Solutions of two equations

[Qwertyuiop[]]

Hi, I have 2 equations here and it's an mcq. In order to get the number of solutions of these two equations, I just multiplied both sides with the denominator to get rid of the fraction and then solved the quadratic. It's the same equation in the numerator so they have the same set of solutions. But is there another way to find the number of solutions of an equation without explicitly solving it ? What if you get an equation that is not easy to solve? What is the right way to answer these type of questions? Thanks

 

[Dr.Peterson]

Hi, I have 2 equations here and it's an mcq. In order to get the number of solutions of these two equations, I just multiplied both sides with the denominator to get rid of the fraction and then solved the quadratic. It's the same equation in the numerator so they have the same set of solutions. But is there another way to find the number of solutions of an equation without explicitly solving it ? What if you get an equation that is not easy to solve? What is the right way to answer these type of questions? Thanks

View attachment 32896

I think you need to explicitly solve it, because the reason they might not have had the same set of solutions would be if one or both were extraneous, due to the denominator. (You did check for that, right?) Please note that it is not about the number of solutions as you suggest, but the actual set of solutions.

If the (common) numerator were such that the equation were hard to solve, you could just find the values that would make the denominators zero, and check to see if they are zeros of the numerator. (That would, in principle, include complex solutions!)

 

[Steven G]

To add to what Dr Peterson said, suppose that x=7 makes the numerator 0. x=7 might not be a solution if x=7 also makes the denominator zero. That is, 0/0 = 0 is NOT true. So once you find a zero for the numerator you must go back and make sure it does not make the denominator 0. If it does, it is an extraneous root.

 

[Qwertyuiop[]]

I think you need to explicitly solve it, because the reason they might not have had the same set of solutions would be if one or both were extraneous, due to the denominator. (You did check for that, right?) Please note that it is not about the number of solutions as you suggest, but the actual set of solutions.

If the (common) numerator were such that the equation were hard to solve, you could just find the values that would make the denominators zero, and check to see if they are zeros of the numerator. (That would, in principle, include complex solutions!)

oh I thought they were asking for the number of solutions. So what is meant be set of solutions , all the x-values, the domain of the functions ? And then check if both functions have the same range of x-values ?

Also what is meant by "the set of solutions of A) is strictly included in set of solution of B)" ? options b and e.

 

[Dr.Peterson]

oh I thought they were asking for the number of solutions. So what is meant [by] set of solutions , all the x-values, the domain of the functions ? And then check if both functions have the same range of x-values ?

Also what is meant by "the set of solutions of A) is strictly included in set of solution of B)" ? options b and e.

Let's look at what it says:

​

Nothing mentions the mere number of solutions. It talks about specific solutions that are "common" to both equations (that is, solutions of both equations), and about the set of solutions of each equation. The set of solutions of A is {1, 2}; these are the two numbers that make A true. This is also the set of solutions of B, so (d) is correct; the two sets are the same.

If, say, the set of solutions of A were instead {1, 2, 3}, then the set of solutions of B, {1, 2} would be strictly included in the set of solutions of A; that is, the former is a proper subset of the latter. Every element of {1, 2} is an element of {1, 2, 3}, and there are others as well.

Also, this is not about domains or ranges; or even, really, about functions! If you define the left-hand side of each equation as a function of x, then it is about the "zeros" of those functions (values that make the functions equal zero). That is very different from domain or range.

 

[Qwertyuiop[]]

Let's look at what it says:

View attachment 32907​

Nothing mentions the mere number of solutions. It talks about specific solutions that are "common" to both equations (that is, solutions of both equations), and about the set of solutions of each equation. The set of solutions of A is {1, 2}; these are the two numbers that make A true. This is also the set of solutions of B, so (d) is correct; the two sets are the same.

If, say, the set of solutions of A were instead {1, 2, 3}, then the set of solutions of B, {1, 2} would be strictly included in the set of solutions of A; that is, the former is a proper subset of the latter. Every element of {1, 2} is an element of {1, 2, 3}, and there are others as well.

Also, this is not about domains or ranges; or even, really, about functions! If you define the left-hand side of each equation as a function of x, then it is about the "zeros" of those functions (values that make the functions equal zero). That is very different from domain or range.

Thanks for clarifying , i got it

 

[Qwertyuiop[]]

Hi, I am back . I have another question that is very similar to the one we discussed. But in this case, we are just told the function in the numerator is a quadratic. I had it wrong at first ( option b) then when i rechecking that question, i noticed the inequality C will have two extraneous solutions x=1 and x=-1. That makes the denominator 0. SO x=1 and x=-1 has to be excluded from the solution set.
Therefore option b is incorrect. option a) is correct that A and B have same solution set. B is defined for all values of x as denominator can not be 0. So A and B have same set of solutions.
Please let me know if my reasoning is correct or if there is anything i missed. THanks

 

[Dr.Peterson]

Hi, I am back . I have another question that is very similar to the one we discussed. But in this case, we are just told the function in the numerator is a quadratic. I had it wrong at first ( option b) then when i rechecking that question, i noticed the inequality C will have two extraneous solutions x=1 and x=-1. That makes the denominator 0. SO x=1 and x=-1 has to be excluded from the solution set.
Therefore option b is incorrect. option a) is correct that A and B have same solution set. B is defined for all values of x as denominator can not be 0. So A and B have same set of solutions.
Please let me know if my reasoning is correct or if there is anything i missed. THanks

The general idea is correct. But since these are inequalities, not equations, it's not just a matter of discarding two extraneous solutions that arize because the denominator can be zero. (You don't know whether 1 and -1 are in the solution set for A anyway; if these were equations, (c) would be correct.)

