Polynomials question (1 Viewer)

calamebe · Nov 28, 2015

Hey guys there's a question that I keep barely getting wrong, but I don't know what I'm doing wrong.
Question:
http://m.imgur.com/YVgWQ9S

Instead of getting 1/24 in front of the stuff for a2, I get 1/12. Any help would be appreciated.

InteGrand · Nov 28, 2015

calamebe said:
Hey guys there's a question that I keep barely getting wrong, but I don't know what I'm doing wrong.
Question:
http://m.imgur.com/YVgWQ9S

Instead of getting 1/24 in front of the stuff for a2, I get 1/12. Any help would be appreciated.

Maybe you should post your working as it could just be a silly mistake in your algebra.

calamebe · Nov 28, 2015

InteGrand said:
Maybe you should post your working as it could just be a silly mistake in your algebra.

Yeah, I was going to it just took a while to make the working neat enough to read. Here's the working: http://m.imgur.com/d01WDJB

calamebe · Nov 29, 2015

Before I go I'll give this a bump. My working is in the post above.

InteGrand · Nov 29, 2015

calamebe said:
Before I go I'll give this a bump. My working is in the post above.

$$The error in your working is that you double counted the terms (which is why you ended up with double the answer). The values of $a_2$ is going to be the sum of pairwise products of the roots, i.e. Take all the pairs of two distinct roots as products and add them up, where \textsl{the order of the pair is unimportant}. In your working, you accidentally counted things like $1\times 3$ and $3\times 1$ twice in your sum. In fact for each pair $ij$, you included both $ij$ and $ji$, so you counted each term in the \textsl{required} sum twice (so you correctly removed the squares by subtracting off the sum of squares, it's just that what is left over by subtracting this is \textsl{double} the actual answer due to this double counting of pairs). So you ended up calculating $2a_2$. Halving your expression would then give you $a_2$.$

calamebe · Nov 29, 2015

Ah, I see. Thanks heaps.

Polynomials question (1 Viewer)

calamebe

Active Member

InteGrand

Well-Known Member

calamebe

Active Member

calamebe

Active Member

InteGrand

Well-Known Member

calamebe

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)