Roots of polynomial equation $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$












4












$begingroup$



If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_i|=2spacespacespaceforall i$




I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?






algebra-precalculus polynomials complex-numbers roots







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    If you let $x=2z$ you can remove a factor of $64$ from each term.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    @lulu What is $z$?
    $endgroup$
    – tatan
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    $z=frac x2$. It's just a change of variables.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:08
















4












$begingroup$



If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_i|=2spacespacespaceforall i$




I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?






algebra-precalculus polynomials complex-numbers roots







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    If you let $x=2z$ you can remove a factor of $64$ from each term.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    @lulu What is $z$?
    $endgroup$
    – tatan
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    $z=frac x2$. It's just a change of variables.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:08














4












4








4


2



$begingroup$



If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_i|=2spacespacespaceforall i$




I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?






algebra-precalculus polynomials complex-numbers roots







share|cite|improve this question











$endgroup$





If $x_1,x_2,...,x_6$ be the roots of $x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$ then I have to show that $|x_i|=2spacespacespaceforall i$




I get that the roots of the equation must be complex, of the form $a+ib$ where $|a+ib|=sqrt{a^2+b^2}=2$ for all $|x_i|$. I don't get how to find the value of $a+ib$. How do I find the roots?






algebra-precalculus polynomials complex-numbers roots




algebra-precalculus polynomials complex-numbers roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 18:20









José Carlos Santos

157k22126227




157k22126227










asked Oct 12 '18 at 11:04









tatantatan

5,63362655




5,63362655








  • 4




    $begingroup$
    If you let $x=2z$ you can remove a factor of $64$ from each term.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    @lulu What is $z$?
    $endgroup$
    – tatan
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    $z=frac x2$. It's just a change of variables.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:08














  • 4




    $begingroup$
    If you let $x=2z$ you can remove a factor of $64$ from each term.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    @lulu What is $z$?
    $endgroup$
    – tatan
    Oct 12 '18 at 11:07






  • 1




    $begingroup$
    $z=frac x2$. It's just a change of variables.
    $endgroup$
    – lulu
    Oct 12 '18 at 11:08








4




4




$begingroup$
If you let $x=2z$ you can remove a factor of $64$ from each term.
$endgroup$
– lulu
Oct 12 '18 at 11:07




$begingroup$
If you let $x=2z$ you can remove a factor of $64$ from each term.
$endgroup$
– lulu
Oct 12 '18 at 11:07




1




1




$begingroup$
@lulu What is $z$?
$endgroup$
– tatan
Oct 12 '18 at 11:07




$begingroup$
@lulu What is $z$?
$endgroup$
– tatan
Oct 12 '18 at 11:07




1




1




$begingroup$
$z=frac x2$. It's just a change of variables.
$endgroup$
– lulu
Oct 12 '18 at 11:08




$begingroup$
$z=frac x2$. It's just a change of variables.
$endgroup$
– lulu
Oct 12 '18 at 11:08










2 Answers
2






active

oldest

votes


















11












$begingroup$

Hint:begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\=64left(left(frac x2right)^6+left(frac x2right)^5+left(frac x2right)^4+left(frac x2right)^3+left(frac x2right)^2+frac x2+1right).end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can't proceed ;-(
    $endgroup$
    – tatan
    Oct 12 '18 at 11:13






  • 1




    $begingroup$
    Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:14






  • 1




    $begingroup$
    Right, but what matters really is that $left(frac x2right)^7=1$.
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:17






  • 2




    $begingroup$
    I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:22






  • 1




    $begingroup$
    Thanks for your advice ;-)
    $endgroup$
    – tatan
    Oct 12 '18 at 11:23



















3












$begingroup$

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z neq 1$
Then $frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh its have already solved. sorry
    $endgroup$
    – ilovass11
    Oct 12 '18 at 11:32











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2952564%2froots-of-polynomial-equation-x62x54x48x316x232x64-0%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

Hint:begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\=64left(left(frac x2right)^6+left(frac x2right)^5+left(frac x2right)^4+left(frac x2right)^3+left(frac x2right)^2+frac x2+1right).end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can't proceed ;-(
    $endgroup$
    – tatan
    Oct 12 '18 at 11:13






  • 1




    $begingroup$
    Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:14






  • 1




    $begingroup$
    Right, but what matters really is that $left(frac x2right)^7=1$.
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:17






  • 2




    $begingroup$
    I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:22






  • 1




    $begingroup$
    Thanks for your advice ;-)
    $endgroup$
    – tatan
    Oct 12 '18 at 11:23
















11












$begingroup$

Hint:begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\=64left(left(frac x2right)^6+left(frac x2right)^5+left(frac x2right)^4+left(frac x2right)^3+left(frac x2right)^2+frac x2+1right).end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I can't proceed ;-(
    $endgroup$
    – tatan
    Oct 12 '18 at 11:13






  • 1




    $begingroup$
    Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:14






  • 1




    $begingroup$
    Right, but what matters really is that $left(frac x2right)^7=1$.
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:17






  • 2




    $begingroup$
    I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:22






  • 1




    $begingroup$
    Thanks for your advice ;-)
    $endgroup$
    – tatan
    Oct 12 '18 at 11:23














11












11








11





$begingroup$

Hint:begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\=64left(left(frac x2right)^6+left(frac x2right)^5+left(frac x2right)^4+left(frac x2right)^3+left(frac x2right)^2+frac x2+1right).end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.






