What is rule for when solving algebraic equations?
$begingroup$
I'm a high school student trying to get critical intuition when learning algebraic equation solving.
For $x$ any complex number and $c$ constant, simple polynomial such as
$x^n -c=0$ are easily solvable for $x$. Then if we know how to solve polynomial $g(x)=k$ for any constant $k$ we can now solve more complex polynomial $[g(x)]^n -c=0$. Hence we say Quadratic polynomial is solved when we deform it into $a(x-p)^2 -q=0$ since linear polynomial is solvable.
The reason for doing this when solving polynomials is because our intuition can only do linear calculations and fails to do calulations of "higher complexity" directly.
(I cant find sufficient word to describe this)
e.g.
$ab=1$ implies $a=b^{-1}$ and
$a+b=0$ implies $a=-b$
but $a^2 + b^2 +ab=0$ is whole new stuff.
Further from polynomials, I think algebra is all about deforming equations into reasonably simpler form and deciding if such progress is doable.
Am I making things right?
algebra-precalculus
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
$endgroup$
add a comment |
$begingroup$
I'm a high school student trying to get critical intuition when learning algebraic equation solving.
For $x$ any complex number and $c$ constant, simple polynomial such as
$x^n -c=0$ are easily solvable for $x$. Then if we know how to solve polynomial $g(x)=k$ for any constant $k$ we can now solve more complex polynomial $[g(x)]^n -c=0$. Hence we say Quadratic polynomial is solved when we deform it into $a(x-p)^2 -q=0$ since linear polynomial is solvable.
The reason for doing this when solving polynomials is because our intuition can only do linear calculations and fails to do calulations of "higher complexity" directly.
(I cant find sufficient word to describe this)
e.g.
$ab=1$ implies $a=b^{-1}$ and
$a+b=0$ implies $a=-b$
but $a^2 + b^2 +ab=0$ is whole new stuff.
Further from polynomials, I think algebra is all about deforming equations into reasonably simpler form and deciding if such progress is doable.
Am I making things right?
algebra-precalculus
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
$endgroup$
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10
add a comment |
$begingroup$
I'm a high school student trying to get critical intuition when learning algebraic equation solving.
For $x$ any complex number and $c$ constant, simple polynomial such as
$x^n -c=0$ are easily solvable for $x$. Then if we know how to solve polynomial $g(x)=k$ for any constant $k$ we can now solve more complex polynomial $[g(x)]^n -c=0$. Hence we say Quadratic polynomial is solved when we deform it into $a(x-p)^2 -q=0$ since linear polynomial is solvable.
The reason for doing this when solving polynomials is because our intuition can only do linear calculations and fails to do calulations of "higher complexity" directly.
(I cant find sufficient word to describe this)
e.g.
$ab=1$ implies $a=b^{-1}$ and
$a+b=0$ implies $a=-b$
but $a^2 + b^2 +ab=0$ is whole new stuff.
Further from polynomials, I think algebra is all about deforming equations into reasonably simpler form and deciding if such progress is doable.
Am I making things right?
algebra-precalculus
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
$endgroup$
I'm a high school student trying to get critical intuition when learning algebraic equation solving.
For $x$ any complex number and $c$ constant, simple polynomial such as
$x^n -c=0$ are easily solvable for $x$. Then if we know how to solve polynomial $g(x)=k$ for any constant $k$ we can now solve more complex polynomial $[g(x)]^n -c=0$. Hence we say Quadratic polynomial is solved when we deform it into $a(x-p)^2 -q=0$ since linear polynomial is solvable.
The reason for doing this when solving polynomials is because our intuition can only do linear calculations and fails to do calulations of "higher complexity" directly.
(I cant find sufficient word to describe this)
e.g.
$ab=1$ implies $a=b^{-1}$ and
$a+b=0$ implies $a=-b$
but $a^2 + b^2 +ab=0$ is whole new stuff.
Further from polynomials, I think algebra is all about deforming equations into reasonably simpler form and deciding if such progress is doable.
Am I making things right?
algebra-precalculus
algebra-precalculus
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
edited Dec 15 '18 at 3:45
Tianlalu
3,08121038
3,08121038
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
asked Dec 15 '18 at 3:30
Solvable PotatoSolvable Potato
364
364
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10
add a comment |
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10
add a comment |
0
active
oldest
votes
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%2f3040138%2fwhat-is-rule-for-when-solving-algebraic-equations%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3040138%2fwhat-is-rule-for-when-solving-algebraic-equations%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
$begingroup$
What you say is what early algebraist have done : you ask good questions. For example, one can, from a third degree equation one can find a quadratic that helps to solve it, Even for a 4th degree equation, it is possible to find a 3rd degree, then a 2nd degree equation, giving generally awfully complicated solutions. Many many mathematicians have worked on the exact solutions of equations with real or complex roots. The problem is that "chain-computing" the roots covers only few cases. A turning point in this study has been when Galois/Abel have shown that the general 5th degree equation...
$endgroup$
– Jean Marie
Dec 15 '18 at 3:59
$begingroup$
... (and the nth degree general equation with $n geq 5$) hasn't any method for solving them. Of course particular equations like $x^5−1=0$ can be solved easily.
$endgroup$
– Jean Marie
Dec 15 '18 at 4:10