Home / A-Level Further Maths 9231 / Roots of polynomial equations / Symmetric functions of roots
Cambridge International AS & A Level · Topic 1.1
A-Level Further Maths 9231 / Topic 1.1 / AS & A Level

Find the sum of the squares without the roots.

You rarely need the roots themselves. From \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\) you can build the sum of squares, the sum of reciprocals, the sum of cubes and more, and a negative sum of squares even proves that some roots are not real.

Topic 1.1All roots of polynomial equations lessons →
The key idea

For \(ax^{3} + bx^{2} + cx + d = 0\) with roots \(\alpha, \beta, \gamma\): \(\alpha^{2} + \beta^{2} + \gamma^{2} = \left(\sum \alpha\right)^{2} - 2\sum \alpha\beta\) and \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\sum \alpha\beta}{\alpha\beta\gamma}\). Writing \(S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}\), every root satisfies the equation, so \(aS_{n+3} + bS_{n+2} + cS_{n+1} + dS_{n} = 0\), with \(S_{0} = 3\).

ax³ + bx² + cx + d = 0 Σα = −b/a Σαβ = c/a αβγ = −d/a S₂ = (Σα)² − 2Σαβ sum of squares Σ 1/α = Σαβ / αβγ sum of reciprocals aS₃ + bS₂ + cS₁ + dS₀ = 0 sum of cubes, S₀ = 3
Fig. 1: the three coefficient relations feed every symmetric function. The sum of squares needs the first two, the sum of reciprocals the last two, and the recurrence uses the coefficients directly
Section 01

Squares and reciprocals.

A symmetric function of the roots is one that does not change when you swap any two roots, such as \(\alpha^{2} + \beta^{2} + \gamma^{2}\). Every symmetric function can be written using \(\sum \alpha\), \(\sum \alpha\beta\) and \(\alpha\beta\gamma\), which you read straight from the coefficients. The trick is always the same: expand something you know and subtract what you do not want.

WantStart fromResult (cubic)
\(\sum \alpha^{2}\)\(\left(\sum \alpha\right)^{2} = \sum \alpha^{2} + 2\sum \alpha\beta\)\(\left(\sum \alpha\right)^{2} - 2\sum \alpha\beta\)
\(\sum \dfrac{1}{\alpha}\)common denominator \(\alpha\beta\gamma\)\(\dfrac{\sum \alpha\beta}{\alpha\beta\gamma}\)
\(\sum \dfrac{1}{\alpha\beta}\)common denominator \(\alpha\beta\gamma\)\(\dfrac{\sum \alpha}{\alpha\beta\gamma}\)
\(\sum \dfrac{1}{\alpha^{2}}\)\(\left(\sum \dfrac{1}{\alpha}\right)^{2}\)\(\left(\sum \dfrac{1}{\alpha}\right)^{2} - 2\sum \dfrac{1}{\alpha\beta}\)

For a quartic the sum of squares formula is the same, with \(\sum \alpha\beta\) now standing for all six pair products. For a quadratic it is \(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\).

