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Show your work means that the process for doing the problem works in a sufficiently generic situation. The student's methodology would not work for solving the

Show your work means that the process for doing the problem works in a sufficiently generic situation. The student's methodology would not work for solving the equation: x^3 = 7 Hence I would mark the student's answer wrong. At least at the level of a college algebra course. Guess and try is fine for getting an intuition for a problem but it is not fine as a solution. EDIT: I don't care about down votes but I'm interested why there is disagreement with what I wrote. I've been teaching mathematics at the college level for 20 years and I believe my answer above fits with what most college level teachers think about the topic. So, why do you disagree with what I wrote?

analog31 10y ago
Note that I didn't downvote you. I actually consider the answer wrong too, and the professors agreed too. One of them said: "Obviously the student doesn't understand the problem." I thought to myself: "You understand the problem because you know the answer." But I held my tongue. But at the present state of the art, the student has no idea why the answer is wrong. That's where I think their high school math background failed them.
- yequalsx 10y ago
  The danger of always using nice numbers! I've taught topics where I thought I had the best lecture, the best examples and that it would be impossible for a student to misinterpret the method. Some still did come up with flawed thinking. It's hard.
AlexCoventry 10y ago
> The student's methodology would not work for solving the equation: x^3 = 7 It would if the results they reported were the initial steps of a binary search which got very lucky.
Eliezer 10y ago
I HATED "show your work". You want a proof, ask me for a proof. It's not my fault I can solve your problem in my head. Especially horrifying is if you want me to solve the problem a particular way. There is no better way of turning the fun and freedom of math into drudgework. You are licensed to ask me to arrive at the correct answer, or to prove a theorem from premises; telling me how to do it is mathematical abuse.
- yequalsx 10y ago
  The answer is not important. What is important is that one knows a process for solving the general situation. Knowing how to get the answer in a problem with easy numbers is not the point. Demonstrating how to find it no matter what numbers I used is the point. There is also the fact that I'm not just teaching the student so they can pass this class. I'm teaching them so that they can pass the next class too. In beginning algebra, our lowest level course, we start with basic problems. For instance, you drive for 3 hours and travel 150 miles. What's your average speed. Almost every person gets this right. But many can't do this problem. You drive a car for 4/3 an hour and travel 100/6 miles. What is your average speed? Now many can't do this problem. They ask, which way does the division occur? Our goal is that they know and understand a general process. We are setting them up to solve more advanced problems. Problems that can't be done in their head. If you can't write the steps out in the simple case you'll never understand the harder problems. Problems that involve quadratic functions or trig functions. Mathematics is a human activity and communication is part of it. Knowing how to communicate what is in your brain to another person in a way that they can understand is very valuable. I give the students nice numbers and in exchange they are expected to tell me a general way of solving that type of problem. Instead of feeding them for a day I want to teach them how to fish. I strongly disagree with your belief that this mathematical abuse.
  - analog31 10y ago
    Interestingly, I think many of your points would seem obvious if couched in computer programming terms. For instance, how do you write a program that's correct, but that another human can understand and adapt? What happens when a program gets too big to understand all at once in your head? How do you know that you've found all of the possible outputs?
  - webmaven 10y ago
    > Knowing how to get the answer in a problem with easy numbers is not the point. So, don't use easy numbers? > I give the students nice numbers and in exchange they are expected to tell me a general way of solving that type of problem. If this quid-pro-quo is explicit, that seems fair. But I still don't see why "nice numbers" are necessary.
