Does the solutions to an equation imply the form of the equation?












0












$begingroup$


I had a discussion which I actually find quite interesting. The problem was, if $x=sqrt4$ then $x=-2$ or $x=2$. But does the implication go the other way too?



The solution from the textbook stated no.






functions







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$endgroup$












  • $begingroup$
    Please state your question clearly in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:19
















0












$begingroup$


I had a discussion which I actually find quite interesting. The problem was, if $x=sqrt4$ then $x=-2$ or $x=2$. But does the implication go the other way too?



The solution from the textbook stated no.






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please state your question clearly in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:19














0












0








0





$begingroup$


I had a discussion which I actually find quite interesting. The problem was, if $x=sqrt4$ then $x=-2$ or $x=2$. But does the implication go the other way too?



The solution from the textbook stated no.






functions







share|cite|improve this question











$endgroup$




I had a discussion which I actually find quite interesting. The problem was, if $x=sqrt4$ then $x=-2$ or $x=2$. But does the implication go the other way too?



The solution from the textbook stated no.






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 23:13









Tesla

885426




885426










asked Jan 22 at 23:09









Mathiaspilot123Mathiaspilot123

596




596












  • $begingroup$
    Please state your question clearly in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:19


















  • $begingroup$
    Please state your question clearly in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:19
















$begingroup$
Please state your question clearly in the body of the question and not just in the title.
$endgroup$
– Rob Arthan
Jan 22 at 23:19




$begingroup$
Please state your question clearly in the body of the question and not just in the title.
$endgroup$
– Rob Arthan
Jan 22 at 23:19










1 Answer
1






active

oldest

votes


















3












$begingroup$

Not at all. For instance, the equation
$$
x^4 - 16 = 0
$$
also has those two as its only real-number solutions.



By the way, although your book perhaps did not say it, most mathematicians** take $sqrt{4}$ to mean "the non-negative number whose square is $4$", i.e., $+2$. So many of them (including me!) would object to the statement that $x = sqrt{4}$ has $x = -2$ as a solution.



**As Blue points out in the comments, "most" is probably not strong enough. I should say "almost every" or "every one I've ever met".



On the other hand, the equation
$$
x^2 = 4
$$
does have two solutions, namely $x = 2$ and $x = -2$.



I want to add that this is a great question --- in mathematics, sometimes we find something that works in one direction, and the natural question to ask is "does it work in the other direction, too?" In this case, the answer is "No, solution sets don't determine the form of equations." But if you're willing to slightly alter the question and ask, "If $p$ and $q$ are two polynomials, and $p(x) = 0$ and $q(x) = 0$ have the same solutions, must $p$ and $q$ be the same?", then the answer is almost yes!



First, you have to allow complex number solutions, and second, one of them might be not equal to the other, but merely a constant multiple of the other, so that
$$
x^2 + 13x + 7 = 0
$$
and
$$
10x^2 + 130x + 70 = 0
$$
end up having the same solution sets.



And finally, you have to allow for something called "multiplicity of a solution," which makes everything a little more messy. But the gist is still there: after a little work, we really do get a correspondence between solutions-with-multiplicity and polynomials (up to nonzero multiples).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How about any other $ax^2=4$ for some real $a$?
    $endgroup$
    – Mathiaspilot123
    Jan 22 at 23:14








  • 1




    $begingroup$
    I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
    $endgroup$
    – John Hughes
    Jan 22 at 23:16










  • $begingroup$
    @Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
    $endgroup$
    – John Hughes
    Jan 22 at 23:17










  • $begingroup$
    @JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:18






  • 1




    $begingroup$
    @Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
    $endgroup$
    – John Hughes
    Jan 22 at 23:26











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

Not at all. For instance, the equation
$$
x^4 - 16 = 0
$$
also has those two as its only real-number solutions.



By the way, although your book perhaps did not say it, most mathematicians** take $sqrt{4}$ to mean "the non-negative number whose square is $4$", i.e., $+2$. So many of them (including me!) would object to the statement that $x = sqrt{4}$ has $x = -2$ as a solution.



**As Blue points out in the comments, "most" is probably not strong enough. I should say "almost every" or "every one I've ever met".



On the other hand, the equation
$$
x^2 = 4
$$
does have two solutions, namely $x = 2$ and $x = -2$.



I want to add that this is a great question --- in mathematics, sometimes we find something that works in one direction, and the natural question to ask is "does it work in the other direction, too?" In this case, the answer is "No, solution sets don't determine the form of equations." But if you're willing to slightly alter the question and ask, "If $p$ and $q$ are two polynomials, and $p(x) = 0$ and $q(x) = 0$ have the same solutions, must $p$ and $q$ be the same?", then the answer is almost yes!