Worked example Sum of squares and of reciprocals
The roots of \(x^{3} - 4x^{2} + x + 6 = 0\) are \(\alpha, \beta, \gamma\). Find \(\alpha^{2} + \beta^{2} + \gamma^{2}\) and \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\).
  1. \(a = 1,\ b = -4,\ c = 1,\ d = 6\), so \(\sum \alpha = 4\), \(\sum \alpha\beta = 1\), \(\alpha\beta\gamma = -6\).
  2. \(\sum \alpha^{2} = 4^{2} - 2(1) = 14\).
  3. \(\sum \dfrac{1}{\alpha} = \dfrac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \dfrac{1}{-6}\).
  4. Check: the roots are \(-1, 2, 3\), and \(1 + 4 + 9 = 14\), \(-1 + \frac{1}{2} + \frac{1}{3} = -\frac{1}{6}\).
Answer: \(\alpha^{2} + \beta^{2} + \gamma^{2} = 14\), \(\sum \dfrac{1}{\alpha} = -\dfrac{1}{6}\)
Worked example Sum of reciprocals of squares
The roots of \(2x^{3} + 3x^{2} - x + 4 = 0\) are \(\alpha, \beta, \gamma\). Find \(\dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} + \dfrac{1}{\gamma^{2}}\).
  1. \(\sum \alpha = -\dfrac{3}{2}\), \(\sum \alpha\beta = -\dfrac{1}{2}\), \(\alpha\beta\gamma = -\dfrac{4}{2} = -2\).
  2. \(\sum \dfrac{1}{\alpha} = \dfrac{-\frac{1}{2}}{-2} = \dfrac{1}{4}\) and \(\sum \dfrac{1}{\alpha\beta} = \dfrac{\sum \alpha}{\alpha\beta\gamma} = \dfrac{-\frac{3}{2}}{-2} = \dfrac{3}{4}\).
  3. \(\sum \dfrac{1}{\alpha^{2}} = \left(\dfrac{1}{4}\right)^{2} - 2\left(\dfrac{3}{4}\right) = \dfrac{1}{16} - \dfrac{24}{16}\).
Answer: \(-\dfrac{23}{16}\)
Your turn
The roots of \(x^{2} - 3x + 5 = 0\) are \(\alpha\) and \(\beta\). Find \(\alpha^{2} + \beta^{2}\).
Examiner trap

\(\sum \alpha^{2}\) is NOT \(\left(\sum \alpha\right)^{2}\). Squaring the sum also produces every cross term twice, so you must subtract \(2\sum \alpha\beta\). In the same way \(\sum \dfrac{1}{\alpha} \ne \dfrac{1}{\sum \alpha}\).

Section 02

Higher powers by recurrence.

Write \(S_{n} = \alpha^{n} + \beta^{n} + \gamma^{n}\). Each root satisfies the equation, so \(a\alpha^{3} + b\alpha^{2} + c\alpha + d = 0\), and the same for \(\beta\) and \(\gamma\). Multiply each by its root to the power \(n\) and add the three equations: the result links four consecutive sums.

\[aS_{n+3} + bS_{n+2} + cS_{n+1} + dS_{n} = 0\]
Worked example Sum of cubes
For \(x^{3} - 4x^{2} + x + 6 = 0\) with roots \(\alpha, \beta, \gamma\), find \(S_{3} = \alpha^{3} + \beta^{3} + \gamma^{3}\).
  1. The recurrence with \(n = 0\): \(S_{3} - 4S_{2} + S_{1} + 6S_{0} = 0\).
  2. \(S_{0} = 3\), \(S_{1} = 4\), and \(S_{2} = 14\) from the previous section.
  3. \(S_{3} = 4(14) - 4 - 6(3) = 56 - 4 - 18\).
  4. Check with the roots \(-1, 2, 3\): \(-1 + 8 + 27 = 34\).
Answer: \(S_{3} = 34\)
Worked example Two steps up the ladder
The roots of \(x^{3} + 2x - 5 = 0\) are \(\alpha, \beta, \gamma\). Find \(S_{3}\) and \(S_{4}\).
  1. \(\sum \alpha = 0\) (there is no \(x^{2}\) term), \(\sum \alpha\beta = 2\), so \(S_{1} = 0\) and \(S_{2} = 0^{2} - 2(2) = -4\).
  2. Recurrence: \(S_{n+3} + 2S_{n+1} - 5S_{n} = 0\), so \(S_{n+3} = 5S_{n} - 2S_{n+1}\).
  3. \(n = 0\): \(S_{3} = 5(3) - 2(0) = 15\).
  4. \(n = 1\): \(S_{4} = 5S_{1} - 2S_{2} = 0 + 8 = 8\).
Answer: \(S_{3} = 15\), \(S_{4} = 8\)
Your turn
Use the recurrence with \(n = -1\) for \(x^{3} - 4x^{2} + x + 6 = 0\) to find \(S_{-1}\).
Examiner trap

\(S_{0}\) is the number of roots, not 1 and not 0: each \(\alpha^{0} = 1\), and there are three of them in a cubic. Also keep the coefficient \(c\) of \(x\) in the recurrence even when \(b = 0\); only a missing term drops out.

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

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For the same equation \(x^{3} - 2x^{2} + 3x - 1 = 0\), find \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma}\).

