Chapter 1: Roots of polynomial equations We are working with Cambridge Assessment International Education towards endorsement of this title. b Considering a 3 (a + b + g ) + b 3 (a + b + g ) Since a, b, g all satisfy x 3 + 5 x 2 + 1 = 0, 2 2 + g 3 (a + b + g ), which is equal to S3S1, we see that 5 5 1 1 a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = S3S1 − S4 a 3 + a 2 + = 0, b 3 + b 2 + = 0 and 2 2 2 2 Since S1 = 0, g 3 + 5g 2 + 1 = 0 a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = −S4 = −2 2 2 Possible alternative wording:
Adding these equations gives S3 + 5 S2 + 3 = 0 2 2 ⇒ S3 = − 5 × 25 − 3 = − 137 2 2 8 4 Recall that Sn = a + b + g n
n
We consider a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) + S4, since this is equal to a 3 (a + b + g ) + b 3 (a + b + g ) + g 3 (a + b + g ) = S3S1 Then a 3 ( b + g ) + b 3 (a + g ) + g 3 (a + b ) = S3S1 − S4 = S3 (0) − 2 = −2
n
6 a a + b + g = −a,
ab + ag + bg = b,
FT
8 a Since 2x 3 − x 2 + x − 5 = 0 has roots a, b, g , we see that 2x n+3 − x n+2 + x n+1 − 5x n = 0 must also have roots a, b, g .
abg = −a
So:
So Σ a = −a
2a n+3 − a n+2 + a n+1 − 5a n = 0 2b n+3 − b n+2 + b n+1 − 5b n = 0 2g n+3 − g n+2 + g n+1 − 5g n = 0
Σ ab b =− b Using Σ a1 = gives Σ a1 = −a a Σ abg
b From part a, Σ a = −a and Σ a1 = − ab . If these are equal, then –a = − ab so that b = a 2
Adding these equations gives 2Sn+3 − Sn+2 + Sn+1 − 5Sn = 0
b Let x 3 − 1 x 2 + 1 x − 5 = 0 2 2 2
A
Using Σ a 2 = ( Σ a ) 2 − 2 Σ ab we can calculate that
Then a + b + g = 1 , ab + ag + bg = 1 , abg = 5 2 2 2
Σ a 2 = (−a) 2 − 2 × b = a 2 − 2b Then Σ a 2 = a 2 − 2a 2 = −a 2
1 Σ ab 2 1 1 1 Since S−1 = a + + g = , we have S−1 = = 1 5 5 b Σ abg 2
R
Therefore a 2 + b 2 + g 2 = −a 2, hence we have complex roots.
Then, using 2Sn+3 − Sn+2 + Sn+1 − 5Sn = 0 with n = −2 gives:
Here we have the sum of the squares of the roots being equal to a value less than 0, and this can only mean complex roots.
2S1 − 3 + S−1 − 5S−2 = 0
(
)
D
So S−2 = 1 (2S1 − 3 + S−1) = 1 2 × 1 − 3 + 1 = − 9 2 5 5 5 25
7 a S1 = Σ a = a + b + g = 0
Σ ab = ab + ag + bg = −1
Note that S0 = 1, but since we have 3 roots this becomes 3.
S2 = ( Σ a ) − 2 Σ ab = (0) − 2 × (−1) = 2 2
2
Multiplying x 3 − x + 3 = 0 by x gives q 9 a Rewrite as x 3 + p x 2 + pr = 0, then q S1 = Σ a = a + b + g = − p
x 4 − x 2 + 3x = 0 for which a , b, g must be the roots. So we get: a 4 − a 2 + 3a = 0 b 4 − b 2 + 3b = 0 g 4 − g 2 + 3g = 0
b First note that Σ ab = ab + ag + bg = 0 as there is no term in x. Then using S2 = ( Σ a ) 2 − 2 Σ ab gives
Adding these equations gives S4 − S2 + 3S1 = 0
( )
q S2 = − p
So S4 = S2 − 3S1 = 2 − 3(0) = 2
3
2
−2×0=
q2 p2
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