divbit 10y ago
I don't necessarily disagree with you, but imo it definitely matters how the problem is worded and presented, or if it is otherwise made clear what type of work will be accepted. Silly examples: suppose someone disproves the Riemann hypothesis through a computer search which finds a zero-crossing counter-example. Their counterexample is not a 'wrong' answer, even though they have essentially just done a more complicated version of 1^3 = 1, 2^3 = 8, ... and merely providing whatever computer code they used would probably be accepted by anyone. Example in the other direction might be the original four-color-theorem proof, which was partially by exhaustive computer aided search.
- Ar-Curunir 10y ago
  This is where complexity theory comes in; you want a proof that is checkable in polynomial time; listing examples one by one to prove something is not such a proof. If instead the question asked, solve x^3 = 8192; a proof iterating from 0 uptil the cube root of 8192 wouldn't work, because that's exponential in the problem size. Brute force search to get a counter example to the Reimann hypothesis produces a solution that is checkable in polynomial time; you just evaluate the Reimann-zeta function at the produced point.
  - divbit 10y ago
    I think, in the case of an algebra class, the student could assume something like an integral / rational solution space. Otherwise, the solution is just to put the opposite symbol on the other side of the equation, e.g. (8192)^(1/3), which is maybe(?) what the teacher is asking, however would the teacher actually accept an answer written in the form 27^(1/3)? I would guess that I would not accept that. I understand what you are saying, but in this case, it could be argued that the 8192 case is actually a different question, because the underlying multiplicative group is different.
  - 21 10y ago
    Why does it matter the complexity of the solution check? If for some reason the counter example of the Reimann hyp would be checkable in exponential time (whatever that means), and if it would have required 10 years of computer time to check, how would that matter? The solution would still stand (assuming that everybody agrees that no mistake was made).
ezequiel-garzon 10y ago
Do you expect three solutions, or just the real one? What about for x^2=-1, two or none? Is \sqrt[3]{7} enough, or is a decimal approximation required? If it isn't (and the student doesn't know how to go about it) one could argue that isn't much of an answer, but rather an existence claim, and that the methodology doesn't amount to much. Personally I feel that you probably wouldn't give a question such as x^3=27 (neither would I), but if you did, marking it as wrong (as in no credit) after seeing the justification 1^3=1, 2^3=8, 3^3=27 would be too harsh. You can't penalize a student for giving out an easy question.
- yequalsx 10y ago
  Cube root of 7 is the right answer. It doesn't have a finite decimal or repeating decimal answer so a decimal answer is wrong. But I wouldn't take off points for giving a decimal approximation as the answer. In college algebra, for most sections, we deal with real number solutions. They haven't reached the point of knowing about non-real solutions. We teach at the level the students are at. Without having had trig finding the roots of unity is hard and not comprehensible to the students so asking them for all three solutions is a bit much. I would not give a student in college algebra credit for solving x^3 = 27 by guess and check. It demonstrates that they really don't understand what is going on. I give credit for demonstrating understanding. Not demonstrating that they are good guessers.
proactive 10y ago
