Hint
The leading coefficient is nonzero. Read minus the next-to-leading coefficient of the monic polynomial.
Worked solution
(a)
\(\sum\alpha=17\)
Answer: \(17\).
Answer: \(17\)
(b)
\(\sum\alpha^2=(\sum\alpha)^2-2\sum_{i<j}\alpha_i\alpha_j=145\)
Answer: \(145\).
Answer: \(145\)
(c)
\(\displaystyle \sum \frac{1}{\alpha}= \frac{e_{n- 1}}{e_{n}}= \frac{17}{72}\)
Answer: \(\displaystyle \frac{17}{72}\).
Answer: \(\displaystyle \frac{17}{72}\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit valid alternative reasoning and consequential working; do not award a reasoning mark for an unexplained final answer.
(a) · 2 marks
- 1 mark: Read minus the next-to-leading coefficient of the monic polynomial.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Keep the minus sign in the root sum.
(b) · 2 marks
- 1 mark: Subtract twice the pairwise-product sum from the squared root sum.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Do not assume roots must be real to apply coefficient relations.
(c) · 2 marks
- 1 mark: Divide the penultimate elementary symmetric sum by the root product.
- 1 mark: Obtain the stated result with the required sign, units and domain.
Alternative methods: Any complete mathematically valid method with the stated assumptions and equivalent result is acceptable.
Common error: Keep the minus sign in the root sum.
Total: 6 marks.