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Error: $x^2 + 1 = 0$ has solution set ${-1;1}$

$lambda-z-e^{-z}=0$ has one solution in the right half planeQuadratic equation - $alpha$ and $beta$ RootsHow are we to correct this error?(from Gelfand's Algebra)Is there a rigorous approach to solving this, instead of trial and error?How to find solution for this equationFinding solution with Lambert functionIf $a>b>0$ and $a^3+b^3+27ab=729$ then $ax^2+bx-9=0$ has roots $alpha,beta,(alpha<beta)$. Find the value of $4beta -aalpha$.Finding integer solution in a setSolution of an equation with polynomialsFind the range of values which has no real solutions

0

$begingroup$

So, is it correct that the solution set of $x^2 + 1 = 0$ is [-1;1] ?

Is the error in equation development or in the solution set?

Help me, I don't know how to proceed in this question.

algebra-precalculus roots

share|cite|improve this question

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can plug each of $1$ and $-1$ into the equation and find it is not satisfied.
  $endgroup$
  – Ross Millikan
  Mar 19 at 23:19
  

  
  
  
  

add a comment | 

0

$begingroup$

So, is it correct that the solution set of $x^2 + 1 = 0$ is [-1;1] ?

Is the error in equation development or in the solution set?

Help me, I don't know how to proceed in this question.

algebra-precalculus roots

share|cite|improve this question

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can plug each of $1$ and $-1$ into the equation and find it is not satisfied.
  $endgroup$
  – Ross Millikan
  Mar 19 at 23:19
  

  
  
  
  

add a comment | 

$begingroup$

So, is it correct that the solution set of $x^2 + 1 = 0$ is [-1;1] ?

Is the error in equation development or in the solution set?

Help me, I don't know how to proceed in this question.

algebra-precalculus roots

share|cite|improve this question

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

$endgroup$

share|cite|improve this question

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

share|cite|improve this question

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

edited Mar 19 at 23:27

rash

576216

edited Mar 19 at 23:27

rash

576216

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

asked Mar 19 at 23:17

Daniel Sehn ColaoDaniel Sehn Colao

314

4 Answers
4

active

oldest

votes

3

$begingroup$

All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.

share|cite|improve this answer

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) What I would have written if I had more time.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 23:21
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The final implication is true only if it's been established that $x$ is restricted to be a real number.
  $endgroup$
  – Barry Cipra
  Mar 19 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra Sure. I am basing my answer on the assumption that $x$ is a real number. I will wait for the OP to clarify if he is allowing $x$ to be complex.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:28
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the correct way would be x^2+1=0 implies x=1 or x=−1 and empty implies {-1;1}?
  $endgroup$
  – Daniel Sehn Colao
  Mar 19 at 23:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSehnColao Since neither $x=1$ nor $x =-1$ satisfies $x^{2}+1=0$ the conclusion is there is no real number $x$ satisfying the given equation.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:51
  

  
  
  
  

 | 
show 4 more comments

3

$begingroup$

When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $pm 1$.

Note that

$$x^4=1$$

has four roots in the complex domain which are $pm 1$ and also $pm i$.

share|cite|improve this answer

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure why the other solutions are not pointing out the erroneous trick that multiplying by $x^2-1$ add extraneous solutions. IMO that is the entire aspect of the problem. So +1 to you. Also note, even if you don't consider complex roots. $x^2 +1=0$ has no real roots. And $x^2 -1=0$ has two. So $x^4 -1$ has $2$ but two came from $x^2 -1$ and zero came from $x^2+1$.
  $endgroup$
  – fleablood
  Mar 20 at 0:19
  

  
  
  
  

add a comment | 

0

$begingroup$

The final implication should be "$implies xin{1,-1,i,-i}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.

share|cite|improve this answer

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

$endgroup$

add a comment | 

0

$begingroup$

No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.

This would b similar to doing this:

Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So

$(x^2 - 5x+6 ) = 0$ so

$(x^2 - 5x + 6)(x+3057) = 0cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.

The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.

So in this "false proof":

So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).

But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.

By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.

share|cite|improve this answer

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

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3

$begingroup$

All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.

share|cite|improve this answer

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) What I would have written if I had more time.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 23:21
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The final implication is true only if it's been established that $x$ is restricted to be a real number.
  $endgroup$
  – Barry Cipra
  Mar 19 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra Sure. I am basing my answer on the assumption that $x$ is a real number. I will wait for the OP to clarify if he is allowing $x$ to be complex.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:28
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the correct way would be x^2+1=0 implies x=1 or x=−1 and empty implies {-1;1}?
  $endgroup$
  – Daniel Sehn Colao
  Mar 19 at 23:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSehnColao Since neither $x=1$ nor $x =-1$ satisfies $x^{2}+1=0$ the conclusion is there is no real number $x$ satisfying the given equation.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:51
  

  
  
  
  

 | 
show 4 more comments

3

$begingroup$

All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.

share|cite|improve this answer

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) What I would have written if I had more time.
  $endgroup$
  – José Carlos Santos
  Mar 19 at 23:21
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The final implication is true only if it's been established that $x$ is restricted to be a real number.
  $endgroup$
  – Barry Cipra
  Mar 19 at 23:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra Sure. I am basing my answer on the assumption that $x$ is a real number. I will wait for the OP to clarify if he is allowing $x$ to be complex.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:28
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the correct way would be x^2+1=0 implies x=1 or x=−1 and empty implies {-1;1}?
  $endgroup$
  – Daniel Sehn Colao
  Mar 19 at 23:48
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DanielSehnColao Since neither $x=1$ nor $x =-1$ satisfies $x^{2}+1=0$ the conclusion is there is no real number $x$ satisfying the given equation.
  $endgroup$
  – Kavi Rama Murthy
  Mar 19 at 23:51
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

All the implications are true but you are drawing the wrong conclusion. You have proved that $x^{2}+1=0$ implies $x=1$ or $x =-1$ but the converse of this implication is not true. The fact is there is no real number $x$ with $x^{2}+1=0$.

share|cite|improve this answer

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

$endgroup$

share|cite|improve this answer

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

answered Mar 19 at 23:20

Kavi Rama MurthyKavi Rama Murthy

72.6k53170

3

$begingroup$

When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $pm 1$.