You have to be very cautious when marking wrong an answer that is correct, because you're at risk of confusing the student. The guess-and-try approach used in this particular example is very brittle, but it is correct. Professional mathematicians use it frequently, especially with the advent of computer-aided mathematics, for instance to find counter-examples to statements. I see at least three issues with claiming that answer as wrong: first, correctness is essential (in the true sense) in mathematics, and therefore should not be carelessly dismissed in front of the student. Second, students should not be made to believe that guess-and-try is always inappropriate, but rather to understand that it won't always work. Finally, in this particular example the approach chosen is arguably (at least from the student's perspective) simpler than the one expected by the professor. Invalidating a "simpler" approach might give the student the impression that you always need to take the complicated route (ie, "math is hard") when the opposite is true. My own take on this example would be to give (partial?) marks, with a lengthy comment of the type "fair enough, in this case, but what about if you wanted to solve x^3 =7? Your method wouldn't work, then!". Alternatively, if you don't want to give marks, it should be justified at length by rules clearly explained before the exam, while acknowledging the correctness of the approach.
- yequalsx 10y ago
  I think the simplest approach is to use the cube root function. This is the simplest approach because it solves, over the reals, any equation of the form x^3 = real number That's the simplest solution. It works in every case. To me the answer is not important. The methodology is important. Giving a counter example is very much a different type of problem. Just about any method is valid in that type of problem. Passing a class should mean more than I got a lot of answers correct. It should mean an understanding of the material. A college algebra student who solves x^3=27 in the aforementioned manner is lacking a fundamental understanding of the material. Now a third grader who reasons thusly, well that is impressive. The goal is not the right answer. It a demonstration of understanding and abilit appropriate to the level of the course.
  - proactive 10y ago
    You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general approach is necessarily the simplest. > Just about any method is valid in that type of problem. Why is that not true for other types of problems? > Passing a class should mean more than I got a lot of answers correct. I agree. But you shouldn't penalise the student if the exam question is poorly framed (and we all make such mistakes). Just take a note for later and don't make the mistake again.
paulsutter 10y ago
If you were teaching a more formal method, I can see marking it wrong. But you're mistaken to think the student can't find the answer that way. If I had to solve x^3 = 7 in my head, I know that 2^3 = 8, and my next guess would be 1.9 1.9^3 = 6.859 (ok, I'd need a pencil for this) Now I know the answer to two digits, "x is just a bit more than 1.9", and one can keep going if more precision is needed.
- yequalsx 10y ago
  The student demonstrated that they can do such problems with positive integer solutions. Possibly even negative integer solutions. But certainly they did not show any sort of reasoning along the lines you wrote. But let's take your method. Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. But this isn't the solution to the equation. The solution to the equation in question is a number and not a sequence. The answer, over the reals, is 7^(1/3).
  - BeetleB 10y ago
    >Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. ... >The answer, over the reals, is 7^(1/3). So if the student did not write 2, but wrote 8^(1/3), it is OK? As for the method being wrong, no - it isn't. The sequence converges. If you want an exact answer, you should specify that you want an exact answer. >The solution to the equation in question is a number and not a sequence. Sorry, but every number is a sequence. You can start with rational numbers, and define every real number as a sequence of rationals. Lots of books actually do this to define what a real number is.
    - yequalsx 10y ago
      8^(1/3) is 2 (when interpreted over the reals). Your method is wrong because it does not produce the answer. One does not, in college algebra, say the solution is: lim a_n as n->infinity For one thing you did not prove convergence. It is understood that solutions to algebraic equations over the reals are numbers and not approximations. Any student who knows about Dedekind cuts or infinite sequences knows to take the cube root of both sides. Most mathematicians consider it silly (strange, wrong) to say that 2 is an element of 3 even though it is. Anyone solving x^3 = 27 in the method specified does not know any of these finer points of mathematics. The method is bad. It's useful for positive integer solutions but not for the general situation.
asd32rt3tades 10y ago