share|cite|improve this answer











$endgroup$



Hint:begin{multline}x^6+2x^5+4x^4+8x^3+16x^2+32x+64=\=64left(left(frac x2right)^6+left(frac x2right)^5+left(frac x2right)^4+left(frac x2right)^3+left(frac x2right)^2+frac x2+1right).end{multline}Also, use the fact that $x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 29 '18 at 11:35

























answered Oct 12 '18 at 11:08









José Carlos SantosJosé Carlos Santos

157k22126227




157k22126227












  • $begingroup$
    I can't proceed ;-(
    $endgroup$
    – tatan
    Oct 12 '18 at 11:13






  • 1




    $begingroup$
    Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:14






  • 1




    $begingroup$
    Right, but what matters really is that $left(frac x2right)^7=1$.
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:17






  • 2




    $begingroup$
    I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:22






  • 1




    $begingroup$
    Thanks for your advice ;-)
    $endgroup$
    – tatan
    Oct 12 '18 at 11:23


















  • $begingroup$
    I can't proceed ;-(
    $endgroup$
    – tatan
    Oct 12 '18 at 11:13






  • 1




    $begingroup$
    Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:14






  • 1




    $begingroup$
    Right, but what matters really is that $left(frac x2right)^7=1$.
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:17






  • 2




    $begingroup$
    I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
    $endgroup$
    – José Carlos Santos
    Oct 12 '18 at 11:22






  • 1




    $begingroup$
    Thanks for your advice ;-)
    $endgroup$
    – tatan
    Oct 12 '18 at 11:23
















$begingroup$
I can't proceed ;-(
$endgroup$
– tatan
Oct 12 '18 at 11:13




$begingroup$
I can't proceed ;-(
$endgroup$
– tatan
Oct 12 '18 at 11:13




1




1




$begingroup$
Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:14




$begingroup$
Now use the fact that$$x^7-1=(x-1)(x^6+x^5+x^4+x^3+x^2+x+1).$$
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:14




1




1




$begingroup$
Right, but what matters really is that $left(frac x2right)^7=1$.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:17




$begingroup$
Right, but what matters really is that $left(frac x2right)^7=1$.
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:17




2




2




$begingroup$
I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:22




$begingroup$
I strongly suggest that you avoid fractional exponents when dealing with complex non-real numbers. What happens is that$$left(frac x2right)^7=1impliesleftlvertfrac x2rightrvert^7=1iffleftlvertfrac x2rightrvert=1ifflvert xrvert=2.$$
$endgroup$
– José Carlos Santos
Oct 12 '18 at 11:22




1




1




$begingroup$
Thanks for your advice ;-)
$endgroup$
– tatan
Oct 12 '18 at 11:23




$begingroup$
Thanks for your advice ;-)
$endgroup$
– tatan
Oct 12 '18 at 11:23











3












$begingroup$

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z neq 1$
Then $frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh its have already solved. sorry
    $endgroup$
    – ilovass11
    Oct 12 '18 at 11:32
















3












$begingroup$

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z neq 1$
Then $frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    oh its have already solved. sorry
    $endgroup$
    – ilovass11
    Oct 12 '18 at 11:32














3












3








3





$begingroup$

$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z neq 1$
Then $frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$






share|cite|improve this answer











$endgroup$



$$x^6+2x^5+4x^4+8x^3+16x^2+32x+64=0$$
Let $z=x/2$, so
$$z^6+z^5+z^4+z^3+z^2+z+1=0$$
from $z neq 1$
Then $frac{1-z^7}{1-z}=0 $
so $1-z^7=0$
so $z$ is the $n$ th root of $7$.
from $x=2z$ ,so $|x|=|2z|$ so $|x|=|2z|=2$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 12 '18 at 11:44









amWhy

1




1










answered Oct 12 '18 at 11:31









ilovass11ilovass11

716




716












  • $begingroup$
    oh its have already solved. sorry
    $endgroup$
    – ilovass11
    Oct 12 '18 at 11:32


















  • $begingroup$
    oh its have already solved. sorry
    $endgroup$
    – ilovass11
    Oct 12 '18 at 11:32
















$begingroup$
oh its have already solved. sorry
$endgroup$
– ilovass11
Oct 12 '18 at 11:32




$begingroup$
oh its have already solved. sorry
$endgroup$
– ilovass11
Oct 12 '18 at 11:32


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2952564%2froots-of-polynomial-equation-x62x54x48x316x232x64-0%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more

Province de Neuquén

Image
Province de Neuquén Provincia del Neuquén .mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Héraldique Drapeau Localisation de la province de Neuquén Administration Pays   Argentine Capitale Neuquén Gouverneur Omar Gutiérrez ISO 3166-2 AR-Q Démographie Gentilé Neuquino/a Population 550 344  hab. (2010) Densité 5,8  hab./km 2 Géographie Superficie 94 078  km 2 Liens Site web http://www.neuquen.gov.ar modifier   La province de Neuquén est une province d'Argentine, qui se trouve à l'extrémité nord-ouest de la Patagonie argentine. Avec une superficie de 94 378  km 2 , elle est la plus petite province des régions australes du pays. Sa géographie s'abîme sur l'imposante Cordillère des Andes. Sommaire 1 Histoire 1.1 Avant la découverte par les Espagnols 1.2 ...
Read more

Journaliste

Image
Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
Read more