9231 P12 - Nov 2025 - Q2 - 8 marks
5889
(a) The cubic \(x^3+bx^2+cx+d=0\) has roots \(\alpha,\beta,\gamma\). Given \(\alpha+\beta+\gamma=2\), \(\alpha^2+\beta^2+\gamma^2=3\), and \(\alpha^4+\beta^4+\gamma^4=5\), find \(b\) and \(c\).
(b) Find the value of \(d\).
Solution
Solution
Checked by expert
Answer: (a) \(b=-2,\quad c=\dfrac12\)
(b) \(d=\dfrac7{16}\)
(a)
For \(x^3+bx^2+cx+d=0\),
\(b=-(\alpha+\beta+\gamma)=-2.\)
Also
\(\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma).\)
Since \(\alpha\beta+\alpha\gamma+\beta\gamma=c\),
\(3=2^2-2c,\)
so
\(c=\frac12.\)
(b)
Let \(S_r=\alpha^r+\beta^r+\gamma^r\). For the cubic, Newton's sums give
\(S_3+bS_2+cS_1+3d=0.\)
Using \(b=-2\), \(c=\frac12\), \(S_1=2\), and \(S_2=3\),
\(S_3-6+1+3d=0,\)
so
\(S_3=5-3d.\)
Next,
\(S_4+bS_3+cS_2+dS_1=0.\)
Substitute \(S_4=5\):
\(5-2(5-3d)+\frac12(3)+2d=0.\)
Thus
\(-\frac72+8d=0,\)
so
\(d=\frac7{16}.\)
Mark scheme Part Marks Expected result Notes (a) B1 \(b=-(\alpha+\beta+\gamma)=-2\) (a) M1 \(3=2^2-2c\) Uses formula for sum of squares. (a) A1 \(c=\dfrac12\) (b) M1 \(S_3+bS_2+cS_1+3d=0\) Uses original equation or formula for sum of cubes. (b) A1FT \(S_3=-3b-2c-3d=5-3d\) Allow follow through from part (a) for their value. (b) M1 \(S_4+bS_3+cS_2+dS_1=0\) Uses original equation. (b) A1FT \(5-2(5-3d)+\frac12(3)+2d=0\), hence \(8d-\frac72=0\) Allow follow through from part (a) and their value of S3. (b) A1 \(d=\dfrac7{16}\) CAO.