I think the problem lies with the question. Simple question gets simple answer. If you want the solution in the general real case, ask x^3 = 7 since it cannot be solved by the easiest method. If you ask to solve x^3 = 8, I firmly believe it is fair game to solve using a method that only works for small integer.
contravariant 10y ago
I disagree. Whether a process is general or not is not important, sometimes problems aren't part of an obvious larger class of problems, and even then using a method that only works for a small set of problems is not technically wrong, and can sometimes even be many times easier than using a fully general method. For example, solving x^3 = 27 by remembering 3^3 = 27 is a perfectly acceptable solution. Trial and error shouldn't so readily dismissed. In my opinion 'show your work' means that you show why you've arrived at a particular answer, why this is a correct solution, and (if necessary) why this is the only solution. Using that criterion, simply writing something like '3^3 = 27', or more properly 'Simple trial and error shows x=3 to be a solution, since 3^3 = 27', would suffice. If the problem was instead x^3 = 7, then sure this method wouldn't work (although they might be able to figure out it's somewhere between 1 and 2) but then again 'By definition x=3√7.' isn't particularly illuminating either, even though it's a full and correct derivation of the answer.
- yequalsx 10y ago
  I'd say 7^(1/3) is illuminating in the sense that 1. Someone understanding this is more likely to understand the concept of inverse functions. 2. Someone understanding this is more likely to understand how to solve x^(3/5) = 7. And then more likely to understand how to solve x^(sqrt(2)) = 7. In a college algebra level type course the ultimate goal is not knowing how to solve x^3 = 27. The ultimate goal is to understand more complicated ideas. The reasoning displayed is a huge red flag. In college algebra a student presenting the solution given will very likely fail the course. That student needs help.
mrob 10y ago
The task given was to solve an equation and to show how it was solved. The student did exactly that. Their answer is obviously correct and they deserve full marks for it. If you want them to complete a different task you should ask them to do a different task. Eg. The question could be "show a general method for solving equations of the form "x^3=n", where n is any real number". Punishing people for obeying your instructions is a great way to make them hate you. I'm not surprised so many people hate math when it's taught in this way.
- yequalsx 10y ago
  A student solving the equation in the way the OP stated will likely fail the course. They are showing they don't understand the concepts. The student needs help. They are unlikely to understand inverse functions when that topic is taught. They are unlikely to understand how to solve x^(2/5) = 7. Or x^(sqrt(2)) = 7. Probably they don't understand exponents. The solution given is a huge red flag.
  - mrob 10y ago
    It's disgustingly arrogant to assume that. They are showing exactly the understanding needed to answer the question. They obeyed the instructions, and they paid good money to have that opportunity, but they are being punished despite doing exactly what was asked of them. Whether the instructor has the legal right to do this or not, it certainly feels like fraud. If the question setter failed to set a good question, that's the question setter's fault, not the student's fault. Punishing the student for somebody else's mistake is injustice.
    - yequalsx 10y ago
      I've been teaching college algebra for 20 years. There is nothing arrogant in what I said. The student needs help and it is clear that this is so to anyone with experience teaching such courses. After many years and thousands of students I've developed an intuition on when a student needs help and when I don't need to worry about them. If you had to bet a million dollars, based solely on the solution presented and knowing the course was college algebra, would you put the money on them passing or failing?
Veratyr 10y ago
I appreciate the sentiment but I'd argue that if you're looking for generic methodology, you should ask for a generic solution. The question shouldn't be "solve x^3 = 7", it should be "solve x^3 = y" or "solve x^y = z". If a student produces a correct answer to a numerical question that's correct by drawing a graph or doing a linear/binary search, that seems fine to me.

