Math Is Fun Forum

glenn101

Member

Registered: 2008-04-02

Posts: 108

help with a polynomial identity and quadratic response question

Ok, so far I have managed to do parts a and b of question 1 but I do not understand c.

Here is the question:

1. A train completes a journey of 240km at a constant speed.

a) If it had travelled 4km/h slower, it would have taken two hours more for the journey. Find the actual speed of the train.

I got this for a)

and so x = 24, so the actual speed of the train is 24km/h

b) If it had travelled a km/h slower, and still taken two hours more for the journey of 240km, what would have been the actual speed?(Answer is terms of a.) Discuss the practical possible values of a and also the possible values for the speed of the train.

I got this for b)

and so I got this. I can't do it in latex, I've tried for 30mins and have had no success,hopefully you can understand.

x= a+√a(a+480) or in another way a+√a(a+480)/2
     ----------------
               2

x>0, if a=60, x=120.

Now for part c.

If the train had travelled a km/h slower, and taken a hours more for the journey of 240km, and if a is an integer and the speed is an integer, find the possible values for a and the speed of the train.

My text book I am using does show an answer in the textbook which appears to be displayed as a table of values of a and speed. But I am unsure of how they derived there answer. Please help.

Regards,
Glenn.

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"If your going through hell, keep going."

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: help with a polynomial identity and quadratic response question

Start off by modifying the equation you had in b:

Now solve it in the same way you did for b:

(You can click on that equation to see the LaTeX behind it if you wish. The a=0 is there because both sides were divided by a during the derivation of this result, so the 0 case needs to be treated separately. It's clear the a=0 works when you think about it, because keeping the speed the same will make your journey time stay the same too.)

Now we need to find the values of a that make x an integer. It's not too hard to see that this will happen iff a² + 960 is a square number.

From here, I can't think of a much better way of getting answers than just trying stuff. There will be a finite amount of solutions, because (k+1)² - k² = 2k+1, and so once a gets large enough, it will be impossible for both a² and a²+960 to be square numbers. a² obviously is square, so the consequence is that a²+960 can't be.

Cheating in Excel now, the values of a that work are 1, 8, 14, 22, 34, 43, 56, 77, 118 and 239.

Edit: Just noticed something quite interesting about those values. Each one of them is produced by dividing 240 by one of its factors, then taking that factor away from the result.

eg. 240/1 - 1 = 240 - 1 = 239
     240/2 - 2 = 120 - 2 = 118
     240/15 - 15 = 16 - 15 = 1

Not sure why that happens though.

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Why did the vector cross the road?
It wanted to be normal.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: help with a polynomial identity and quadratic response question

Last edited by Dragonshade (2008-04-25 04:18:22)

glenn101

Member

Registered: 2008-04-02

Posts: 108

Re: help with a polynomial identity and quadratic response question

Thanks so much for your time guys.
That helps alot,
thanks,
Glenn.

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"If your going through hell, keep going."