Variables n and m

harpazo

harpazo

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The straight line y = 3x + m is tangent to the curve y = 10x + n. Express n in terms of m.
 
You apparently have copied the problem wrong since the "curve", which you give as y= 10x+ m, is also a line! And two lines are "tangent" if and only if they are the same line! Since y= 3x+ m and y= 10x+ n have different slopes, they are neither tangent nor parallel.

I am going to assume the curve was actually \(\displaystyle y= 10x^2+ n\). If it was something different, please let us know.

There are two ways to do this. If you have taken Calculus then a line is tangent to a curve at a point if and only if (1) they both pass through that point and (2) the slope of the line is equal to the derivative of the curve at that point. Here we are given the line 3x+ m and the curve \(\displaystyle y= 10x^2+ n\). In order to pass through the same point we must have \(\displaystyle y= 3x+ m= 10x^2+ n\). In order to have the same derivative there, we must have \(\displaystyle y'= 3= 20x\). From the second equation we must have x= 3/20 which makes the first equation 3(3/20)+ m= 9/40+ n so that n- m= 9/20- 9/40= 9/40.

The second way is due to Fermat and predates Calculus. Since you posted this under "Intermediate/Advanced Algebra" perhaps this is appropriate. Two lines or curves cross where the two y values are equal for the same x. That is, \(\displaystyle 3x+ m= 10x^2+ n\), as before, has a root x. If the two lines are tangent there then x is a double root. The quadratic equation, \(\displaystyle 10x^2- 3x+ n-m= 0\) has roots, by the quadratic formula, \(\displaystyle x= \frac{3\pm\sqrt{9- 40(n- m)}}{20}\). x is a double root if and only if \(\displaystyle 9- 40(n-m)= 0\) so \(\displaystyle n- m= 9/40\).
 
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HallsofIvy said:
You apparently have copied the problem wrong since the "curve", which you give as y= 10x+ m, is also a line! And two lines are "tangent" if and only if they are the same line! Since y= 3x+ m and y= 10x+ n have different slopes, they are neither tangent nor parallel.

I am going to assume the curve was actually \(\displaystyle y= 10x^2+ n\). If it was something different, please let us know.

There are two ways to do this. If you have taken Calculus then a line is tangent to a curve at a point if and only if (1) they both pass through that point and (2) the slope of the line is equal to the derivative of the curve at that point. Here we are given the line 3x+ m and the curve \(\displaystyle y= 10x^2+ n\). In order to pass through the same point we must have \(\displaystyle y= 3x+ m= 10x^2+ n\). In order to have the same derivative there, we must have \(\displaystyle y'= 3= 20x\). From the second equation we must have x= 3/20 which makes the first equation 3(3/20)+ m= 9/40+ n so that n- m= 9/20- 9/40= 9/40.

The second way is due to Fermat and predates Calculus. Since you posted this under "Intermediate/Advanced Algebra" perhaps this is appropriate. Two lines or curves cross where the two y values are equal for the same x. That is, \(\displaystyle 3x+ m= 10x^2+ n\), as before, has a root x. If the two lines are tangent there then x is a double root. The quadratic equation, \(\displaystyle 10x^2- 3x+ n-m= 0\) has roots, by the quadratic formula, \(\displaystyle x= \frac{3\pm\sqrt{9- 40(n- m)}}{20}\). x is a double root if and only if \(\displaystyle 9- 40(n-m)= 0\) so \(\displaystyle n- m= 9/40\).
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I copied and pasted the question from another site.
 
It's clearly a bad problem as posted, for several reasons. (It has no solution, for one thing.)

If you were the original poster, I would ask for the source and the context, to find out whether it was copied wrong, or was perhaps meant as a trick question. As it is, all I can do is to ignore it.

Why did you choose to ask it here?
 
Dr.Peterson said:
It's clearly a bad problem as posted, for several reasons. (It has no solution, for one thing.)

If you were the original poster, I would ask for the source and the context, to find out whether it was copied wrong, or was perhaps meant as a trick question. As it is, all I can do is to ignore it.

Why did you choose to ask it here?
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1. I search math sites for questions that are interesting to me.