analog31 10y ago

Note that I didn't downvote you. I actually consider the answer wrong too, and the professors agreed too. One of them said: "Obviously the student doesn't understand the problem." I thought to myself: "You understand the problem because you know the answer." But I held my tongue. But at the present state of the art, the student has no idea why the answer is wrong. That's where I think their high school math background failed them.

yequalsx 10y ago
The danger of always using nice numbers! I've taught topics where I thought I had the best lecture, the best examples and that it would be impossible for a student to misinterpret the method. Some still did come up with flawed thinking. It's hard.

AlexCoventry 10y ago

> The student's methodology would not work for solving the equation: x^3 = 7 It would if the results they reported were the initial steps of a binary search which got very lucky.

Eliezer 10y ago

I HATED "show your work". You want a proof, ask me for a proof. It's not my fault I can solve your problem in my head. Especially horrifying is if you want me to solve the problem a particular way. There is no better way of turning the fun and freedom of math into drudgework. You are licensed to ask me to arrive at the correct answer, or to prove a theorem from premises; telling me how to do it is mathematical abuse.

yequalsx 10y ago
The answer is not important. What is important is that one knows a process for solving the general situation. Knowing how to get the answer in a problem with easy numbers is not the point. Demonstrating how to find it no matter what numbers I used is the point. There is also the fact that I'm not just teaching the student so they can pass this class. I'm teaching them so that they can pass the next class too. In beginning algebra, our lowest level course, we start with basic problems. For instance, you drive for 3 hours and travel 150 miles. What's your average speed. Almost every person gets this right. But many can't do this problem. You drive a car for 4/3 an hour and travel 100/6 miles. What is your average speed? Now many can't do this problem. They ask, which way does the division occur? Our goal is that they know and understand a general process. We are setting them up to solve more advanced problems. Problems that can't be done in their head. If you can't write the steps out in the simple case you'll never understand the harder problems. Problems that involve quadratic functions or trig functions. Mathematics is a human activity and communication is part of it. Knowing how to communicate what is in your brain to another person in a way that they can understand is very valuable. I give the students nice numbers and in exchange they are expected to tell me a general way of solving that type of problem. Instead of feeding them for a day I want to teach them how to fish. I strongly disagree with your belief that this mathematical abuse.
- analog31 10y ago
  Interestingly, I think many of your points would seem obvious if couched in computer programming terms. For instance, how do you write a program that's correct, but that another human can understand and adapt? What happens when a program gets too big to understand all at once in your head? How do you know that you've found all of the possible outputs?
- webmaven 10y ago
  > Knowing how to get the answer in a problem with easy numbers is not the point. So, don't use easy numbers? > I give the students nice numbers and in exchange they are expected to tell me a general way of solving that type of problem. If this quid-pro-quo is explicit, that seems fair. But I still don't see why "nice numbers" are necessary.

divbit 10y ago

I don't necessarily disagree with you, but imo it definitely matters how the problem is worded and presented, or if it is otherwise made clear what type of work will be accepted. Silly examples: suppose someone disproves the Riemann hypothesis through a computer search which finds a zero-crossing counter-example. Their counterexample is not a 'wrong' answer, even though they have essentially just done a more complicated version of 1^3 = 1, 2^3 = 8, ... and merely providing whatever computer code they used would probably be accepted by anyone. Example in the other direction might be the original four-color-theorem proof, which was partially by exhaustive computer aided search.

Ar-Curunir 10y ago
This is where complexity theory comes in; you want a proof that is checkable in polynomial time; listing examples one by one to prove something is not such a proof. If instead the question asked, solve x^3 = 8192; a proof iterating from 0 uptil the cube root of 8192 wouldn't work, because that's exponential in the problem size. Brute force search to get a counter example to the Reimann hypothesis produces a solution that is checkable in polynomial time; you just evaluate the Reimann-zeta function at the produced point.
- divbit 10y ago
  I think, in the case of an algebra class, the student could assume something like an integral / rational solution space. Otherwise, the solution is just to put the opposite symbol on the other side of the equation, e.g. (8192)^(1/3), which is maybe(?) what the teacher is asking, however would the teacher actually accept an answer written in the form 27^(1/3)? I would guess that I would not accept that. I understand what you are saying, but in this case, it could be argued that the 8192 case is actually a different question, because the underlying multiplicative group is different.
- 21 10y ago
  Why does it matter the complexity of the solution check? If for some reason the counter example of the Reimann hyp would be checkable in exponential time (whatever that means), and if it would have required 10 years of computer time to check, how would that matter? The solution would still stand (assuming that everybody agrees that no mistake was made).

ezequiel-garzon 10y ago

Do you expect three solutions, or just the real one? What about for x^2=-1, two or none? Is \sqrt[3]{7} enough, or is a decimal approximation required? If it isn't (and the student doesn't know how to go about it) one could argue that isn't much of an answer, but rather an existence claim, and that the methodology doesn't amount to much. Personally I feel that you probably wouldn't give a question such as x^3=27 (neither would I), but if you did, marking it as wrong (as in no credit) after seeing the justification 1^3=1, 2^3=8, 3^3=27 would be too harsh. You can't penalize a student for giving out an easy question.

yequalsx 10y ago
Cube root of 7 is the right answer. It doesn't have a finite decimal or repeating decimal answer so a decimal answer is wrong. But I wouldn't take off points for giving a decimal approximation as the answer. In college algebra, for most sections, we deal with real number solutions. They haven't reached the point of knowing about non-real solutions. We teach at the level the students are at. Without having had trig finding the roots of unity is hard and not comprehensible to the students so asking them for all three solutions is a bit much. I would not give a student in college algebra credit for solving x^3 = 27 by guess and check. It demonstrates that they really don't understand what is going on. I give credit for demonstrating understanding. Not demonstrating that they are good guessers.

