An interesting property of the roots [closed]

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Let $x^2+x+1=0$ be a quadratic equation having two roots $x_1$ and $x_2$.

How would you prove that $(x_1)^2=x_2$ and $(x_2)^2=x_1$?

P.S. I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.

roots roots-of-unity

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edited Jul 14 at 14:46

asked Jul 14 at 14:37

user571036

27818

closed as off-topic by Alex Francisco, abiessu, user223391, amWhy, Leucippus Jul 15 at 0:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, abiessu, Community, amWhy, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
  – Bob Krueger
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What did you try?
  – tien lee
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
  – user571036
  Jul 14 at 14:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user571036 But at least show your work and someone could probably help you with that approach.
  – tien lee
  Jul 14 at 14:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
  – achille hui
  Jul 14 at 14:50
  

  
  
  
  

 | 
show 1 more comment

up vote
-2
down vote

favorite

Let $x^2+x+1=0$ be a quadratic equation having two roots $x_1$ and $x_2$.

How would you prove that $(x_1)^2=x_2$ and $(x_2)^2=x_1$?

P.S. I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.

roots roots-of-unity

share|cite|improve this question

edited Jul 14 at 14:46

asked Jul 14 at 14:37

user571036

27818

closed as off-topic by Alex Francisco, abiessu, user223391, amWhy, Leucippus Jul 15 at 0:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, abiessu, Community, amWhy, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
  – Bob Krueger
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What did you try?
  – tien lee
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
  – user571036
  Jul 14 at 14:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user571036 But at least show your work and someone could probably help you with that approach.
  – tien lee
  Jul 14 at 14:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
  – achille hui
  Jul 14 at 14:50
  

  
  
  
  

 | 
show 1 more comment

up vote
-2
down vote

favorite

up vote
-2
down vote

favorite

Let $x^2+x+1=0$ be a quadratic equation having two roots $x_1$ and $x_2$.

How would you prove that $(x_1)^2=x_2$ and $(x_2)^2=x_1$?

P.S. I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.

roots roots-of-unity

share|cite|improve this question

edited Jul 14 at 14:46

asked Jul 14 at 14:37

user571036

27818

Let $x^2+x+1=0$ be a quadratic equation having two roots $x_1$ and $x_2$.

How would you prove that $(x_1)^2=x_2$ and $(x_2)^2=x_1$?

P.S. I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.

roots roots-of-unity

share|cite|improve this question

edited Jul 14 at 14:46

asked Jul 14 at 14:37

user571036

27818

share|cite|improve this question

share|cite|improve this question

edited Jul 14 at 14:46

edited Jul 14 at 14:46

edited Jul 14 at 14:46

asked Jul 14 at 14:37

user571036

27818

asked Jul 14 at 14:37

user571036

27818

asked Jul 14 at 14:37

user571036

27818

27818

closed as off-topic by Alex Francisco, abiessu, user223391, amWhy, Leucippus Jul 15 at 0:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, abiessu, Community, amWhy, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Alex Francisco, abiessu, user223391, amWhy, Leucippus Jul 15 at 0:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex Francisco, abiessu, Community, amWhy, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
  – Bob Krueger
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What did you try?
  – tien lee
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
  – user571036
  Jul 14 at 14:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user571036 But at least show your work and someone could probably help you with that approach.
  – tien lee
  Jul 14 at 14:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
  – achille hui
  Jul 14 at 14:50
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
  – Bob Krueger
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What did you try?
  – tien lee
  Jul 14 at 14:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
  – user571036
  Jul 14 at 14:41
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user571036 But at least show your work and someone could probably help you with that approach.
  – tien lee
  Jul 14 at 14:43
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
  – achille hui
  Jul 14 at 14:50
  

  
  
  
  

As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
– Bob Krueger
Jul 14 at 14:39

As you tagged this “roots-of-unity,” I’m sure this would be an easy exercise if you knew complex numbers.
– Bob Krueger
Jul 14 at 14:39

1

1

What did you try?
– tien lee
Jul 14 at 14:39

What did you try?
– tien lee
Jul 14 at 14:39

@tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
– user571036
Jul 14 at 14:41

@tienlee I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.
– user571036
Jul 14 at 14:41

1

1

@user571036 But at least show your work and someone could probably help you with that approach.
– tien lee
Jul 14 at 14:43

@user571036 But at least show your work and someone could probably help you with that approach.
– tien lee
Jul 14 at 14:43

1

1

Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
– achille hui
Jul 14 at 14:50

Vieta's formula tell us $x_1 + x_2 = -1$, so $x_1^2 = -1 - x_1 = x_2$ ...
– achille hui
Jul 14 at 14:50

 | 
show 1 more comment

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great answer. Thanks.
  – user571036
  Jul 14 at 15:32
  

  
  