2. I copied and pasted here. I did not TYPE the question here.

3. Why do I ask any questions here?

A. This is a math site.
B. I am revisiting math learned more than 20 years ago.
C. I love math.
 
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I asked a specific question. It was not offensive. It was not a challenge. I wasn't saying you shouldn't ask questions, only that more information would be helpful.

Did you post the question here because you had the same concerns about it that we have expressed? Or because an answer was given on the other site that you disagreed with? Or because for some reason you couldn't ask about it there? (I have no idea what kind of site it came from, and a search didn't locate it.)

The specific reason you shared it with us can be relevant. In any case, I'm not sure there's anything more to be said, if you have nothing more to add.
 
Harpazo, you need to follow the forum's guidelines and stop posting exercises with no work shown or no questions asked. You're wasting time, again, because you've given us no clue about what you need

In particular, you need to include a statement in your OPs informing the membership from where the question comes. Some of us have no interest in participating with exercises from those poorly-sourced web sites you continue to visit.

:(
 
Dr.Peterson said:
I asked a specific question. It was not offensive. It was not a challenge. I wasn't saying you shouldn't ask questions, only that more information would be helpful.

Did you post the question here because you had the same concerns about it that we have expressed? Or because an answer was given on the other site that you disagreed with? Or because for some reason you couldn't ask about it there? (I have no idea what kind of site it came from, and a search didn't locate it.)

The specific reason you shared it with us can be relevant. In any case, I'm not sure there's anything more to be said, if you have nothing more to add.
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I posted the questions because it is interesting to me. I also could not make sense as typed.
 
Interesting ... I searched for the exact wording you entered, because you explicitly said "I copied and pasted here. I did not TYPE the question here." (And I hadn't mentioned typing.) But it turns out that you did modify the question, from

"Given that the straight line y = 3x + m is a tangent to the curve y = x + 9x + n. Express n in terms of m."​

to

"The straight line y = 3x + m is tangent to the curve y = 10x + n. Express n in terms of m."​

I hesitate to call this lying; but it did cause me not to be able to find the source, and also changed what my answer would be if you had asked whether the question made sense. I would have said that it was obviously a typo for "y = x^2 + 9x + n".

Are you still interested in how to answer that question (which requires beginning calculus, even though it's listed under Middle School)?

In any case, I have to say that this is a very poor example of how to use our site.
 
Dr.Peterson said:
… I hesitate to call this lying …
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Speaking for myself only, I'm unable to take at face value anything that harpazo posts, anymore.

Too much fibbing and disingenuousness! His argumentativeness certainly isn't helping, either.

:(
 
Thanks. Dr. Peterson! If the problem is to find m and n such that y= 3x+ m is tangent to the curve \(\displaystyle y= x^2+ 9x+ n\) then we don't have to use Calculus. As I said before, y=3x+ m will be tangent to the parabola \(\displaystyle y= x^2+ 9x+ n\) if the quadratic equation \(\displaystyle y= 3x+ m= x^2+ 9x+ n\) or \(\displaystyle x^2+6x+ n- m=0\) has a double root. That is true if the "discriminant", \(\displaystyle 36- 4(n- m)\) is 0, So \(\displaystyle n- m= 9\).
 
HallsofIvy said:
Thanks. Dr. Peterson! If the problem is to find m and n such that y= 3x+ m is tangent to the curve \(\displaystyle y= x^2+ 9x+ n\) then we don't have to use Calculus. As I said before, y=3x+ m will be tangent to the parabola \(\displaystyle y= x^2+ 9x+ n\) if the quadratic equation \(\displaystyle y= 3x+ m= x^2+ 9x+ n\) or \(\displaystyle x^2+6x+ n- m=0\) has a double root. That is true if the "discriminant", \(\displaystyle 36- 4(n- m)\) is 0, So \(\displaystyle n- m= 9\).
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True -- this is a much nicer problem, and more likely to have been intended to be solved by such a method.

Of course, if the original poster had written to us (with the correct problem), I would first have asked what he has been learning, and what he understands to be the meaning of "tangent", to get a better sense of what method is appropriate. Context is so important in asking questions, as is stating the exact problem.

Another method, besides the discriminant, is completing the square: n-m has to be 9 in order for the LHS to be a perfect square and have one zero.
 
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