AS & A Level · Practice questions · Symmetric functions of roots

Build it from three numbers.

Six original Cambridge-style questions on symmetric functions of the roots of cubic and quartic equations, each answered from \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\) without finding a single root.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Read the three sums, then expand and subtract.

Read \(\sum \alpha = -\dfrac{b}{a}\), \(\sum \alpha\beta = \dfrac{c}{a}\), \(\alpha\beta\gamma = -\dfrac{d}{a}\), with signs.
Squares: \(\sum \alpha^{2} = \left(\sum \alpha\right)^{2} - 2\sum \alpha\beta\).
Reciprocals: put over the common denominator \(\alpha\beta\gamma\).
Powers: \(aS_{n+3} + bS_{n+2} + cS_{n+1} + dS_{n} = 0\) with \(S_{0} = 3\).
Non-real roots: \(\sum \alpha^{2} < 0\) with real coefficients means at least two non-real roots.

Recall

[3 marks]

The roots of \(x^{3} - 5x^{2} + 2x + 7 = 0\) are \(\alpha, \beta, \gamma\). Find the values of \(\alpha^{2} + \beta^{2} + \gamma^{2}\) and \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\).

\(\sum \alpha = 5\), \(\sum \alpha\beta = 2\), \(\alpha\beta\gamma = -7\). ✓
\(\alpha^{2} + \beta^{2} + \gamma^{2} = 25 - 2(2) = 21\). ✓
\(\sum \dfrac{1}{\alpha} = \dfrac{\sum \alpha\beta}{\alpha\beta\gamma} = \dfrac{2}{-7} = -\dfrac{2}{7}\). ✓

Calculation

[4 marks]

The roots of \(3x^{3} + 2x^{2} - x + 6 = 0\) are \(\alpha, \beta, \gamma\). Find the value of \(\dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} + \dfrac{1}{\gamma^{2}}\).

\(\sum \alpha = -\dfrac{2}{3}\), \(\sum \alpha\beta = -\dfrac{1}{3}\), \(\alpha\beta\gamma = -2\). ✓
\(\sum \dfrac{1}{\alpha} = \dfrac{-\frac{1}{3}}{-2} = \dfrac{1}{6}\). ✓
\(\sum \dfrac{1}{\alpha\beta} = \dfrac{\sum \alpha}{\alpha\beta\gamma} = \dfrac{-\frac{2}{3}}{-2} = \dfrac{1}{3}\). ✓
\(\sum \dfrac{1}{\alpha^{2}} = \left(\dfrac{1}{6}\right)^{2} - 2\left(\dfrac{1}{3}\right) = \dfrac{1}{36} - \dfrac{24}{36} = -\dfrac{23}{36}\). ✓

Calculation

[4 marks]

The roots of \(x^{3} - 3x^{2} + 4x - 1 = 0\) are \(\alpha, \beta, \gamma\), and \(S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}\).

(a) Find \(S_{2}\).

(b) Use a recurrence to find \(S_{3}\) and \(S_{4}\).

(a) \(\sum \alpha = 3\), \(\sum \alpha\beta = 4\), so \(S_{2} = 9 - 8 = 1\). ✓
(b) Each root satisfies the equation, so \(S_{n+3} - 3S_{n+2} + 4S_{n+1} - S_{n} = 0\), i.e. \(S_{n+3} = 3S_{n+2} - 4S_{n+1} + S_{n}\). ✓
\(n = 0\): \(S_{3} = 3(1) - 4(3) + 3 = -6\). ✓
\(n = 1\): \(S_{4} = 3(-6) - 4(1) + 3 = -19\). ✓

Proof

[3 marks]

Show that the equation \(x^{4} + 2x^{3} + 3x^{2} - 5x + 1 = 0\) has at least two roots that are not real.

\(\sum \alpha = -2\) and \(\sum \alpha\beta = 3\). ✓
\(\sum \alpha^{2} = (-2)^{2} - 2(3) = -2\). ✓
If all the roots were real, \(\sum \alpha^{2} \ge 0\); since \(\sum \alpha^{2} < 0\), at least one root is non-real. ✓
The coefficients are real, so non-real roots occur in conjugate pairs: at least two roots are not real. ✓

Analysis

[5 marks]

The roots of \(x^{3} + kx^{2} + 2x - 3 = 0\) are \(\alpha, \beta, \gamma\), where \(k\) is a constant. It is given that \(\alpha^{2} + \beta^{2} + \gamma^{2} = 5\).

(a) Find the possible values of \(k\).

(b) For the positive value of \(k\), find \(\alpha^{3} + \beta^{3} + \gamma^{3}\).

(a) \(\sum \alpha = -k\), \(\sum \alpha\beta = 2\), so \(k^{2} - 4 = 5\). ✓
\(k^{2} = 9\), so \(k = 3\) or \(k = -3\). ✓
(b) With \(k = 3\): \(S_{1} = -3\), \(S_{2} = 5\), \(S_{0} = 3\). ✓
\(S_{3} + 3S_{2} + 2S_{1} - 3S_{0} = 0\), so \(S_{3} = -3(5) - 2(-3) + 3(3)\). ✓
\(S_{3} = -15 + 6 + 9 = 0\). ✓

Analysis

[5 marks]

The roots of \(x^{3} - 2x^{2} - 5x + 6 = 0\) are \(\alpha, \beta, \gamma\).

(a) Find \(\sum \alpha^{2}\beta\), the sum of the six terms \(\alpha^{2}\beta + \alpha^{2}\gamma + \beta^{2}\alpha + \beta^{2}\gamma + \gamma^{2}\alpha + \gamma^{2}\beta\).

(b) By writing \(\alpha + \beta = 2 - \gamma\), or otherwise, find \((\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)\).

\(\sum \alpha = 2\), \(\sum \alpha\beta = -5\), \(\alpha\beta\gamma = -6\). ✓
(a) \(\sum \alpha^{2}\beta = \left(\sum \alpha\right)\left(\sum \alpha\beta\right) - 3\alpha\beta\gamma = (2)(-5) + 18 = 8\). ✓
(b) \((\alpha + \beta)(\beta + \gamma)(\gamma + \alpha) = (2 - \gamma)(2 - \alpha)(2 - \beta)\). ✓
Since \(x^{3} - 2x^{2} - 5x + 6 = (x - \alpha)(x - \beta)(x - \gamma)\), this is the polynomial at \(x = 2\): \(8 - 8 - 10 + 6 = -4\). ✓
Check: \(\left(\sum \alpha\right)\left(\sum \alpha\beta\right) - \alpha\beta\gamma = -10 + 6 = -4\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.