Rather, the fact that the denominator is not always positive is the main issue. The entire interval from -1 to 1 behaves differently in C; regardless of the solution of A, the solution of C will be different because any solution of A in [-1,1] will not be a solution of C, and any non-solution of A in (-1,1) will be a solution of C!

But the problem does not bring out those subtleties.

 

[JeffM]

If P(x) > 0, it is positive by definition. And [imath]x^4 + 1[/imath] is > 0 for all x and so is positive by definition everywhere. So wherever P(x) is positive, [imath]P(x) \div (x^4 + 1)[/imath] is positive.

 

[JeffM]

Dr. Peterson posted his excellent answer while I was writing mine. My answer merely explained why choice a is correct and ignored the other options. That was perhaps negligent, but it is the nature of multiple choice answers.

The essential point is that we need think only about those values of x that make P(x) positive. We have no clue what those values of x are. Thus, the only way that the fractions can be positive when P(x) is positive is if the denominator is also positive at the same values of x. But we know that [imath]x^4 - 1[/imath] is positive only if |x| > 1. Not knowing where P(x) is positive means that we cannot know where, if anywhere, [imath]P(x) \div (x^4 - 1)[/imath] is positive.

Moreover, it is a badly worded problem. Expressions A, B, and C MAY HAVE the same solutions sets. The problem should have asked which expressions MUST have the same solution sets.

 

[Qwertyuiop[]]

Dr. Peterson posted his excellent answer while I was writing mine. My answer merely explained why choice a is correct and ignored the other options. That was perhaps negligent, but it is the nature of multiple choice answers.

The essential point is that we need think only about those values of x that make P(x) positive. We have no clue what those values of x are. Thus, the only way that the fractions can be positive when P(x) is positive is if the denominator is also positive at the same values of x. But we know that [imath]x^4 - 1[/imath] is positive only if |x| > 1. Not knowing where P(x) is positive means that we cannot know where, if anywhere, [imath]P(x) \div (x^4 - 1)[/imath] is positive.

Moreover, it is a badly worded problem. Expressions A, B, and C MAY HAVE the same solutions sets. The problem should have asked which expressions MUST have the same solution sets.

hmm yes i agree all those questions i get from uni are badly worded. Maybe they use google translate : P

 

[Dr.Peterson]

Moreover, it is a badly worded problem. Expressions A, B, and C MAY HAVE the same solutions sets. The problem should have asked which expressions MUST have the same solution sets.

As I see it, it is not badly worded, if one understands the mathematical way of thinking; the only trouble is that students need more help in seeing things right than the author and I do!

It asks "which of the following is true". In math, "is true" means "can be proved to be true using only the information available"; that is, it means the same as "must be true". And choice (c) supports that understanding, in reminding us that we can't say something is true if we don't know it, because we lack information required to prove it.

Pedagogically, I would agree that it would be wiser to word the problem in a way that makes that clear to students!

 

[JeffM]

As I see it, it is not badly worded, if one understands the mathematical way of thinking; the only trouble is that students need more help in seeing things right than the author and I do!

It asks "which of the following is true". In math, "is true" means "can be proved to be true using only the information available"; that is, it means the same as "must be true". And choice (c) supports that understanding, in reminding us that we can't say something is true if we don't know it, because we lack information required to prove it.

Pedagogically, I would agree that it would be wiser to word the problem in a way that makes that clear to students!

@Dr.Peterson

I concede the point. Even outside mathematical usage, “is true” never means “may be true.” I agree.

The problem lies in that ”is true” in popular usage means is “true in all cases.” Here, the meaning is “is always true in that set of cases that we are limiting discussion to.” But saying ”is true” in popular usage does not mean “is true logically demonstrable what we know.” It is an absolute statement. Qualifications must be made clear. “What does ‘is’ mean” is a byword for a deceptive answer.

So my view is that no matter how great a mathematician may be, he must, for questions not directed at professional mathematicians, pose them in regular usage or at least disclose that a usage is specialized. People who are not mathematicians should not be expected to know what specialists in algebra mean by a “ring.”

And of course you are correct that we are talking about pedagogical error. Unfortunately, my experience with students has been that students have many problems because teachers of mathematics sometimes forget that students do not know the specialized usages of mathematics. To help students frequently requires the teacher to forget what has become second nature.