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Why isn't substituting a variable from the same equation not valid?
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There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + frac1x = 0$
$x = -1 - frac1x$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-frac1x) + 1$
$x^2-frac1x = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
algebra-precalculus proof-verification proof-writing proof-explanation
edited 1 hour ago
Carl Mummert
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asked 1 hour ago
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up vote
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There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + frac1x = 0$
$x = -1 - frac1x$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-frac1x) + 1$
$x^2-frac1x = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
algebra-precalculus proof-verification proof-writing proof-explanation
edited 1 hour ago
Carl Mummert
64.4k7128240
asked 1 hour ago
Strikers
345
New contributor
Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
1
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + frac1x = 0$
$x = -1 - frac1x$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-frac1x) + 1$
$x^2-frac1x = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
algebra-precalculus proof-verification proof-writing proof-explanation
edited 1 hour ago
Carl Mummert
64.4k7128240
asked 1 hour ago
Strikers
345
New contributor
Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + frac1x = 0$
$x = -1 - frac1x$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-frac1x) + 1$
$x^2-frac1x = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
algebra-precalculus proof-verification proof-writing proof-explanation
algebra-precalculus proof-verification proof-writing proof-explanation
edited 1 hour ago
Carl Mummert
64.4k7128240
asked 1 hour ago
Strikers
345
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Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
Carl Mummert
64.4k7128240
asked 1 hour ago
Strikers
345
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edited 1 hour ago
Carl Mummert
64.4k7128240
edited 1 hour ago
Carl Mummert
64.4k7128240
edited 1 hour ago
Carl Mummert
64.4k7128240
64.4k7128240
asked 1 hour ago
Strikers
345
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Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago
Strikers
345
asked 1 hour ago
Strikers
345
345
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Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Strikers is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
1
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago
add a comment |Â
1
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
1
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago
1
1
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
1
1
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago
add a comment |Â
4 Answers
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up vote
3
down vote
You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
answered 1 hour ago
Carl Mummert
64.4k7128240
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
 |Â
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up vote
2
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For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $,x=1,$. We can substitute $,1 mapsto x,$ on the RHS and end up with $,x=x,$. Of course that $,x=1 implies x=x,$, but the converse is not true.
In OP's case, the original equation is quadratic in $,x,$ which has $2$ roots in $Bbb C,$, while the derived equation is a cubic which has $3$ roots in $Bbb C,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $,x=1,$ as noted already, which does not satisfy the original quadratic.
answered 1 hour ago
dxiv
56.5k64798
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.why does it have to satisfy the original equation?It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.
– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
 |Â
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1
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The higher level description of your work is:
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$
$1$ is not a solution to the original equation.
And therefore we conclude the assumption is false: that is,
- Therefore, the original equation has no solutions.
Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$ or either $-frac12 pm fracsqrt32 i$ (because those are the three cube roots of $1$)
$1$ is not a solution to the original equation.
$-frac12 pm fracsqrt32 i$ are solutions to the original equation
and therefore
- The solutions to the equation are $-frac12 pm fracsqrt32 i$
answered 1 hour ago
Hurkyl
110k9114257
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
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1
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Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$beginalign
x_1 &= 1, & textor \
x_2 &= -frac12 + fracsqrt 32 i, & textor \
x_3 &= -frac12 - fracsqrt 32 i
endalign$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
answered 1 hour ago
Trevor Kafka
1714
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Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
answered 1 hour ago
Carl Mummert
64.4k7128240
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
 |Â
show 3 more comments
up vote
3
down vote
You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
answered 1 hour ago
Carl Mummert
64.4k7128240
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
 |Â
show 3 more comments
up vote
3
down vote
up vote
3
down vote
You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
answered 1 hour ago
Carl Mummert
64.4k7128240
You can indeed substitute. First, though, note that $1$ is not a solution to $x = -1 - 1/x$. So, by making that substitution, we are excluding $x = 1$ as a solution to our equation. In a sense, we are looking for a solution of $x^2 +x + 1 = 0$ that is also a solution to $x = -1 - 1/x$.
