Is there a general formula for any polynomials of rational, non-integer degree?
$begingroup$
There are formulae out there for certain integer-degree polynomials; for instance, the general solution for $ax^2+bx+c$ is $x=frac{-bpmsqrt{b^2-4ac}}{2a}$.
Is there a general form equation for, say, $asqrt{x}+bx+c$? Or $asqrt[3]{x}+bsqrt{x}+cx+d$? And so on?
algebra-precalculus
asked Jan 9 at 3:16
DonielFDonielF
515515
$endgroup$
add a comment |
$begingroup$
There are formulae out there for certain integer-degree polynomials; for instance, the general solution for $ax^2+bx+c$ is $x=frac{-bpmsqrt{b^2-4ac}}{2a}$.
Is there a general form equation for, say, $asqrt{x}+bx+c$? Or $asqrt[3]{x}+bsqrt{x}+cx+d$? And so on?
algebra-precalculus
asked Jan 9 at 3:16
DonielFDonielF
515515
$endgroup$
3
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
1
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24
add a comment |
$begingroup$
There are formulae out there for certain integer-degree polynomials; for instance, the general solution for $ax^2+bx+c$ is $x=frac{-bpmsqrt{b^2-4ac}}{2a}$.
Is there a general form equation for, say, $asqrt{x}+bx+c$? Or $asqrt[3]{x}+bsqrt{x}+cx+d$? And so on?
algebra-precalculus
asked Jan 9 at 3:16
DonielFDonielF
515515
$endgroup$
There are formulae out there for certain integer-degree polynomials; for instance, the general solution for $ax^2+bx+c$ is $x=frac{-bpmsqrt{b^2-4ac}}{2a}$.
Is there a general form equation for, say, $asqrt{x}+bx+c$? Or $asqrt[3]{x}+bsqrt{x}+cx+d$? And so on?
algebra-precalculus
algebra-precalculus
asked Jan 9 at 3:16
DonielFDonielF
515515
asked Jan 9 at 3:16
DonielFDonielF
515515
asked Jan 9 at 3:16
DonielFDonielF
515515
asked Jan 9 at 3:16
DonielFDonielF
515515
asked Jan 9 at 3:16
DonielFDonielF
515515
515515
3
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
1
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24
add a comment |
3
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
1
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24
3
3
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
1
1
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are only general formulas for finding the roots of polynomials up to the $4$th degree. It's been proven that there are no such formulas for all $5$th or higher degree polynomials. The Wikipedia entry Polynomial says:
In 1824, Niels Henrik Abel proved the striking result that there are equations of degree 5 whose solutions cannot be expressed by a (finite) formula, involving only arithmetic operations and radicals (see Abel–Ruffini theorem). In 1830, Évariste Galois proved that most equations of degree higher than four cannot be solved by radicals, and showed that for each equation, one may decide whether it is solvable by radicals, and, if it is, solve it.
As mentioned by myself and other commenters, such as hardmath, you can convert equations involving rational roots of a variable by using a substitution of the smallest power which will cause all of the powers of this transformed equation to be integral.
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
$endgroup$
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
|
show 6 more comments
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%2f3067030%2fis-there-a-general-formula-for-any-polynomials-of-rational-non-integer-degree%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
$begingroup$
There are only general formulas for finding the roots of polynomials up to the $4$th degree. It's been proven that there are no such formulas for all $5$th or higher degree polynomials. The Wikipedia entry Polynomial says:
In 1824, Niels Henrik Abel proved the striking result that there are equations of degree 5 whose solutions cannot be expressed by a (finite) formula, involving only arithmetic operations and radicals (see Abel–Ruffini theorem). In 1830, Évariste Galois proved that most equations of degree higher than four cannot be solved by radicals, and showed that for each equation, one may decide whether it is solvable by radicals, and, if it is, solve it.
As mentioned by myself and other commenters, such as hardmath, you can convert equations involving rational roots of a variable by using a substitution of the smallest power which will cause all of the powers of this transformed equation to be integral.
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
$endgroup$
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
|
show 6 more comments
$begingroup$
There are only general formulas for finding the roots of polynomials up to the $4$th degree. It's been proven that there are no such formulas for all $5$th or higher degree polynomials. The Wikipedia entry Polynomial says:
In 1824, Niels Henrik Abel proved the striking result that there are equations of degree 5 whose solutions cannot be expressed by a (finite) formula, involving only arithmetic operations and radicals (see Abel–Ruffini theorem). In 1830, Évariste Galois proved that most equations of degree higher than four cannot be solved by radicals, and showed that for each equation, one may decide whether it is solvable by radicals, and, if it is, solve it.
