplicate by lulu, Winther, Holo, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Oct 4 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by lulu, Winther, Holo, amWhy algebra-precalculus
Users with the  algebra-precalculus badge can single-handedly close algebra-precalculus questions as duplicates and reopen them as needed.

StackExchange.ready(function()
if (StackExchange.options.isMobile) return;

$('.dupe-hammer-message-hover:not(.hover-bound)').each(function()
var $hover = $(this).addClass('hover-bound'),
$msg = $hover.siblings('.dupe-hammer-message');

$hover.hover(
function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
,
function()
StackExchange.helpers.removeMessages();

);
);
);
Oct 4 at 16:24


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • @ChinnapparajR, two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:44







  • 2




    Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
    – lulu
    Oct 4 at 12:44










  • @Edi with rational coefficients?
    – tarit goswami
    Oct 4 at 12:44






  • 10




    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
    Oct 4 at 12:45







  • 4




    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
    Oct 4 at 12:51

















  • @ChinnapparajR, two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:44







  • 2




    Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
    – lulu
    Oct 4 at 12:44










  • @Edi with rational coefficients?
    – tarit goswami
    Oct 4 at 12:44






  • 10




    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
    Oct 4 at 12:45







  • 4




    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
    Oct 4 at 12:51
















@ChinnapparajR, two rationals one irrational, is my question
– Edi
Oct 4 at 12:44





@ChinnapparajR, two rationals one irrational, is my question
– Edi
Oct 4 at 12:44





2




2




Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
Oct 4 at 12:44




Note: I assume you are requiring that all roots are real. Otherwise $x^3-2$ does the job.
– lulu
Oct 4 at 12:44












@Edi with rational coefficients?
– tarit goswami
Oct 4 at 12:44




@Edi with rational coefficients?
– tarit goswami
Oct 4 at 12:44




10




10




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
Oct 4 at 12:45





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
Oct 4 at 12:45





4




4




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
Oct 4 at 12:51





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
Oct 4 at 12:51











4 Answers
4






active

oldest

votes

















up vote
1
down vote



accepted










Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.






share|cite|improve this answer
















  • 1




    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
    Oct 4 at 12:58






  • 1




    @GEdgar: that's quite right. Of course, you can infer which case is targeted here.
    – Yves Daoust
    Oct 4 at 12:58






  • 1




    @GEdgar When I posted my answer, there was only one question.
    – José Carlos Santos
    Oct 4 at 13:13










  • Adding my 2nd question was almost simultaneous with your answer! :)
    – Edi
    Oct 4 at 14:17

















up vote
10
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



$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$






share|cite|improve this answer




















  • Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
    – Calvin Godfrey
    Oct 4 at 13:29

















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.






share|cite|improve this answer


















  • 1




    two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:51






  • 1




    This case is already known by the OP.
    – Yves Daoust
    Oct 4 at 12:52










  • Note $0$ is rational. So it is better to use the $x^2$ coefficient.
    – GEdgar
    Oct 4 at 12:56










  • @GEdgar You are correct, I fixed it.
    – 5xum
    Oct 4 at 13:05

















up vote
2
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.






share|cite|improve this answer


















  • 2




    See José's answer for three irrational roots with sum $0$ and product $-1$.
    – GEdgar
    Oct 4 at 13:02


















4 Answers
4






active

oldest

votes








4 Answers
4






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.






share|cite|improve this answer
















  • 1




    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
    Oct 4 at 12:58






  • 1




    @GEdgar: that's quite right. Of course, you can infer which case is targeted here.
    – Yves Daoust
    Oct 4 at 12:58






  • 1




    @GEdgar When I posted my answer, there was only one question.
    – José Carlos Santos
    Oct 4 at 13:13










  • Adding my 2nd question was almost simultaneous with your answer! :)
    – Edi
    Oct 4 at 14:17














up vote
1
down vote



accepted










Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.






share|cite|improve this answer
















  • 1




    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
    Oct 4 at 12:58






  • 1




    @GEdgar: that's quite right. Of course, you can infer which case is targeted here.
    – Yves Daoust
    Oct 4 at 12:58






  • 1




    @GEdgar When I posted my answer, there was only one question.
    – José Carlos Santos
    Oct 4 at 13:13










  • Adding my 2nd question was almost simultaneous with your answer! :)
    – Edi
    Oct 4 at 14:17












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.






share|cite|improve this answer












Yes, it is possible. Take $p(x)=x^3-3x+1$, for instance. By the rational root theorem, it has no rational root.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 4 at 12:42









José Carlos Santos

127k17102189




127k17102189







  • 1




    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
    Oct 4 at 12:58






  • 1




    @GEdgar: that's quite right. Of course, you can infer which case is targeted here.
    – Yves Daoust
    Oct 4 at 12:58






  • 1




    @GEdgar When I posted my answer, there was only one question.
    – José Carlos Santos
    Oct 4 at 13:13










  • Adding my 2nd question was almost simultaneous with your answer! :)
    – Edi
    Oct 4 at 14:17












  • 1




    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
    Oct 4 at 12:58






  • 1




    @GEdgar: that's quite right. Of course, you can infer which case is targeted here.
    – Yves Daoust
    Oct 4 at 12:58






