Mfcttrf

Why is a PhD thesis typically 150 pages?

What would be the effects of (relatively) widespread precognition on the stock market?

Is it better to have a 10 year gap or a bad reference?

How old is the Italian word "malandrino"?

How can I deal with someone that wants to kill something that isn't supposed to be killed?

My current job follows "worst practices". How can I talk about my experience in an interview without giving off red flags?

Why Lie algebras if what we care about in physics are groups?

Adding gears to my grandson's 12" bike

Is it better to deliver many low-value stories or few high-value stories?

Acoustic guitar chords' positions vs those of a Bass guitar

Can a warlock shoot multiple beams from the Eldritch Blast cantrip with only a single free hand?

Why is the UH-60 tail rotor canted?

Can a creature sustain itself by eating its own severed body parts?

Source for "everyone has a specific area of Torah that they're naturally drawn to"

Can I make Ubuntu 18.04 switch between multiple windows of the program by just clicking the icon?

Function pointer parameter without asterisk

Which dice game has a board with 9x9 squares that has different colors on the diagonals and midway on some edges?

As the Ferris wheel turns

I have a domain, static IP and many devices I'd like to access outside my house. How to route them?

Found old paper shares of Motorola Inc that has since been broken up

What does the following chess proverb mean: "Chess is a sea where a gnat may drink from and an elephant may bathe in."

What is the best word describing the nature of expiring in a short amount of time, connoting "losing public attention"?

If hash functions append the length, why does length extension attack work?

What are "the high ends of castles" called?

ambiguity in solving algebraic equations.

Is there a name for this strange solution to a quadratic equation involving a square root?Solving $|x^2 - 1| - 1 = 3x - 2$ without graphingSimultaneous Equations…Solving Yoshida equationsSolve easy equationsWhere is the mistake in solving the inequality?Solving Radical InequalitiesHow to get rid of the absolute value when solving a system of complex equations?Can the Elimination Method be used to solve a system of polynomial equations?Conflicting situations arising while solving for $x$

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

2

0

$begingroup$

Given $x+frac1x=2 tag1$
$x^3+frac1x^4=1 tag2$

Find the value of $$x^4+frac1x^3$$

If equation $bf2$ is multiplied by $bf x$ we get $x^4+frac1x^3=x$ now, we just need to find the value of $x$ which from $bf1$ is equal to $1$.
But the answer to this question is $bf3$.
Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.

Another approach
Given, $(x + 1/x) = 2$

Squaring both sides,

$$x^2 + 1/x^2 + 2 = 4$$

$$ x^2 + 1/x^2 = 2 $$

Let,

$A = x^3 + 1/x^4$

$B = x^4 + 1/x^3$

Now, add A and B,

$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$

$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$

$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$

$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $

$A + B = 2 + (2)2 – 2$

$ A + B = 4$

Given that $A = 1$

Then, $B = 4 – 1 = 3$

$ x^4 + 1/x^3 = 3$

algebra-precalculus analysis systems-of-equations

share|cite|improve this question

edited 6 hours ago

Nebo Alex

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $x+frac1x=2$ then $x=1$. Are you sure you have the question right?
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  

add a comment | 

2

0

$begingroup$

Given $x+frac1x=2 tag1$
$x^3+frac1x^4=1 tag2$

Find the value of $$x^4+frac1x^3$$

If equation $bf2$ is multiplied by $bf x$ we get $x^4+frac1x^3=x$ now, we just need to find the value of $x$ which from $bf1$ is equal to $1$.
But the answer to this question is $bf3$.
Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.

Another approach
Given, $(x + 1/x) = 2$

Squaring both sides,

$$x^2 + 1/x^2 + 2 = 4$$

$$ x^2 + 1/x^2 = 2 $$

Let,

$A = x^3 + 1/x^4$

$B = x^4 + 1/x^3$

Now, add A and B,

$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$

$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$

$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$

$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $

$A + B = 2 + (2)2 – 2$

$ A + B = 4$

Given that $A = 1$

Then, $B = 4 – 1 = 3$

$ x^4 + 1/x^3 = 3$

algebra-precalculus analysis systems-of-equations

share|cite|improve this question

edited 6 hours ago

Nebo Alex

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  If $x+frac1x=2$ then $x=1$. Are you sure you have the question right?
  $endgroup$
  – Lord Shark the Unknown
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Given $x+frac1x=2 tag1$
$x^3+frac1x^4=1 tag2$

Find the value of $$x^4+frac1x^3$$

If equation $bf2$ is multiplied by $bf x$ we get $x^4+frac1x^3=x$ now, we just need to find the value of $x$ which from $bf1$ is equal to $1$.
But the answer to this question is $bf3$.
Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.

Another approach
Given, $(x + 1/x) = 2$

Squaring both sides,

$$x^2 + 1/x^2 + 2 = 4$$

$$ x^2 + 1/x^2 = 2 $$

Let,

$A = x^3 + 1/x^4$

$B = x^4 + 1/x^3$

Now, add A and B,

$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$

$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$

$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$

$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $

$A + B = 2 + (2)2 – 2$

$ A + B = 4$

Given that $A = 1$

Then, $B = 4 – 1 = 3$

$ x^4 + 1/x^3 = 3$

algebra-precalculus analysis systems-of-equations

share|cite|improve this question

edited 6 hours ago

Nebo Alex

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

$endgroup$

share|cite|improve this question

edited 6 hours ago

Nebo Alex

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

share|cite|improve this question

edited 6 hours ago

Nebo Alex

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

asked 8 hours ago

Nebo AlexNebo Alex

1,1641111 silver badges2828 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.

Thus the system does not have a solution.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thank you for figuring it out I missed it. I will delete this question because its wrong
  $endgroup$
  – Nebo Alex
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a good question. Keep it alive
  $endgroup$
  – Mohammad Riazi-Kermani
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + frac1x = 2 iff 3 + frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.

share|cite|improve this answer

answered 8 hours ago

MonadologieMonadologie

5961313 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%2f3300686%2fambiguity-in-solving-algebraic-equations%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.

Thus the system does not have a solution.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thank you for figuring it out I missed it. I will delete this question because its wrong
  $endgroup$
  – Nebo Alex
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a good question. Keep it alive
  $endgroup$
  – Mohammad Riazi-Kermani
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.

Thus the system does not have a solution.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh thank you for figuring it out I missed it. I will delete this question because its wrong
  $endgroup$
  – Nebo Alex
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a good question. Keep it alive
  $endgroup$
  – Mohammad Riazi-Kermani
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.

Thus the system does not have a solution.

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

answered 8 hours ago

Mohammad Riazi-KermaniMohammad Riazi-Kermani

49.4k44 gold badges2727 silver badges7171 bronze badges

2

$begingroup$

If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + frac1x = 2 iff 3 + frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.

share|cite|improve this answer

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + frac1x = 2 iff 3 + frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.

share|cite|improve this answer

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $x=3$ and $x$ is solution to $(1)$, we have
$$
x + frac1x = 2 iff 3 + frac 1 3 = 2
$$
which is obviously wrong. There must be some mistake in the solution.

share|cite|improve this answer

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges

answered 8 hours ago

MonadologieMonadologie

5961313 bronze badges