Here is what we get by substituting:
$$ x^2 + x + 1 = 0$$
$$ x^2 + (-1 - 1/x) + 1 = 0$$
$$ x^2 - 1/x = 0 $$
$$ x^2 = 1/x$$
$$ x^3 = 1 $$
There are three complex solutions to that equation. We have to exclude the "false solution" $x =1$ because the substitution $x = -1 - 1/x$ already prevented $x$ from being $1$. Either of the other two complex number solutions to $x^3 = 1$ are solutions of the original equation $x^2 + x + 1$.
This can also be seen because $x^3 -1 = (x-1)(x^2 + x + 1)$. So there are three complex solutions to $x^3 - 1 = 0$, and by removing the $x-1$ term we leave behind two complex number solutions to $x^2 + x + 1 = 0$.
answered 1 hour ago
Carl Mummert
64.4k7128240
edited 1 hour ago
answered 1 hour ago
Carl Mummert
64.4k7128240
answered 1 hour ago
Carl Mummert
64.4k7128240
answered 1 hour ago
Carl Mummert
64.4k7128240
64.4k7128240
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
 |Â
show 3 more comments
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
Hi, thank you for your response. I have not learned about complex numbers yet and am confused as to why $x^3 = 1$ to $x = 1$ is not valid. Also, why is it that just because $x = 1$ does not satisfy the previous equation have to matter if the original equation yields $x = 1$. Thanks.
– Strikers
55 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
In general, once you learn about the complex numbers, any equation of the form $x^n = 1$ has $n$ different complex number roots. For example $x^3 = 1$ has three different complex number roots, one of which is $x = 1$ and the other two are not real numbers.
– Carl Mummert
34 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Here is a different analogy. If you start with $x = 1$ and square both sides you get $x^2 = 1$. This has two roots, $1$ and $-1$, even though the original equation only had one root. This is because squaring both sides is not a reversible operation. In your case, replacing $x$ with $-1-1/x$ is also not a reversible operation. So even though the final equation has $x = 1$ as a solution, the original one doesn't.
– Carl Mummert
31 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
Hi. What do you specifically mean by it is not a "reversible operation"? Thanks.
– Strikers
28 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
It is common in algebra to do something that is not reversible, like squaring both sides of an equation, which can lead to false solutions. For example if you square both sides of an equation, the new equation may have more solutions than the original equation did. An operation is called reversible if every solution to the new equation has to be a solution to the original equation. For example, adding a constant to both sides of an equation is a reversible operation.
– Carl Mummert
25 mins ago
 |Â
show 3 more comments
up vote
2
down vote
For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $,x=1,$. We can substitute $,1 mapsto x,$ on the RHS and end up with $,x=x,$. Of course that $,x=1 implies x=x,$, but the converse is not true.
In OP's case, the original equation is quadratic in $,x,$ which has $2$ roots in $Bbb C,$, while the derived equation is a cubic which has $3$ roots in $Bbb C,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $,x=1,$ as noted already, which does not satisfy the original quadratic.
answered 1 hour ago
dxiv
56.5k64798
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.why does it have to satisfy the original equation?It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.
– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
 |Â
show 2 more comments
up vote
2
down vote
For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $,x=1,$. We can substitute $,1 mapsto x,$ on the RHS and end up with $,x=x,$. Of course that $,x=1 implies x=x,$, but the converse is not true.
In OP's case, the original equation is quadratic in $,x,$ which has $2$ roots in $Bbb C,$, while the derived equation is a cubic which has $3$ roots in $Bbb C,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $,x=1,$ as noted already, which does not satisfy the original quadratic.
answered 1 hour ago
dxiv
56.5k64798
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.why does it have to satisfy the original equation?It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.
– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
 |Â
show 2 more comments
up vote
2
down vote
up vote
2
down vote
For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $,x=1,$. We can substitute $,1 mapsto x,$ on the RHS and end up with $,x=x,$. Of course that $,x=1 implies x=x,$, but the converse is not true.