proactive 10y ago

You have to be very cautious when marking wrong an answer that is correct, because you're at risk of confusing the student. The guess-and-try approach used in this particular example is very brittle, but it is correct. Professional mathematicians use it frequently, especially with the advent of computer-aided mathematics, for instance to find counter-examples to statements. I see at least three issues with claiming that answer as wrong: first, correctness is essential (in the true sense) in mathematics, and therefore should not be carelessly dismissed in front of the student. Second, students should not be made to believe that guess-and-try is always inappropriate, but rather to understand that it won't always work. Finally, in this particular example the approach chosen is arguably (at least from the student's perspective) simpler than the one expected by the professor. Invalidating a "simpler" approach might give the student the impression that you always need to take the complicated route (ie, "math is hard") when the opposite is true. My own take on this example would be to give (partial?) marks, with a lengthy comment of the type "fair enough, in this case, but what about if you wanted to solve x^3 =7? Your method wouldn't work, then!". Alternatively, if you don't want to give marks, it should be justified at length by rules clearly explained before the exam, while acknowledging the correctness of the approach.

yequalsx 10y ago
I think the simplest approach is to use the cube root function. This is the simplest approach because it solves, over the reals, any equation of the form x^3 = real number That's the simplest solution. It works in every case. To me the answer is not important. The methodology is important. Giving a counter example is very much a different type of problem. Just about any method is valid in that type of problem. Passing a class should mean more than I got a lot of answers correct. It should mean an understanding of the material. A college algebra student who solves x^3=27 in the aforementioned manner is lacking a fundamental understanding of the material. Now a third grader who reasons thusly, well that is impressive. The goal is not the right answer. It a demonstration of understanding and abilit appropriate to the level of the course.
- proactive 10y ago
  You and I think that the cube root function is the simplest approach. The student doesn't necessarily agree. And I disagree with you that the most general approach is necessarily the simplest. > Just about any method is valid in that type of problem. Why is that not true for other types of problems? > Passing a class should mean more than I got a lot of answers correct. I agree. But you shouldn't penalise the student if the exam question is poorly framed (and we all make such mistakes). Just take a note for later and don't make the mistake again.

paulsutter 10y ago

If you were teaching a more formal method, I can see marking it wrong. But you're mistaken to think the student can't find the answer that way. If I had to solve x^3 = 7 in my head, I know that 2^3 = 8, and my next guess would be 1.9 1.9^3 = 6.859 (ok, I'd need a pencil for this) Now I know the answer to two digits, "x is just a bit more than 1.9", and one can keep going if more precision is needed.

yequalsx 10y ago
The student demonstrated that they can do such problems with positive integer solutions. Possibly even negative integer solutions. But certainly they did not show any sort of reasoning along the lines you wrote. But let's take your method. Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. But this isn't the solution to the equation. The solution to the equation in question is a number and not a sequence. The answer, over the reals, is 7^(1/3).
- BeetleB 10y ago
  >Your method is wrong because you aren't finding the solution. You are finding a sequence of numbers that converges to the solution. ... >The answer, over the reals, is 7^(1/3). So if the student did not write 2, but wrote 8^(1/3), it is OK? As for the method being wrong, no - it isn't. The sequence converges. If you want an exact answer, you should specify that you want an exact answer. >The solution to the equation in question is a number and not a sequence. Sorry, but every number is a sequence. You can start with rational numbers, and define every real number as a sequence of rationals. Lots of books actually do this to define what a real number is.
  - yequalsx 10y ago
    8^(1/3) is 2 (when interpreted over the reals). Your method is wrong because it does not produce the answer. One does not, in college algebra, say the solution is: lim a_n as n->infinity For one thing you did not prove convergence. It is understood that solutions to algebraic equations over the reals are numbers and not approximations. Any student who knows about Dedekind cuts or infinite sequences knows to take the cube root of both sides. Most mathematicians consider it silly (strange, wrong) to say that 2 is an element of 3 even though it is. Anyone solving x^3 = 27 in the method specified does not know any of these finer points of mathematics. The method is bad. It's useful for positive integer solutions but not for the general situation.

