Discussing a parametric equation

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I'd like to apologize for my awful grammar, but I hope someone can help me figure this out
I wrote the exercise down with my pen, and then the pencil is my reasoning for the exercise.
31960160_955944287908495_7245848173267124224_n.jpg
I tried solving the equation for m = -1/2, and it gives me no solutions.
I tried with m = 1, and there I get 1 solution : x = ln 2.
I have no idea why I'm not getting any solutions for m = -1/2, and I don't know what to discuss further.
If someone could please help me find the answers, or at least shove me in the right direction, because I'm lost at this point.

Thank you very much, I sincerely hope you can help me figure this out.
Good day.
 
I'd like to apologize for my awful grammar, but I hope someone can help me figure this out
I wrote the exercise down with my pen, and then the pencil is my reasoning for the exercise.
View attachment 9445
I tried solving the equation for m = -1/2, and it gives me no solutions.
I tried with m = 1, and there I get 1 solution : x = ln 2.
I have no idea why I'm not getting any solutions for m = -1/2, and I don't know what to discuss further.
If someone could please help me find the answers, or at least shove me in the right direction, because I'm lost at this point.

Thank you very much, I sincerely hope you can help me figure this out.
Good day.

You found the values of m for which the equation y^2 - 4my + 2m + 2 = 0 has a solution. But in order for the given equation, e^(2x) - 4me^x + 2m + 2 = 0, to have a solution, y must be positive; e^x can't be negative.

So what you should have been looking for are values of m for which y^2 - 4my + 2m + 2 = 0 has a positive solution.
 
You found the values of m for which the equation y^2 - 4my + 2m + 2 = 0 has a solution. But in order for the given equation, e^(2x) - 4me^x + 2m + 2 = 0, to have a solution, y must be positive; e^x can't be negative.

So what you should have been looking for are values of m for which y^2 - 4my + 2m + 2 = 0 has a positive solution.
yes you are right, that is a rookie mistake.I forgot exponentials are > 0. How can I know if the value I chose for m will give me a NEGATIVE "y"? I understand that many of my solutions will be rejected because y has to be > 0 but I don't understand how I can find the values for y, positive or negative, I thought I was looking for a generalized solution of the equation, using the parameter.
 
yes you are right, that is a rookie mistake.I forgot exponentials are > 0. How can I know if the value I chose for m will give me a NEGATIVE "y"? I understand that many of my solutions will be rejected because y has to be > 0 but I don't understand how I can find the values for y, positive or negative, I thought I was looking for a generalized solution of the equation, using the parameter.

I myself just continued what you started: don't stop with the discriminant, but actually solve for y (using the whole quadratic formula), and then solve the inequality that says y > 0. This turns out to be easier than you might think.

There may well be a more direct way, but I haven't looked further.
 
I myself just continued what you started: don't stop with the discriminant, but actually solve for y (using the whole quadratic formula), and then solve the inequality that says y > 0. This turns out to be easier than you might think.

There may well be a more direct way, but I haven't looked further.
I have done this,

31946618_956010794568511_5311750848563380224_n.jpg
But is this the end of the exercise? I got this question as a preparation for my end year math exam in senior year, and I have 8 hours of math every week, the question might be looking for a more detailed and complex answer? I don't want to sound over confident, i'm just not quite sure what the teacher is looking for by asking this question. I have never come across a parametric equation exercise with only 1 equation. Usually it's 3 equations and I can use "Cramer's method" to solve the equations after discussing parameter m.
 
But is this the end of the exercise? I got this question as a preparation for my end year math exam in senior year, and I have 8 hours of math every week, the question might be looking for a more detailed and complex answer? I don't want to sound over confident, i'm just not quite sure what the teacher is looking for by asking this question. I have never come across a parametric equation exercise with only 1 equation. Usually it's 3 equations and I can use "Cramer's method" to solve the equations after discussing parameter m.

You've only checked two values of m. What is your final answer, and on what grounds?

What I suggested was to solve for y for any m, and determine which values of m give positive solutions. Don't plug in a value for m.

Now, I have no idea what your teacher may expect, because I don't know what you've learned. The way you copied the problem, you were told only to "Solve for x, discuss m". That is rather cryptic, but presumably means at least that you are to state all values of m for which there is a solution, and to show those solutions. Have you had other problems with similar instructions? What was expected then?

Cramer's method is for a system of linear equations. This is entirely different, being non-linear. But I don't know what sort of "parametric equation exercise" you are used to, and whether there is any similarity.
 
You've only checked two values of m. What is your final answer, and on what grounds?

What I suggested was to solve for y for any m, and determine which values of m give positive solutions. Don't plug in a value for m.

Now, I have no idea what your teacher may expect, because I don't know what you've learned. The way you copied the problem, you were told only to "Solve for x, discuss m". That is rather cryptic, but presumably means at least that you are to state all values of m for which there is a solution, and to show those solutions. Have you had other problems with similar instructions? What was expected then?

