Mixing
What's the general rule for when applying f(x) = x^2 introduces a “extraneous solution” to a equation?...
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This question already has an answer here:
When do we get extraneous roots?
3 answers
For example consider the equation
$x = 3$
$implies x^2 = 9 $
$implies x = 3 ;or; x = -3$
In this case an extraneous solution is introduced i.e. $ x = -3$
Now for this equation
$sqrt x = 3 $
$implies x = 9$
In this case, squaring both sides of the equation doesn't introduce an extraneous solution. I want to ask whats the difference between the first case and the second case and why sometimes squaring both sides can lead to an extraneous solution and why somtimes it doesn't.
algebra-precalculus
asked Jan 22 at 0:36
Mark ChenMark Chen
91
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marked as duplicate by Xander Henderson, Lord Shark the Unknown, onurcanbektas, A. Pongrácz, Did Jan 22 at 7:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
When do we get extraneous roots?
3 answers
For example consider the equation
$x = 3$
$implies x^2 = 9 $
$implies x = 3 ;or; x = -3$
In this case an extraneous solution is introduced i.e. $ x = -3$
Now for this equation
$sqrt x = 3 $
$implies x = 9$
In this case, squaring both sides of the equation doesn't introduce an extraneous solution. I want to ask whats the difference between the first case and the second case and why sometimes squaring both sides can lead to an extraneous solution and why somtimes it doesn't.
algebra-precalculus
asked Jan 22 at 0:36
Mark ChenMark Chen
91
$endgroup$
marked as duplicate by Xander Henderson, Lord Shark the Unknown, onurcanbektas, A. Pongrácz, Did Jan 22 at 7:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47
add a comment |
$begingroup$
This question already has an answer here:
When do we get extraneous roots?
3 answers
For example consider the equation
$x = 3$
$implies x^2 = 9 $
$implies x = 3 ;or; x = -3$
In this case an extraneous solution is introduced i.e. $ x = -3$
Now for this equation
$sqrt x = 3 $
$implies x = 9$
In this case, squaring both sides of the equation doesn't introduce an extraneous solution. I want to ask whats the difference between the first case and the second case and why sometimes squaring both sides can lead to an extraneous solution and why somtimes it doesn't.
algebra-precalculus
asked Jan 22 at 0:36
Mark ChenMark Chen
91
$endgroup$
This question already has an answer here:
When do we get extraneous roots?
3 answers
For example consider the equation
$x = 3$
$implies x^2 = 9 $
$implies x = 3 ;or; x = -3$
In this case an extraneous solution is introduced i.e. $ x = -3$
Now for this equation
$sqrt x = 3 $
$implies x = 9$
In this case, squaring both sides of the equation doesn't introduce an extraneous solution. I want to ask whats the difference between the first case and the second case and why sometimes squaring both sides can lead to an extraneous solution and why somtimes it doesn't.
This question already has an answer here:
When do we get extraneous roots?
3 answers
algebra-precalculus
algebra-precalculus
asked Jan 22 at 0:36
Mark ChenMark Chen
91
asked Jan 22 at 0:36
Mark ChenMark Chen
91
asked Jan 22 at 0:36
Mark ChenMark Chen
91
asked Jan 22 at 0:36
Mark ChenMark Chen
91
asked Jan 22 at 0:36
Mark ChenMark Chen
91
91
marked as duplicate by Xander Henderson, Lord Shark the Unknown, onurcanbektas, A. Pongrácz, Did Jan 22 at 7:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, Lord Shark the Unknown, onurcanbektas, A. Pongrácz, Did Jan 22 at 7:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47
add a comment |
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47
add a comment |
3 Answers
3
active
oldest
votes
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In the second case, $sqrt{x}ge 0$ must hold, which automatically eliminates the extraneous solution for you.
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
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add a comment |
$begingroup$
The difference really lies in whether the function you're applying is injective. Essentially, a function $f:Ato B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,yin A$, i.e. each output only has one corresponding input. Now, the function $f:mathbb Rtomathbb R, xmapsto x^2$ is not injective, because $(-a)^2=a^2$ but $aneq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:mathbb R_{geq 0}tomathbb R,xmapsto x^2$ is injective, so if we know that $a,bgeq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.
Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
add a comment |
$begingroup$
The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $sqrt x =-3$
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In the second case, $sqrt{x}ge 0$ must hold, which automatically eliminates the extraneous solution for you.
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
$endgroup$
add a comment |
$begingroup$
In the second case, $sqrt{x}ge 0$ must hold, which automatically eliminates the extraneous solution for you.
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
$endgroup$
add a comment |
$begingroup$
In the second case, $sqrt{x}ge 0$ must hold, which automatically eliminates the extraneous solution for you.
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
$endgroup$
In the second case, $sqrt{x}ge 0$ must hold, which automatically eliminates the extraneous solution for you.
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
answered Jan 22 at 0:40
vadim123vadim123
76.3k897191
76.3k897191
add a comment |
add a comment |
$begingroup$
The difference really lies in whether the function you're applying is injective. Essentially, a function $f:Ato B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,yin A$, i.e. each output only has one corresponding input. Now, the function $f:mathbb Rtomathbb R, xmapsto x^2$ is not injective, because $(-a)^2=a^2$ but $aneq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:mathbb R_{geq 0}tomathbb R,xmapsto x^2$ is injective, so if we know that $a,bgeq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.
Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
add a comment |
$begingroup$
The difference really lies in whether the function you're applying is injective. Essentially, a function $f:Ato B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,yin A$, i.e. each output only has one corresponding input. Now, the function $f:mathbb Rtomathbb R, xmapsto x^2$ is not injective, because $(-a)^2=a^2$ but $aneq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:mathbb R_{geq 0}tomathbb R,xmapsto x^2$ is injective, so if we know that $a,bgeq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.
Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
$endgroup$
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
add a comment |
$begingroup$
The difference really lies in whether the function you're applying is injective. Essentially, a function $f:Ato B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,yin A$, i.e. each output only has one corresponding input. Now, the function $f:mathbb Rtomathbb R, xmapsto x^2$ is not injective, because $(-a)^2=a^2$ but $aneq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:mathbb R_{geq 0}tomathbb R,xmapsto x^2$ is injective, so if we know that $a,bgeq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.
Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
$endgroup$
The difference really lies in whether the function you're applying is injective. Essentially, a function $f:Ato B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,yin A$, i.e. each output only has one corresponding input. Now, the function $f:mathbb Rtomathbb R, xmapsto x^2$ is not injective, because $(-a)^2=a^2$ but $aneq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:mathbb R_{geq 0}tomathbb R,xmapsto x^2$ is injective, so if we know that $a,bgeq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.
Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
answered Jan 22 at 1:19
YiFanYiFan
4,5261727
4,5261727
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
add a comment |
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x?
$endgroup$
– Mark Chen
Jan 25 at 3:08
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
$begingroup$
No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:mathbb Ctomathbb C,zmapsto z^2$ as well, which is similarly not injective.
$endgroup$
– YiFan
Jan 25 at 3:36
add a comment |
$begingroup$
The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $sqrt x =-3$
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $sqrt x =-3$
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $sqrt x =-3$
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
$endgroup$
The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $sqrt x =-3$
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 0:48
Rhys HughesRhys Hughes
6,9741530
6,9741530
add a comment |
add a comment |
$begingroup$
See also - Why do extraneous solutions exist? and Why do I keep getting this incorrect solution?
$endgroup$
– Devashish Kaushik
Jan 22 at 0:47