asd32rt3tades 10y ago

I think the problem lies with the question. Simple question gets simple answer. If you want the solution in the general real case, ask x^3 = 7 since it cannot be solved by the easiest method. If you ask to solve x^3 = 8, I firmly believe it is fair game to solve using a method that only works for small integer.

contravariant 10y ago

I disagree. Whether a process is general or not is not important, sometimes problems aren't part of an obvious larger class of problems, and even then using a method that only works for a small set of problems is not technically wrong, and can sometimes even be many times easier than using a fully general method. For example, solving x^3 = 27 by remembering 3^3 = 27 is a perfectly acceptable solution. Trial and error shouldn't so readily dismissed. In my opinion 'show your work' means that you show why you've arrived at a particular answer, why this is a correct solution, and (if necessary) why this is the only solution. Using that criterion, simply writing something like '3^3 = 27', or more properly 'Simple trial and error shows x=3 to be a solution, since 3^3 = 27', would suffice. If the problem was instead x^3 = 7, then sure this method wouldn't work (although they might be able to figure out it's somewhere between 1 and 2) but then again 'By definition x=3√7.' isn't particularly illuminating either, even though it's a full and correct derivation of the answer.

yequalsx 10y ago
I'd say 7^(1/3) is illuminating in the sense that 1. Someone understanding this is more likely to understand the concept of inverse functions. 2. Someone understanding this is more likely to understand how to solve x^(3/5) = 7. And then more likely to understand how to solve x^(sqrt(2)) = 7. In a college algebra level type course the ultimate goal is not knowing how to solve x^3 = 27. The ultimate goal is to understand more complicated ideas. The reasoning displayed is a huge red flag. In college algebra a student presenting the solution given will very likely fail the course. That student needs help.

mrob 10y ago

The task given was to solve an equation and to show how it was solved. The student did exactly that. Their answer is obviously correct and they deserve full marks for it. If you want them to complete a different task you should ask them to do a different task. Eg. The question could be "show a general method for solving equations of the form "x^3=n", where n is any real number". Punishing people for obeying your instructions is a great way to make them hate you. I'm not surprised so many people hate math when it's taught in this way.

yequalsx 10y ago
A student solving the equation in the way the OP stated will likely fail the course. They are showing they don't understand the concepts. The student needs help. They are unlikely to understand inverse functions when that topic is taught. They are unlikely to understand how to solve x^(2/5) = 7. Or x^(sqrt(2)) = 7. Probably they don't understand exponents. The solution given is a huge red flag.
- mrob 10y ago
  It's disgustingly arrogant to assume that. They are showing exactly the understanding needed to answer the question. They obeyed the instructions, and they paid good money to have that opportunity, but they are being punished despite doing exactly what was asked of them. Whether the instructor has the legal right to do this or not, it certainly feels like fraud. If the question setter failed to set a good question, that's the question setter's fault, not the student's fault. Punishing the student for somebody else's mistake is injustice.
  - yequalsx 10y ago
    I've been teaching college algebra for 20 years. There is nothing arrogant in what I said. The student needs help and it is clear that this is so to anyone with experience teaching such courses. After many years and thousands of students I've developed an intuition on when a student needs help and when I don't need to worry about them. If you had to bet a million dollars, based solely on the solution presented and knowing the course was college algebra, would you put the money on them passing or failing?

Veratyr 10y ago

I appreciate the sentiment but I'd argue that if you're looking for generic methodology, you should ask for a generic solution. The question shouldn't be "solve x^3 = 7", it should be "solve x^3 = y" or "solve x^y = z". If a student produces a correct answer to a numerical question that's correct by drawing a graph or doing a linear/binary search, that seems fine to me.