In OP's case, the original equation is quadratic in $,x,$ which has $2$ roots in $Bbb C,$, while the derived equation is a cubic which has $3$ roots in $Bbb C,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $,x=1,$ as noted already, which does not satisfy the original quadratic.
answered 1 hour ago
dxiv
56.5k64798
For a different angle, substituting a variable from the same equation is valid, but not reversible. Doing such a substitution can introduce extraneous solutions that do not necessarily satisfy the original equation.
A trivial example of such a case is the equation $,x=1,$. We can substitute $,1 mapsto x,$ on the RHS and end up with $,x=x,$. Of course that $,x=1 implies x=x,$, but the converse is not true.
In OP's case, the original equation is quadratic in $,x,$ which has $2$ roots in $Bbb C,$, while the derived equation is a cubic which has $3$ roots in $Bbb C,$. It is quite clear that the two solution sets cannot be identical, and in fact the cubic has the extraneous root $,x=1,$ as noted already, which does not satisfy the original quadratic.
answered 1 hour ago
dxiv
56.5k64798
answered 1 hour ago
dxiv
56.5k64798
answered 1 hour ago
dxiv
56.5k64798
answered 1 hour ago
dxiv
56.5k64798
56.5k64798
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.why does it have to satisfy the original equation?It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.
– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
 |Â
show 2 more comments
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.why does it have to satisfy the original equation?It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.
– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
Hi. I am confused to why $x = x$ is not valid? Thanks.
– Strikers
52 mins ago
1
1
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
@Strikers It is valid, as I wrote, but it is no longer equivalent to the original equation. The roots of $,x=x,$ do indeed include the root $,x=1,$ of the original equation, but also include (infinitely) many other extraneous roots. In your case, the steps all the way to $,x^3=1,$ are correct, but at this point it is not guaranteed that all roots of $,x^3=1,$ also satisfy the original equation $,x^2+x+1=0,$, and in fact $,x=1,$ does not. So if you are looking for real roots, only, what you can conclude is that the original equation has no real roots.
– dxiv
47 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
Hi. So why does it have to satisfy the original equation? Thanks.
– Strikers
38 mins ago
@Strikers Please re-read my answer and comment more carefully.
why does it have to satisfy the original equation? It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.– dxiv
32 mins ago
@Strikers Please re-read my answer and comment more carefully.
why does it have to satisfy the original equation? It does not have to in general, and $,x=1,$ does not in your particular case. What you proved is that if there is a real root, then that root must be $,x=1,$. But $,x=1,$ is not a root of the original equation, and therefore the original equation has no real roots. Well, you almost proved it, only thing that's missing is the correct conclusion at the end.– dxiv
32 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
Sorry. If you are still answering, I'd like to ask: you mentioned that I proved that if there is a real root. However, I did not prove if there is a real root I proved there is a real root. I get that $x = 1$ is not a root of the original equation, but where does the conclusion that the original equation has no real roots come from? Thanks.
– Strikers
26 mins ago
 |Â
show 2 more comments
up vote
1
down vote
The higher level description of your work is:
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$
$1$ is not a solution to the original equation.
And therefore we conclude the assumption is false: that is,
- Therefore, the original equation has no solutions.
Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$ or either $-frac12 pm fracsqrt32 i$ (because those are the three cube roots of $1$)
$1$ is not a solution to the original equation.
$-frac12 pm fracsqrt32 i$ are solutions to the original equation
and therefore
- The solutions to the equation are $-frac12 pm fracsqrt32 i$
answered 1 hour ago
Hurkyl
110k9114257
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
add a comment |Â
up vote
1
down vote
The higher level description of your work is:
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$
$1$ is not a solution to the original equation.
And therefore we conclude the assumption is false: that is,
- Therefore, the original equation has no solutions.
Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$ or either $-frac12 pm fracsqrt32 i$ (because those are the three cube roots of $1$)
$1$ is not a solution to the original equation.
