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# another sum of roots question... watch

1. ok.

x^2 - x + k = 0

given one root is double the other, find them, and k.

now...

s + 2s = 3s = -b/a = -(-1)/1 = 1

s = 1/3

but the book says s = -1/3.

i'm probably making a really stupid mistake... but tell me where i'm going wrong
2. Just done the problem, the solution's quote long but I basically got k=2/9 and the roots as x=1/3 and x=2/3. PM me and I'll try and write the solution up for you.
3. (Original post by distortedgav)
Just done the problem, the solution's quote long but I basically got k=2/9 and the roots as x=1/3 and x=2/3. PM me and I'll try and write the solution up for you.
that's what i got. (from what i did above, just use st = c/a and plug s in...). the answer must be a misprint
4. It probably is.
5. ∑s = b/a
Consider a polynomial π(x-sn)
If you take the expansion for the coefficient of xn-1 (or b) you would find that it is simply ∑sn .
Notice that in our polynomial that I defined it will have roots at each sn and it corisponds to the polynomial with the same roots where the a =1.
Simply using an arguement where we multiply the above polynomial by a we find ∑sn = b/a.
Using a similar arguement one can prove the general result that ∑(-s)n = -an + 1 /a1
where sn = the product of n different roots in every combination and an is the coefficient of xorder of the polynomial -n

Come to think of it, why are you doing this after GCSEs? I did sum of polynomial roots in P5, which is an A2 Further Maths Module.
Without results proved in P5 its quite hard to work out.
i'm just working from a random textobook over summer, and this is near the front. at the moment i'm only going through an exercise on sums of roots of quadratics, but the next exercise is about polynomials in general.

uh, I kinda lost all track of what you were talking about around the third line, sorry
6. fine, delete your post
7. (Original post by chewwy)
ok.

x^2 - x + k = 0

given one root is double the other, find them, and k.

now...

s + 2s = 3s = -b/a = -(-1)/1 = 1

s = 1/3

but the book says s = -1/3.

i'm probably making a really stupid mistake... but tell me where i'm going wrong
The book is definately wrong.
You are correct in that 3s =1. I am however somewhat stumped at why you are doing this stuff. This is clearly P5 work, which doesn't exist anymore and was a Further Maths A2 module when I did it. But your sig seems to signify you have just finished GCSEs.
However once you have worked out that alpha = 1/3 and beta = 2/3 you can simply use the formula ∑alphabeta = c/a. In this case since there is only two roots ∑alpha beta = alpha beta = 1/3 x 2/3 which is simply 2/9
8. (Original post by Mehh)
The book is definately wrong.
You are correct in that 3s =1. I am however somewhat stumped at why you are doing this stuff. This is clearly P5 work, which doesn't exist anymore and was a Further Maths A2 module when I did it. But your sig seems to signify you have just finished GCSEs.
However once you have worked out that alpha = 1/3 and beta = 2/3 you can simply use the formula ∑alphabeta = c/a. In this case since there is only two roots ∑alpha beta = alpha beta = 1/3 x 2/3 which is simply 2/9
well. it was p2 & p4 when i did it 6 months ago. maybe he's doing it to get a headstart for his AS's?
9. (Original post by chewwy)
fine, delete your post
Sorry got really confused over my working. Its about the origin of sum of roots = -b/a malarky.

The thing is this.
Concider a polynomial. f(x)
Now it has an order n, thus n number of roots.
We can write it out as a1 (x-alpha)(x-beta)(x-gamma)etc etc.
Or in the same notation as sigma aπ(x-sn)
[π is to multiply as ∑ is to +]

So we concider what happens when we try to find the first coeffiecient of the expanded polynomial. It is simply a. Simple enough to work out right?

We concider the second co-efficient. This time we need to think. The second co-efficient is for xn-1. Now we can think of adding all the ways of getting xn-1 up. Or we can spot that to get xn-1, you simply take n-1 x's multiply them together and then multiply it by one of the -s's. Right? And the sum of all them should make the second coefficient. But then we need to multiply that by a. So we get -a∑sn as the second co-efficient.

We can keep on going and work out a pattern. But at P5 we only did this for cubics, as it got really annoying and tedius.
10. (Original post by Mehh)
The book is definately wrong.
You are correct in that 3s =1. I am however somewhat stumped at why you are doing this stuff. This is clearly P5 work, which doesn't exist anymore and was a Further Maths A2 module when I did it. But your sig seems to signify you have just finished GCSEs.
However once you have worked out that alpha = 1/3 and beta = 2/3 you can simply use the formula ∑alphabeta = c/a. In this case since there is only two roots ∑alpha beta = alpha beta = 1/3 x 2/3 which is simply 2/9
P5 does exist. It' just called FP2 now. FP1 = P4, FP2=P5, FP3 = P6
11. (Original post by El Chueco)
well. it was p2 & p4 when i did it 6 months ago. maybe he's doing it to get a headstart for his AS's?
Really? Cuz when it comes down to it when I say P5 I actually mean P5±1 (you know I am a Physicist). But its been awhile and I really didnt' care for which modules each topic was in. Still doesn't stop the fact P4 is Further Maths. Besides they made Maths easier not harder.

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