First, you have to allow complex number solutions, and second, one of them might be not equal to the other, but merely a constant multiple of the other, so that
$$
x^2 + 13x + 7 = 0
$$
and
$$
10x^2 + 130x + 70 = 0
$$
end up having the same solution sets.



And finally, you have to allow for something called "multiplicity of a solution," which makes everything a little more messy. But the gist is still there: after a little work, we really do get a correspondence between solutions-with-multiplicity and polynomials (up to nonzero multiples).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How about any other $ax^2=4$ for some real $a$?
    $endgroup$
    – Mathiaspilot123
    Jan 22 at 23:14








  • 1




    $begingroup$
    I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
    $endgroup$
    – John Hughes
    Jan 22 at 23:16










  • $begingroup$
    @Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
    $endgroup$
    – John Hughes
    Jan 22 at 23:17










  • $begingroup$
    @JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:18






  • 1




    $begingroup$
    @Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
    $endgroup$
    – John Hughes
    Jan 22 at 23:26
















3












$begingroup$

Not at all. For instance, the equation
$$
x^4 - 16 = 0
$$
also has those two as its only real-number solutions.



By the way, although your book perhaps did not say it, most mathematicians** take $sqrt{4}$ to mean "the non-negative number whose square is $4$", i.e., $+2$. So many of them (including me!) would object to the statement that $x = sqrt{4}$ has $x = -2$ as a solution.



**As Blue points out in the comments, "most" is probably not strong enough. I should say "almost every" or "every one I've ever met".



On the other hand, the equation
$$
x^2 = 4
$$
does have two solutions, namely $x = 2$ and $x = -2$.



I want to add that this is a great question --- in mathematics, sometimes we find something that works in one direction, and the natural question to ask is "does it work in the other direction, too?" In this case, the answer is "No, solution sets don't determine the form of equations." But if you're willing to slightly alter the question and ask, "If $p$ and $q$ are two polynomials, and $p(x) = 0$ and $q(x) = 0$ have the same solutions, must $p$ and $q$ be the same?", then the answer is almost yes!



First, you have to allow complex number solutions, and second, one of them might be not equal to the other, but merely a constant multiple of the other, so that
$$
x^2 + 13x + 7 = 0
$$
and
$$
10x^2 + 130x + 70 = 0
$$
end up having the same solution sets.



And finally, you have to allow for something called "multiplicity of a solution," which makes everything a little more messy. But the gist is still there: after a little work, we really do get a correspondence between solutions-with-multiplicity and polynomials (up to nonzero multiples).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How about any other $ax^2=4$ for some real $a$?
    $endgroup$
    – Mathiaspilot123
    Jan 22 at 23:14








  • 1




    $begingroup$
    I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
    $endgroup$
    – John Hughes
    Jan 22 at 23:16










  • $begingroup$
    @Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
    $endgroup$
    – John Hughes
    Jan 22 at 23:17










  • $begingroup$
    @JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:18






  • 1




    $begingroup$
    @Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
    $endgroup$
    – John Hughes
    Jan 22 at 23:26














3












3








3





$begingroup$

Not at all. For instance, the equation
$$
x^4 - 16 = 0
$$
also has those two as its only real-number solutions.



By the way, although your book perhaps did not say it, most mathematicians** take $sqrt{4}$ to mean "the non-negative number whose square is $4$", i.e., $+2$. So many of them (including me!) would object to the statement that $x = sqrt{4}$ has $x = -2$ as a solution.



**As Blue points out in the comments, "most" is probably not strong enough. I should say "almost every" or "every one I've ever met".



On the other hand, the equation
$$
x^2 = 4
$$
does have two solutions, namely $x = 2$ and $x = -2$.



I want to add that this is a great question --- in mathematics, sometimes we find something that works in one direction, and the natural question to ask is "does it work in the other direction, too?" In this case, the answer is "No, solution sets don't determine the form of equations." But if you're willing to slightly alter the question and ask, "If $p$ and $q$ are two polynomials, and $p(x) = 0$ and $q(x) = 0$ have the same solutions, must $p$ and $q$ be the same?", then the answer is almost yes!



First, you have to allow complex number solutions, and second, one of them might be not equal to the other, but merely a constant multiple of the other, so that
$$
x^2 + 13x + 7 = 0
$$
and
$$
10x^2 + 130x + 70 = 0
$$
end up having the same solution sets.



And finally, you have to allow for something called "multiplicity of a solution," which makes everything a little more messy. But the gist is still there: after a little work, we really do get a correspondence between solutions-with-multiplicity and polynomials (up to nonzero multiples).






share|cite|improve this answer











$endgroup$



Not at all. For instance, the equation
$$
x^4 - 16 = 0
$$
also has those two as its only real-number solutions.