$-frac12 pm fracsqrt32 i$ are solutions to the original equation
and therefore
- The solutions to the equation are $-frac12 pm fracsqrt32 i$
answered 1 hour ago
Hurkyl
110k9114257
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The higher level description of your work is:
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$
$1$ is not a solution to the original equation.
And therefore we conclude the assumption is false: that is,
- Therefore, the original equation has no solutions.
Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$ or either $-frac12 pm fracsqrt32 i$ (because those are the three cube roots of $1$)
$1$ is not a solution to the original equation.
$-frac12 pm fracsqrt32 i$ are solutions to the original equation
and therefore
- The solutions to the equation are $-frac12 pm fracsqrt32 i$
answered 1 hour ago
Hurkyl
110k9114257
The higher level description of your work is:
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$
$1$ is not a solution to the original equation.
And therefore we conclude the assumption is false: that is,
- Therefore, the original equation has no solutions.
Incidentally, if you allow complex numbers then $x^3 = 1$ has three solutions, and you'd have to modify your work to
- Assume $x$ is a solution to the original equation.
- Then $x$ has to be $1$ or either $-frac12 pm fracsqrt32 i$ (because those are the three cube roots of $1$)
$1$ is not a solution to the original equation.
$-frac12 pm fracsqrt32 i$ are solutions to the original equation
and therefore
- The solutions to the equation are $-frac12 pm fracsqrt32 i$
answered 1 hour ago
Hurkyl
110k9114257
edited 1 hour ago
answered 1 hour ago
Hurkyl
110k9114257
answered 1 hour ago
Hurkyl
110k9114257
answered 1 hour ago
Hurkyl
110k9114257
110k9114257
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
add a comment |Â
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
Hi. I have learned about complex numbers and confused to why you mentioned I "assumed x is a solution to the original equation". I used clear mathematical reason to get to the final answer. Thanks.
– Strikers
53 mins ago
add a comment |Â
up vote
1
down vote
Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$beginalign
x_1 &= 1, & textor \
x_2 &= -frac12 + fracsqrt 32 i, & textor \
x_3 &= -frac12 - fracsqrt 32 i
endalign$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
answered 1 hour ago
Trevor Kafka
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
add a comment |Â
up vote
1
down vote
Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$beginalign
x_1 &= 1, & textor \
x_2 &= -frac12 + fracsqrt 32 i, & textor \
x_3 &= -frac12 - fracsqrt 32 i
endalign$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
answered 1 hour ago
Trevor Kafka
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$beginalign
x_1 &= 1, & textor \
x_2 &= -frac12 + fracsqrt 32 i, & textor \
x_3 &= -frac12 - fracsqrt 32 i
endalign$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
answered 1 hour ago
Trevor Kafka
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$beginalign
x_1 &= 1, & textor \
x_2 &= -frac12 + fracsqrt 32 i, & textor \
x_3 &= -frac12 - fracsqrt 32 i
endalign$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
answered 1 hour ago
Trevor Kafka
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Trevor Kafka
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Trevor Kafka
1714
answered 1 hour ago
Trevor Kafka
1714
1714
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Trevor Kafka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
add a comment |Â
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
Hi. What is the $i$ in your solution? Also $x^3 = 1$ to $x = 1$, I don't see why that is not allowed. Thanks.
– Strikers
51 mins ago
add a comment |Â
Strikers is a new contributor. Be nice, and check out our Code of Conduct.
Strikers is a new contributor. Be nice, and check out our Code of Conduct.
Strikers is a new contributor. Be nice, and check out our Code of Conduct.
Strikers is a new contributor. Be nice, and check out our Code of Conduct.
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1
Can you describe how you arrived at $x=1$ more precisely?
– Jakobian
1 hour ago
@Jakobian Okay, I have edited it.
– Strikers
1 hour ago
1
The key point, as in my answer, is that there are three solutions to $x^3 = 1$.
– Carl Mummert
1 hour ago