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For a quartic equation with roots \(\alpha, \beta, \gamma, \delta\), what is \(S_{0}\)?

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The roots of \(x^{3} - 2x - 1 = 0\) have \(S_{1} = 0\) and \(S_{2} = 4\). Find \(S_{3}\).

Section 03

Mixed sums and non-real roots.

Sums such as \(\alpha^{2}\beta + \alpha^{2}\gamma + \beta^{2}\alpha + \beta^{2}\gamma + \gamma^{2}\alpha + \gamma^{2}\beta\) (six terms, written \(\sum \alpha^{2}\beta\)) come from multiplying two known sums and removing the extra terms. Expanding \(\left(\sum \alpha\right)\left(\sum \alpha\beta\right)\) gives all six terms of \(\sum \alpha^{2}\beta\) plus \(3\alpha\beta\gamma\).

\[\sum \alpha^{2}\beta = \left(\sum \alpha\right)\left(\sum \alpha\beta\right) - 3\alpha\beta\gamma\]

The sum of squares also gives a classic deduction. If all the roots of an equation with real coefficients were real, each \(\alpha^{2} \ge 0\), so \(\sum \alpha^{2} \ge 0\). Hence if \(\sum \alpha^{2} < 0\), at least one root is not real, and because non-real roots of a real polynomial come in conjugate pairs, at least two are not real.

Worked example A six-term sum
For \(x^{3} - 4x^{2} + x + 6 = 0\) with roots \(\alpha, \beta, \gamma\), find \(\sum \alpha^{2}\beta\).
  1. \(\sum \alpha = 4\), \(\sum \alpha\beta = 1\), \(\alpha\beta\gamma = -6\).
  2. \(\sum \alpha^{2}\beta = (4)(1) - 3(-6) = 4 + 18\).
  3. Check with \(-1, 2, 3\): \(\alpha^{2}(\beta + \gamma) + \ldots = 1(5) + 4(2) + 9(1) = 22\).
Answer: \(\sum \alpha^{2}\beta = 22\)
Worked example Proving roots are not real
Show that the equation \(x^{4} - 2x^{3} + 5x^{2} + x - 3 = 0\) has at least two non-real roots.
  1. \(\sum \alpha = -\dfrac{-2}{1} = 2\) and \(\sum \alpha\beta = 5\) (all six pair products).
  2. \(\sum \alpha^{2} = 2^{2} - 2(5) = -6\).
  3. If all four roots were real, \(\sum \alpha^{2}\) would be at least 0, but it is \(-6 < 0\), so at least one root is non-real.
  4. The coefficients are real, so non-real roots occur in conjugate pairs.
Answer: At least two roots are non-real
Your turn
For \(x^{3} - 4x^{2} + x + 6 = 0\), find \((\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)\). (Hint: \(\alpha + \beta = 4 - \gamma\).)
Examiner trap

The deduction only works one way. \(\sum \alpha^{2} < 0\) proves some roots are non-real, but \(\sum \alpha^{2} \ge 0\) proves nothing: \(x^{3} - 1 = 0\) has \(\sum \alpha^{2} = 0\) yet two of its roots are non-real.

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The roots of \(2x^{3} + x^{2} - 4x + 3 = 0\) are \(\alpha, \beta, \gamma\). Find the value of \(\dfrac{1}{\alpha^{2}} + \dfrac{1}{\beta^{2}} + \dfrac{1}{\gamma^{2}}\).

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

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The roots of \(2x^{3} - 6x^{2} + 4x - 1 = 0\) are \(\alpha, \beta, \gamma\). Find \(\sum \alpha^{2}\beta\), the sum of the six terms \(\alpha^{2}\beta + \alpha^{2}\gamma + \ldots\).

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For \(x^{4} + 3x^{3} + 5x^{2} - x + 2 = 0\), which conclusion follows from the value of \(\sum \alpha^{2}\)?

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Symmetric functions built from the coefficients.

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Six original Cambridge-style questions
Sums of squares and reciprocals, reciprocals of squares, \(S_{3}\) and \(S_{4}\) by recurrence, a proof of non-real roots, an unknown coefficient from \(\sum \alpha^{2}\), and six-term sums.