By the way, although your book perhaps did not say it, most mathematicians** take $sqrt{4}$ to mean "the non-negative number whose square is $4$", i.e., $+2$. So many of them (including me!) would object to the statement that $x = sqrt{4}$ has $x = -2$ as a solution.



**As Blue points out in the comments, "most" is probably not strong enough. I should say "almost every" or "every one I've ever met".



On the other hand, the equation
$$
x^2 = 4
$$
does have two solutions, namely $x = 2$ and $x = -2$.



I want to add that this is a great question --- in mathematics, sometimes we find something that works in one direction, and the natural question to ask is "does it work in the other direction, too?" In this case, the answer is "No, solution sets don't determine the form of equations." But if you're willing to slightly alter the question and ask, "If $p$ and $q$ are two polynomials, and $p(x) = 0$ and $q(x) = 0$ have the same solutions, must $p$ and $q$ be the same?", then the answer is almost yes!



First, you have to allow complex number solutions, and second, one of them might be not equal to the other, but merely a constant multiple of the other, so that
$$
x^2 + 13x + 7 = 0
$$
and
$$
10x^2 + 130x + 70 = 0
$$
end up having the same solution sets.



And finally, you have to allow for something called "multiplicity of a solution," which makes everything a little more messy. But the gist is still there: after a little work, we really do get a correspondence between solutions-with-multiplicity and polynomials (up to nonzero multiples).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 22 at 23:24

























answered Jan 22 at 23:12









John HughesJohn Hughes

64.4k24191




64.4k24191












  • $begingroup$
    How about any other $ax^2=4$ for some real $a$?
    $endgroup$
    – Mathiaspilot123
    Jan 22 at 23:14








  • 1




    $begingroup$
    I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
    $endgroup$
    – John Hughes
    Jan 22 at 23:16










  • $begingroup$
    @Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
    $endgroup$
    – John Hughes
    Jan 22 at 23:17










  • $begingroup$
    @JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:18






  • 1




    $begingroup$
    @Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
    $endgroup$
    – John Hughes
    Jan 22 at 23:26


















  • $begingroup$
    How about any other $ax^2=4$ for some real $a$?
    $endgroup$
    – Mathiaspilot123
    Jan 22 at 23:14








  • 1




    $begingroup$
    I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
    $endgroup$
    – John Hughes
    Jan 22 at 23:16










  • $begingroup$
    @Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
    $endgroup$
    – John Hughes
    Jan 22 at 23:17










  • $begingroup$
    @JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
    $endgroup$
    – Rob Arthan
    Jan 22 at 23:18






  • 1




    $begingroup$
    @Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
    $endgroup$
    – John Hughes
    Jan 22 at 23:26
















$begingroup$
How about any other $ax^2=4$ for some real $a$?
$endgroup$
– Mathiaspilot123
Jan 22 at 23:14






$begingroup$
How about any other $ax^2=4$ for some real $a$?
$endgroup$
– Mathiaspilot123
Jan 22 at 23:14






1




1




$begingroup$
I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
$endgroup$
– John Hughes
Jan 22 at 23:16




$begingroup$
I was answering the question in the title: does the set of solutions to an equation imply the form of the equation? I was pointing out that over the reals, there are many different equations whose solution sets are all ${2, -2}$, including $x^2 = 4, x^4 = 16, (x^2 - 4)(x^2 + 1) = 0,$ etc. I'm sorry I wasn't clear.
$endgroup$
– John Hughes
Jan 22 at 23:16












$begingroup$
@Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
$endgroup$
– John Hughes
Jan 22 at 23:17




$begingroup$
@Mathiaspilot123: unless $a = 1$, this equation does not have ${2, -2}$ as its solution set. For instance, if $a = 4$, then the only solutions to $4x^2 = 4$ are $x = pm 1$.
$endgroup$
– John Hughes
Jan 22 at 23:17












$begingroup$
@JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
$endgroup$
– Rob Arthan
Jan 22 at 23:18




$begingroup$
@JohnHughes: thanks for the clarification. I think it would have been a good idea to get the OP to state the question in the body of the question and not just in the title.
$endgroup$
– Rob Arthan
Jan 22 at 23:18




1




1




$begingroup$
@Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
$endgroup$
– John Hughes
Jan 22 at 23:26




$begingroup$
@Mathiaspilot123: yes, $a(x^2 - 4) = 0$ has the same solutions as $x^2 - 4 = 0$, for any nonzero value of $a$. If $a$ is zero, however, the solution sets are wildly different. This is an example of the phenomenon I was describing in my last paragraph.
$endgroup$
– John Hughes
Jan 22 at 23:26


















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What does “Dominus providebit” mean?

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5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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