Note that

$$x^4=1$$

has four roots in the complex domain which are $pm 1$ and also $pm i$.

share|cite|improve this answer

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure why the other solutions are not pointing out the erroneous trick that multiplying by $x^2-1$ add extraneous solutions. IMO that is the entire aspect of the problem. So +1 to you. Also note, even if you don't consider complex roots. $x^2 +1=0$ has no real roots. And $x^2 -1=0$ has two. So $x^4 -1$ has $2$ but two came from $x^2 -1$ and zero came from $x^2+1$.
  $endgroup$
  – fleablood
  Mar 20 at 0:19
  

  
  
  
  

add a comment | 

3

$begingroup$

When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $pm 1$.

Note that

$$x^4=1$$

has four roots in the complex domain which are $pm 1$ and also $pm i$.

share|cite|improve this answer

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure why the other solutions are not pointing out the erroneous trick that multiplying by $x^2-1$ add extraneous solutions. IMO that is the entire aspect of the problem. So +1 to you. Also note, even if you don't consider complex roots. $x^2 +1=0$ has no real roots. And $x^2 -1=0$ has two. So $x^4 -1$ has $2$ but two came from $x^2 -1$ and zero came from $x^2+1$.
  $endgroup$
  – fleablood
  Mar 20 at 0:19
  

  
  
  
  

add a comment | 

$begingroup$

When they multiplied by $(1-x^2)$, they introduced two additional roots for the equation $pm 1$.

Note that

$$x^4=1$$

has four roots in the complex domain which are $pm 1$ and also $pm i$.

share|cite|improve this answer

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

$endgroup$

share|cite|improve this answer

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

answered Mar 19 at 23:19

MachineLearnerMachineLearner

1,319112

0

$begingroup$

The final implication should be "$implies xin{1,-1,i,-i}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.

share|cite|improve this answer

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

$endgroup$

add a comment | 

0

$begingroup$

The final implication should be "$implies xin{1,-1,i,-i}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.

share|cite|improve this answer

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

$endgroup$

add a comment | 

$begingroup$

The final implication should be "$implies xin{1,-1,i,-i}$." That is, the solutions to $x^2+1=0$ (if any) are among the elements of this set, not that all the elements of the set are automatically solutions of the equation.

share|cite|improve this answer

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

$endgroup$

share|cite|improve this answer

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

answered Mar 19 at 23:25

Barry CipraBarry Cipra

60.6k655129

0

$begingroup$

No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.

This would b similar to doing this:

Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So

$(x^2 - 5x+6 ) = 0$ so

$(x^2 - 5x + 6)(x+3057) = 0cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.

The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.

So in this "false proof":

So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).

But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.

By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.

share|cite|improve this answer

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

0

$begingroup$

No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.

This would b similar to doing this:

Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So

$(x^2 - 5x+6 ) = 0$ so

$(x^2 - 5x + 6)(x+3057) = 0cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.

The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.

So in this "false proof":

So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).

But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.

By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.

share|cite|improve this answer

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

$begingroup$

No. It is not as plugging in $1$ and $-1$ will give you $(-1)^2 + 1 = 2$ and $1^2 +1 =2$.

The problem is that multiplying by $x^2 -1$ gives extraneous solutions and $1, -1$ are the solutions to $x^2 -1 =0$ which was brought in from nowhere.

This would b similar to doing this:

Suppose $(x -3)(x-2) = x^2 -5x + 6 = 0$ (So the solutions are $x = 3$ or $x = 2$. As that is zero we multiply it by $x+3057$ So

$(x^2 - 5x+6 ) = 0$ so

$(x^2 - 5x + 6)(x+3057) = 0cdot (x+3057)=0$ so

$x^3 - 3052x^2 -1579x +18342 = 0$

If we tried to solve $x^3 - 3052x^2 -1579x +18342 = 0$ we would get $x = 3$ or $x = 2$ or $x =-3057$.

The third solution came in when we multiplied by $x+3057$. That is because $x = -3057$ is a solution to $x+3057=0$. So by multiplying $0$ by $x+3057$ we add a new solution to the problem.

So in this "false proof":

So $x^2 + 1 = 0$ has no real solutions. (It has complex solutions, $x = i$ or $x = -i$ but no real solutions).

But $x^2 -1=0$ has two real solutions; $x = 1$ and $x = -1$.

By mulitplying both sides of the equation $x^2 + 1= 0$ by $x^2 -1$ we are taking all the original solutions (there are no real solutions but there were complex $x=i$ and $x=-i$) and adding the solutions $x = 1$ and $x =-1$. So we end up with solutions $x =1$ and $x = -1$ but they were both artificially added when there no real solutions in the first place.

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answered Mar 19 at 23:45

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share|cite|improve this answer

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891

answered Mar 19 at 23:45

fleabloodfleablood

73.8k22891