As mentioned by myself and other commenters, such as hardmath, you can convert equations involving rational roots of a variable by using a substitution of the smallest power which will cause all of the powers of this transformed equation to be integral.
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
$endgroup$
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
|
show 6 more comments
$begingroup$
There are only general formulas for finding the roots of polynomials up to the $4$th degree. It's been proven that there are no such formulas for all $5$th or higher degree polynomials. The Wikipedia entry Polynomial says:
In 1824, Niels Henrik Abel proved the striking result that there are equations of degree 5 whose solutions cannot be expressed by a (finite) formula, involving only arithmetic operations and radicals (see Abel–Ruffini theorem). In 1830, Évariste Galois proved that most equations of degree higher than four cannot be solved by radicals, and showed that for each equation, one may decide whether it is solvable by radicals, and, if it is, solve it.
As mentioned by myself and other commenters, such as hardmath, you can convert equations involving rational roots of a variable by using a substitution of the smallest power which will cause all of the powers of this transformed equation to be integral.
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
$endgroup$
There are only general formulas for finding the roots of polynomials up to the $4$th degree. It's been proven that there are no such formulas for all $5$th or higher degree polynomials. The Wikipedia entry Polynomial says:
In 1824, Niels Henrik Abel proved the striking result that there are equations of degree 5 whose solutions cannot be expressed by a (finite) formula, involving only arithmetic operations and radicals (see Abel–Ruffini theorem). In 1830, Évariste Galois proved that most equations of degree higher than four cannot be solved by radicals, and showed that for each equation, one may decide whether it is solvable by radicals, and, if it is, solve it.
As mentioned by myself and other commenters, such as hardmath, you can convert equations involving rational roots of a variable by using a substitution of the smallest power which will cause all of the powers of this transformed equation to be integral.
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
edited Jan 9 at 6:13
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
answered Jan 9 at 3:18
John OmielanJohn Omielan
3,1951214
3,1951214
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
|
show 6 more comments
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
$begingroup$
Among integers, yes, but what about rationals in general?
$endgroup$
– DonielF
Jan 9 at 3:19
1
1
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
@DonielF Your rational form equations are basically the same as ones with integers when you make the appropriate substitution. For example, with your $asqrt{x} + bx + c$, you can have $y = sqrt{x}$ to have the equivalent polynomial of $ay + by^2 + c$.
$endgroup$
– John Omielan
Jan 9 at 3:23
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
So... You're saying that the solution to $asqrt{x}+bx+c$ is $x^2=frac{-apmsqrt{a^2-4bc}}{2b}$?
$endgroup$
– DonielF
Jan 9 at 3:25
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
@DonielF Actually, by "solution", if you mean the roots, then it would be $sqrt{x}$ on the left side, not $x^2$. Otherwise, you get the idea.
$endgroup$
– John Omielan
Jan 9 at 3:26
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
$begingroup$
All $n$th degree polynomials have $n$ roots, although not necessarily real and/or unique. In the case of a quadratic equation, both possibilities are roots. However, some questions have limitations, usually such as requiring integral or positive values, which could mean that one, or even both, of the roots are not valid.
$endgroup$
– John Omielan
Jan 9 at 3:32
|
show 6 more comments
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%2f3067030%2fis-there-a-general-formula-for-any-polynomials-of-rational-non-integer-degree%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
3
$begingroup$
$asqrt{x}+bx+c$ technically isn't a polynomial. See <en.wikipedia.org/wiki/Polynomial> for a definition.
$endgroup$
– Gnumbertester
Jan 9 at 3:20
$begingroup$
@Gnumbertester So what would it be? Just a function?
$endgroup$
– DonielF
Jan 9 at 3:21
1
$begingroup$
Sometimes it makes sense to rewrite the kinds of equations you illustrate by making a change of variables that gives us a polynomial. As far as equations involving a sum of powers that are not whole numbers, there is a generalization of Descartes rule of signs for these, but no general formula for roots.
$endgroup$
– hardmath
Jan 9 at 3:24