Mdthbs

Does the length of a password for Wi-Fi affect speed?

Need help understanding lens reach

Decode a variable-length quantity

Are any British or English Acts of Parliament still in force in the United States?

What are these silver stripes on Cosmic Girl for?

How do I say "Outdoor Pre-show"

Double blind peer review when paper cites author's GitHub repo for code

Does the Voyager team use a wrapper (Fortran(77?) to Python) to transmit current commands?

Validation and verification of mathematical models

Finish the Mastermind

How can I tell if a flight itinerary is fake

In Pokémon Go, why does one of my Pikachu have an option to evolve, but another one doesn't?

Is it true that control+alt+delete only became a thing because IBM would not build Bill Gates a computer with a task manager button?

Was there ever a difference between 'volo' and 'volo'?

Why should public servants be apolitical?

Casting Goblin Matron with Plague Engineer on the battlefield

Why are the inside diameters of some pipe larger than the stated size?

Why is there a need to prevent a racist, sexist, or otherwise bigoted vendor from discriminating who they sell to?

What can make Linux unresponsive for minutes when browsing certain websites?

Whats the name of this projection?

Why do implementations of "stdint.h" disagree on the definition of UINT8_C?

How would I as a DM create a smart phone-like spell/device my players could use?

If there were no space agencies, could a person go to space?

Is this cheap "air conditioner" able to cool a room?

polynomial, find the sum of the inverse roots of this equation.

Finding the sum of non-real roots of a polynomial.find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$Find the sum of real roots of a biquadratic equationMost efficient way to find polynomial rootsPolynomial with real rootsFind all the roots of this polynomialHelp with sum and product of roots.Quadratic equation with integral coefficients but no roots are givenFinding the sum of squares of roots of a quartic polynomial.Using binomial coefficients to find sum of roots of a polynomial.

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

4

$begingroup$

Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.

My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?

polynomials

share|cite|improve this question

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.

My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?

polynomials

share|cite|improve this question

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

$endgroup$

add a comment | 

$begingroup$

Calculate the sum of the inverse roots of the equation $x^3–7x^2+4x–1=0$.

My development was:
sum = -7
product = 1
Thus, I believe that to find the inverse roots one only has to share the sum with the product, ie (-7/1) = 7, however, the answer given in the question is 4.
what do you think? do you have any formula?

polynomials

share|cite|improve this question

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

share|cite|improve this question

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

edited 8 hours ago

Jean Marie

34.2k44 gold badges2626 silver badges6060 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

asked 9 hours ago

funfun funfun

3044 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac{4}{-1}=4$.

share|cite|improve this answer

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For those who may not see why reversing the coefficients gives a polynomial with inverse roots, divide the original equation by its largest power of $x$.
  $endgroup$
  – Paul Sinclair
  1 hour ago
  

  
  
  
  

add a comment | 

7

$begingroup$

Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1{x_1}+frac1{x_2}+frac1{x_3}=frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=frac{-c/a}{d/a}=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.

share|cite|improve this answer

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

$endgroup$

add a comment | 

Your Answer

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

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3317349%2fpolynomial-find-the-sum-of-the-inverse-roots-of-this-equation%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

6

$begingroup$

If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac{4}{-1}=4$.

share|cite|improve this answer

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For those who may not see why reversing the coefficients gives a polynomial with inverse roots, divide the original equation by its largest power of $x$.
  $endgroup$
  – Paul Sinclair
  1 hour ago
  

  
  
  
  

add a comment | 

6

$begingroup$

If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac{4}{-1}=4$.

share|cite|improve this answer

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For those who may not see why reversing the coefficients gives a polynomial with inverse roots, divide the original equation by its largest power of $x$.
  $endgroup$
  – Paul Sinclair
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

If we reverse the coefficients of a polynomial with roots $r_i$, we get a polynomial with roots $1/r_i$. In this case, the reversal is $-x^3+4x^2-7x+1=0$. By Viète's formulas, we see that the sum of reciprocals of roots of the original polynomial is $-frac{4}{-1}=4$.

share|cite|improve this answer

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

answered 9 hours ago

Parcly TaxelParcly Taxel

49.5k1313 gold badges7979 silver badges117117 bronze badges

7

$begingroup$

Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1{x_1}+frac1{x_2}+frac1{x_3}=frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=frac{-c/a}{d/a}=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.

share|cite|improve this answer

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

$endgroup$

add a comment | 

7

$begingroup$

Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1{x_1}+frac1{x_2}+frac1{x_3}=frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=frac{-c/a}{d/a}=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.

share|cite|improve this answer

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let the roots be $x_1,x_2,x_3$, all non-zero. Then you need$$frac1{x_1}+frac1{x_2}+frac1{x_3}=frac{x_1x_2+x_2x_3+x_3x_1}{x_1x_2x_3}=frac{-c/a}{d/a}=-frac cd$$where $x^3-7x^2+4x-1=ax^3+bx^2+cx+d$.

share|cite|improve this answer

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges

answered 9 hours ago

Shubham JohriShubham Johri

6,85899 silver badges1818 bronze badges