  • 1




    @GEdgar When I posted my answer, there was only one question.
    – José Carlos Santos
    Oct 4 at 13:13










  • Adding my 2nd question was almost simultaneous with your answer! :)
    – Edi
    Oct 4 at 14:17







1




1




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
Oct 4 at 12:58




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
Oct 4 at 12:58




1




1




@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
Oct 4 at 12:58




@GEdgar: that's quite right. Of course, you can infer which case is targeted here.
– Yves Daoust
Oct 4 at 12:58




1




1




@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
Oct 4 at 13:13




@GEdgar When I posted my answer, there was only one question.
– José Carlos Santos
Oct 4 at 13:13












Adding my 2nd question was almost simultaneous with your answer! :)
– Edi
Oct 4 at 14:17




Adding my 2nd question was almost simultaneous with your answer! :)
– Edi
Oct 4 at 14:17










up vote
10
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



$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$






share|cite|improve this answer




















  • Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
    – Calvin Godfrey
    Oct 4 at 13:29














up vote
10
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



$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$






share|cite|improve this answer




















  • Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
    – Calvin Godfrey
    Oct 4 at 13:29












up vote
10
down vote










up vote
10
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



$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$






share|cite|improve this answer












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



$$cosfracpi9, cosfrac7pi9, cosfrac13pi9.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 4 at 13:04









Yves Daoust

117k667213




117k667213











  • Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
    – Calvin Godfrey
    Oct 4 at 13:29
















  • Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
    – Calvin Godfrey
    Oct 4 at 13:29















Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
– Calvin Godfrey
Oct 4 at 13:29




Sure it is, the roots of that are 0, sqrt(2), and -sqrt(2). The two irrational roots are sqrt(2) and -sqrt(2).
– Calvin Godfrey
Oct 4 at 13:29










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.






share|cite|improve this answer


















  • 1




    two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:51






  • 1




    This case is already known by the OP.
    – Yves Daoust
    Oct 4 at 12:52










  • Note $0$ is rational. So it is better to use the $x^2$ coefficient.
    – GEdgar
    Oct 4 at 12:56










  • @GEdgar You are correct, I fixed it.
    – 5xum
    Oct 4 at 13:05














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.






share|cite|improve this answer


















  • 1




    two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:51






  • 1




    This case is already known by the OP.
    – Yves Daoust
    Oct 4 at 12:52










  • Note $0$ is rational. So it is better to use the $x^2$ coefficient.
    – GEdgar
    Oct 4 at 12:56










  • @GEdgar You are correct, I fixed it.
    – 5xum
    Oct 4 at 13:05












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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 4 at 13:05

























answered Oct 4 at 12:49









5xum

84.7k388153




84.7k388153







  • 1




    two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:51






  • 1




    This case is already known by the OP.
    – Yves Daoust
    Oct 4 at 12:52










  • Note $0$ is rational. So it is better to use the $x^2$ coefficient.
    – GEdgar
    Oct 4 at 12:56










  • @GEdgar You are correct, I fixed it.
    – 5xum
    Oct 4 at 13:05












  • 1




    two rationals one irrational, is my question
    – Edi
    Oct 4 at 12:51






  • 1




    This case is already known by the OP.
    – Yves Daoust
    Oct 4 at 12:52










  • Note $0$ is rational. So it is better to use the $x^2$ coefficient.
    – GEdgar
    Oct 4 at 12:56










  • @GEdgar You are correct, I fixed it.
    – 5xum
    Oct 4 at 13:05







1




1




two rationals one irrational, is my question
– Edi
Oct 4 at 12:51




two rationals one irrational, is my question
– Edi
Oct 4 at 12:51




1




1




This case is already known by the OP.
– Yves Daoust
Oct 4 at 12:52




This case is already known by the OP.
– Yves Daoust
Oct 4 at 12:52












Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
Oct 4 at 12:56




Note $0$ is rational. So it is better to use the $x^2$ coefficient.
– GEdgar
Oct 4 at 12:56












@GEdgar You are correct, I fixed it.
– 5xum
Oct 4 at 13:05




@GEdgar You are correct, I fixed it.
– 5xum
Oct 4 at 13:05










up vote
2
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.






share|cite|improve this answer


















  • 2




    See José's answer for three irrational roots with sum $0$ and product $-1$.
    – GEdgar
    Oct 4 at 13:02















up vote
2
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.






share|cite|improve this answer


















  • 2




    See José's answer for three irrational roots with sum $0$ and product $-1$.
    – GEdgar
    Oct 4 at 13:02













up vote
2
down vote










up vote
2
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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 4 at 13:13

























answered Oct 4 at 12:49









tarit goswami

1,501219




1,501219







  • 2




    See José's answer for three irrational roots with sum $0$ and product $-1$.
    – GEdgar
    Oct 4 at 13:02













  • 2




    See José's answer for three irrational roots with sum $0$ and product $-1$.
    – GEdgar
    Oct 4 at 13:02








2




2




See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
Oct 4 at 13:02





See José's answer for three irrational roots with sum $0$ and product $-1$.
– GEdgar
Oct 4 at 13:02



Popular posts from this blog

How to check contact read email or not when send email to Individual?

Displaying single band from multi-band raster using QGIS

How many registers does an x86_64 CPU actually have?