  
  

add a comment | 

up vote
1
down vote

There are a few ways. Notice that
$, x^3 - 1 = (x - 1)(x^2 + x + 1). ,$ If $,x_1,$ is a root of the equation
$, E_1!:, x^2 + x + 1 = 0,$ then this implies $, x_1^3 = 1 ,$ and $,x_1,$ is a root of the equation $, E_2!:, x^3 -1 = 0.,$ Define
$, x_2 := x_1^2. ,$ Now $, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 ,$ which implies that $, x_2 ,$ is also a root of equation $, E_2, ,$
Since $,x_1 ne 1, ,$ $, x_2 ne 1 ,$ and $, x_1 ne x_2, ,$ then $,x_2,$ is the other root of the equation $, E_1.,$

share|cite|improve this answer

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great answer. Thanks.
  – user571036
  Jul 14 at 15:32
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great answer. Thanks.
  – user571036
  Jul 14 at 15:32
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

answered Jul 14 at 15:02

Mohammad Riazi-Kermani

27.7k41852

27.7k41852

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great answer. Thanks.
  – user571036
  Jul 14 at 15:32
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Great answer. Thanks.
  – user571036
  Jul 14 at 15:32
  

  
  
  
  

Great answer. Thanks.
– user571036
Jul 14 at 15:32

Great answer. Thanks.
– user571036
Jul 14 at 15:32

add a comment | 

up vote
1
down vote

There are a few ways. Notice that
$, x^3 - 1 = (x - 1)(x^2 + x + 1). ,$ If $,x_1,$ is a root of the equation
$, E_1!:, x^2 + x + 1 = 0,$ then this implies $, x_1^3 = 1 ,$ and $,x_1,$ is a root of the equation $, E_2!:, x^3 -1 = 0.,$ Define
$, x_2 := x_1^2. ,$ Now $, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 ,$ which implies that $, x_2 ,$ is also a root of equation $, E_2, ,$
Since $,x_1 ne 1, ,$ $, x_2 ne 1 ,$ and $, x_1 ne x_2, ,$ then $,x_2,$ is the other root of the equation $, E_1.,$

share|cite|improve this answer

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

add a comment | 

up vote
1
down vote

There are a few ways. Notice that
$, x^3 - 1 = (x - 1)(x^2 + x + 1). ,$ If $,x_1,$ is a root of the equation
$, E_1!:, x^2 + x + 1 = 0,$ then this implies $, x_1^3 = 1 ,$ and $,x_1,$ is a root of the equation $, E_2!:, x^3 -1 = 0.,$ Define
$, x_2 := x_1^2. ,$ Now $, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 ,$ which implies that $, x_2 ,$ is also a root of equation $, E_2, ,$
Since $,x_1 ne 1, ,$ $, x_2 ne 1 ,$ and $, x_1 ne x_2, ,$ then $,x_2,$ is the other root of the equation $, E_1.,$

share|cite|improve this answer

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

add a comment | 

up vote
1
down vote

up vote
1
down vote

There are a few ways. Notice that
$, x^3 - 1 = (x - 1)(x^2 + x + 1). ,$ If $,x_1,$ is a root of the equation
$, E_1!:, x^2 + x + 1 = 0,$ then this implies $, x_1^3 = 1 ,$ and $,x_1,$ is a root of the equation $, E_2!:, x^3 -1 = 0.,$ Define
$, x_2 := x_1^2. ,$ Now $, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 ,$ which implies that $, x_2 ,$ is also a root of equation $, E_2, ,$
Since $,x_1 ne 1, ,$ $, x_2 ne 1 ,$ and $, x_1 ne x_2, ,$ then $,x_2,$ is the other root of the equation $, E_1.,$

share|cite|improve this answer

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

There are a few ways. Notice that
$, x^3 - 1 = (x - 1)(x^2 + x + 1). ,$ If $,x_1,$ is a root of the equation
$, E_1!:, x^2 + x + 1 = 0,$ then this implies $, x_1^3 = 1 ,$ and $,x_1,$ is a root of the equation $, E_2!:, x^3 -1 = 0.,$ Define
$, x_2 := x_1^2. ,$ Now $, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 ,$ which implies that $, x_2 ,$ is also a root of equation $, E_2, ,$
Since $,x_1 ne 1, ,$ $, x_2 ne 1 ,$ and $, x_1 ne x_2, ,$ then $,x_2,$ is the other root of the equation $, E_1.,$

share|cite|improve this answer

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

share|cite|improve this answer

share|cite|improve this answer

edited Jul 14 at 21:26

edited Jul 14 at 21:26

edited Jul 14 at 21:26

answered Jul 14 at 15:03

Somos

11.7k11033

answered Jul 14 at 15:03

Somos

11.7k11033

answered Jul 14 at 15:03

Somos

11.7k11033

11.7k11033

add a comment | 

add a comment |