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# question Watch

1. I'll post the solution now.
2. (Original post by DomStaff)
...
Have you managed to do it? I finally got it by letting:

Comparing the coefficients of the two forms of gives:

The task is to now eliminate from the equations, since you already have in terms of and .
3. This literally took forever to type out in Latex! I have put a spoiler on just in case anyone is still working on the problem. If so, your first hint is to write the roots as k, n-k and n.

There may be a much more elegant way to do this, but my method is certainly correct.

Spoiler:
Show
Let the roots be

Therefore:

We can rearrange the bottom two equations for
gives us:

gives us:

Equating these two gives:

Now, we can eliminate n:

Our previous equation then, after some rearranging becomes:

Multiplying out denominators:

as desired.

4. (Original post by DomStaff)
...
That's a pretty nice method! What did you think of my one?
(They seem pretty similar.)
5. (Original post by Khallil)
That's a pretty nice method! What did you think of my one?

I personally would say that our methods are similar. As a matter of fact, they are identical once you eliminate your a_2 * a_3 (I just tried your method).

In terms of length, I'd say they are relatively similar, yours may be slightly shorter. Still, on paper, mine was only half a page.

Your method is nice. Although, at first I was puzzled as to where your 1=1 came from, but then I realised haha!
6. (Original post by DomStaff)
...
Haha, yep! I tried to make it clear that I didn't get this time around.
7. i have no idea how people do that latex stuff and write solutions in that neat way.

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8. (Original post by physicsmaths)
...
Magic!
9. (Original post by physicsmaths)
i have no idea how people do that latex stuff and write solutions in that neat way.

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Did you do the solution in the end?
10. (Original post by DomStaff)
Did you do the solution in the end?
yh i did it a didfferent way. i used factorisation etc

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11. (Original post by physicsmaths)
yh i did it a didfferent way. i used factorisation etc

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Post your solution so we don't have to guess your solution! I always like to see an alternative method.

You'll be fine with Latex, if you really can't be bothered with it, just write it normally, like (a/b) or b^(3) or whatever.
12. (Original post by DomStaff)
Post your solution so we don't have to guess your solution! I always like to see an alternative method.

You'll be fine with Latex, if you really can't be bothered with it, just write it normally, like (a/b) or b^(3) or whatever.
basically looking at my solution you guys have used standard results where i have had to derive those results. So i have had a hefty amount of working before hand. Then used them results tonprove the rest.

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13. basically
Let the equation be rewritten as an equation with roots alpha, beta , alpha+beta.
hence it can be written as
a(x-alpha)(x-beta)(x-(alpha+beta))
upon expansion prove what alpha and beta must equal i e (-b/a) then its simple from there on. My solution is more along the lones of the other guy. The first solution that was put up.

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14. (Original post by physicsmaths)
basically
Let the equation be rewritten as an equation with roots alpha, beta , alpha+beta.
hence it can be written as
a(x-alpha)(x-beta)(x-(alpha+beta))
upon expansion prove what alpha and beta must equal i e (-b/a) then its simple from there on. My solution is more along the lones of the other guy. The first solution that was put up.

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khalil is his name.

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