Does $x^3 - frac{m}{n}sqrt{5}x - 1$ has rational root?Expressing a root of a polynomial as a rational...
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Does $x^3 - frac{m}{n}sqrt{5}x - 1$ has rational root?
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I am trying to show whether $p(x) = x^3 - frac{m}{n}sqrt{5}x - 1$ has a rational root or not, where $frac{m}{n}$ is rational. My attempt so far is to turn $p(x)$ into another polynomial $q(x) = ( - frac{m}{n}sqrt{5}x - (1 - x^3))(- frac{m}{n}sqrt{5}x + (1 - x^3))= -n^2x^6 + 2n^2x^3 + 5m^2x^2 - n^2$, that has all the roots of $p(x)$, and trying to show $p(x)$ has (or has no) rational roots. But I am having trouble determining this for $q(x)$ since the coefficients of the highest and lowest ordered term is an arbitrary integer, which makes applying rational root theorem difficult. So I am wondering whether there is a way to answer the question.
elementary-number-theory polynomials roots rational-numbers cubic-equations
edited yesterday
Michael Rozenberg
107k1895199
asked yesterday
PsychoComPsychoCom
1227
$endgroup$
add a comment |
$begingroup$
I am trying to show whether $p(x) = x^3 - frac{m}{n}sqrt{5}x - 1$ has a rational root or not, where $frac{m}{n}$ is rational. My attempt so far is to turn $p(x)$ into another polynomial $q(x) = ( - frac{m}{n}sqrt{5}x - (1 - x^3))(- frac{m}{n}sqrt{5}x + (1 - x^3))= -n^2x^6 + 2n^2x^3 + 5m^2x^2 - n^2$, that has all the roots of $p(x)$, and trying to show $p(x)$ has (or has no) rational roots. But I am having trouble determining this for $q(x)$ since the coefficients of the highest and lowest ordered term is an arbitrary integer, which makes applying rational root theorem difficult. So I am wondering whether there is a way to answer the question.
elementary-number-theory polynomials roots rational-numbers cubic-equations
edited yesterday
Michael Rozenberg
107k1895199
asked yesterday
PsychoComPsychoCom
1227
$endgroup$
add a comment |
$begingroup$
I am trying to show whether $p(x) = x^3 - frac{m}{n}sqrt{5}x - 1$ has a rational root or not, where $frac{m}{n}$ is rational. My attempt so far is to turn $p(x)$ into another polynomial $q(x) = ( - frac{m}{n}sqrt{5}x - (1 - x^3))(- frac{m}{n}sqrt{5}x + (1 - x^3))= -n^2x^6 + 2n^2x^3 + 5m^2x^2 - n^2$, that has all the roots of $p(x)$, and trying to show $p(x)$ has (or has no) rational roots. But I am having trouble determining this for $q(x)$ since the coefficients of the highest and lowest ordered term is an arbitrary integer, which makes applying rational root theorem difficult. So I am wondering whether there is a way to answer the question.
elementary-number-theory polynomials roots rational-numbers cubic-equations
edited yesterday
Michael Rozenberg
107k1895199
asked yesterday
PsychoComPsychoCom
1227
$endgroup$
I am trying to show whether $p(x) = x^3 - frac{m}{n}sqrt{5}x - 1$ has a rational root or not, where $frac{m}{n}$ is rational. My attempt so far is to turn $p(x)$ into another polynomial $q(x) = ( - frac{m}{n}sqrt{5}x - (1 - x^3))(- frac{m}{n}sqrt{5}x + (1 - x^3))= -n^2x^6 + 2n^2x^3 + 5m^2x^2 - n^2$, that has all the roots of $p(x)$, and trying to show $p(x)$ has (or has no) rational roots. But I am having trouble determining this for $q(x)$ since the coefficients of the highest and lowest ordered term is an arbitrary integer, which makes applying rational root theorem difficult. So I am wondering whether there is a way to answer the question.
elementary-number-theory polynomials roots rational-numbers cubic-equations
elementary-number-theory polynomials roots rational-numbers cubic-equations
edited yesterday
Michael Rozenberg
107k1895199
asked yesterday
PsychoComPsychoCom
1227
edited yesterday
Michael Rozenberg
107k1895199
asked yesterday
PsychoComPsychoCom
1227
edited yesterday
Michael Rozenberg
107k1895199
edited yesterday
Michael Rozenberg
107k1895199
edited yesterday
Michael Rozenberg
107k1895199
107k1895199
asked yesterday
PsychoComPsychoCom
1227
asked yesterday
PsychoComPsychoCom
1227
asked yesterday
PsychoComPsychoCom
1227
1227
add a comment |
add a comment |
1 Answer
1
active
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$begingroup$
Let $x$ be a rational root.
Thus, $xneq0$ and $$sqrt5=frac{x^3-1}{frac{m}{n}x},$$ which is a contradiction for $frac{m}{n}neq0$.
Can you end it now?
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $x$ be a rational root.
Thus, $xneq0$ and $$sqrt5=frac{x^3-1}{frac{m}{n}x},$$ which is a contradiction for $frac{m}{n}neq0$.
Can you end it now?
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
add a comment |
$begingroup$
Let $x$ be a rational root.
Thus, $xneq0$ and $$sqrt5=frac{x^3-1}{frac{m}{n}x},$$ which is a contradiction for $frac{m}{n}neq0$.
Can you end it now?
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
add a comment |
$begingroup$
Let $x$ be a rational root.
Thus, $xneq0$ and $$sqrt5=frac{x^3-1}{frac{m}{n}x},$$ which is a contradiction for $frac{m}{n}neq0$.
Can you end it now?
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
$endgroup$
Let $x$ be a rational root.
Thus, $xneq0$ and $$sqrt5=frac{x^3-1}{frac{m}{n}x},$$ which is a contradiction for $frac{m}{n}neq0$.
Can you end it now?
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
answered yesterday
Michael RozenbergMichael Rozenberg
107k1895199
107k1895199
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
add a comment |
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
good catch....!
$endgroup$
– Chinnapparaj R
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
$begingroup$
Sorry but I can't see why this contradicts with $frac{m}{n} ne 0$?
$endgroup$
– PsychoCom
yesterday
1
1
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
$begingroup$
@PsychoCom We got that $sqrt5$ is a rational number, which is not so.
$endgroup$
– Michael Rozenberg
yesterday
add a comment |
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