Cramer's method is for a system of linear equations. This is entirely different, being non-linear. But I don't know what sort of "parametric equation exercise" you are used to, and whether there is any similarity.
Sadly this is the first time I've come across this specific type of exercise.
I know this is wrong, and it's because I squared each side of the inequality when I can know for certain -4m is positive
View attachment 944931948679_956050291231228_2630054705815879680_n.jpg
 
I have never come across a parametric equation exercise with only 1 equation. Usually it's 3 equations and I can use "Cramer's method" to solve the equations after discussing parameter m.

Sadly this is the first time I've come across this specific type of exercise.

So you've see other sorts of "parametric equation exercise", but not this specific type. I was hoping you'd say more about what type you have seen, and how they were worded, because that might help me guess what is intended. But I guess you're saying, at least, that you have no evidence what "discuss m" means. So ask your teacher! Teachers are there to help you learn, not to make things hard.

I know this is wrong, and it's because I squared each side of the inequality when I can know for certain -4m is positive

You've done a nice job, with just a couple points where I might differ with you. One important thing to come away with is that "discuss m" probably needs to include intervals where there is no solution, or two solutions, or only one solution, which we haven't dealt with yet.

In your left column (the + case), you say, "only if m<0". Actually, if m>=0, the inequality is always true, since the RHS is negative. But you solved -8m - 8 > 0 incorrectly; it should be m < -1. Combining with what we already know about when the discriminant is positive, this case gives a positive solution for all m < -1.

In your right column, you're right that there is no solution if m < 0; so this solution also exists for all m >= 1.

So what can you say about when there are 0, 1, or 2 solutions? After you answer that, you might want to use a graphing calculator or a site like desmos.com to graph the two solutions (treating m like x) and compare what you see.

(I haven't worked all this out myself on paper, just following along in my head, so I may be missing some errors; check everything I say!)
 
So you've see other sorts of "parametric equation exercise", but not this specific type. I was hoping you'd say more about what type you have seen, and how they were worded, because that might help me guess what is intended. But I guess you're saying, at least, that you have no evidence what "discuss m" means. So ask your teacher! Teachers are there to help you learn, not to make things hard.



You've done a nice job, with just a couple points where I might differ with you. One important thing to come away with is that "discuss m" probably needs to include intervals where there is no solution, or two solutions, or only one solution, which we haven't dealt with yet.

In your left column (the + case), you say, "only if m<0". Actually, if m>=0, the inequality is always true, since the RHS is negative. But you solved -8m - 8 > 0 incorrectly; it should be m < -1. Combining with what we already know about when the discriminant is positive, this case gives a positive solution for all m < -1.

In your right column, you're right that there is no solution if m < 0; so this solution also exists for all m >= 1.

So what can you say about when there are 0, 1, or 2 solutions? After you answer that, you might want to use a graphing calculator or a site like desmos.com to graph the two solutions (treating m like x) and compare what you see.

(I haven't worked all this out myself on paper, just following along in my head, so I may be missing some errors; check everything I say!)
This is all I could get after thinking about it all day :
31959857_956448987858025_6271382658593325056_n.jpg31947626_956448874524703_3680457144953470976_n.jpg
sorry it's in french, I need to explain this to a classmate on tuesday, but there is still something missing...
I tried with values of m between 0 and 1 and there still are some solutions I don't understand why.
 
This is all I could get after thinking about it all day :

sorry it's in french, I need to explain this to a classmate on tuesday, but there is still something missing...
I tried with values of m between 0 and 1 and there still are some solutions I don't understand why.

The only part I see wrong in your work is that for m<-1/2, in the negative radical case, you made a sign error; that solution never exists for m<-1/2. So for m<-1, there is only one solution, not two. Have you tried checking this by taking, say, m = -2?

I thought we had shown that there can't be any solutions in 0<m<1, because the discriminant is negative. For what values did you get solutions?
 
The only part I see wrong in your work is that for m<-1/2, in the negative radical case, you made a sign error; that solution never exists for m<-1/2. So for m<-1, there is only one solution, not two. Have you tried checking this by taking, say, m = -2?

I thought we had shown that there can't be any solutions in 0<m<1, because the discriminant is negative. For what values did you get solutions?
I tried m = 2/3 to check, and I get S = ln(2+-(sqrt(10))/2)
(2-(sqrt(10))/2 is positive, so it gives 2 solutions

My reasoning was that Delta (b2-4ac) was positive, so I was checking for m<-1/2, and then my calculations showed that m < -1 for solutions
there are indeed no solutions for values of m between -1/2 and -1 ( -1<m<-1/2) because e^x > 0
I checked for m = -2; and I get 1 solution.

woah this is way harder than I expected haha, so much fun! Thanks a lot for helping me out btw
 
I tried m = 2/3 to check, and I get S = ln(2+-(sqrt(10))/2)
(2-(sqrt(10))/2 is positive, so it gives 2 solutions

My reasoning was that Delta (b2-4ac) was positive, so I was checking for m<-1/2, and then my calculations showed that m < -1 for solutions
there are indeed no solutions for values of m between -1/2 and -1 ( -1<m<-1/2) because e^x > 0
I checked for m = -2; and I get 1 solution.

woah this is way harder than I expected haha, so much fun! Thanks a lot for helping me out btw

Something is wrong here, because you showed clearly that the discriminant, 16m^2 - 8m - 8, is negative between -1/2 and 1. I certainly don't get 10 for m = 2/3. Can you show me the details?
 
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