Roots and coefficients: symmetric functions

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This section opens Paper 1 (Further Pure Mathematics 1), the compulsory AS Level paper.

A polynomial equation hides its roots, but its coefficients already tell you their sum, their product and other combinations.

You can find many expressions in the roots without ever solving the equation.

The syllabus asks you to recall and use the relations between the roots and coefficients of polynomial equations, e.g. to evaluate symmetric functions of the roots or to solve problems involving unknown coefficients in equations.

How the marks work

B1: an independent mark for a correct fact or result, such as a relation between roots and coefficients.

M1: a method mark for a correct method applied to this problem. Slips do not lose it. An asterisk, M1*, means later marks depend on it.

A1: an accuracy mark for a correct answer. It needs the M1 before it. DM1 is a method mark that needs the starred M1.

AG: answer given. The result is printed, so every step to it must be shown.

FT: follow through. You keep the mark when you use your own earlier wrong value correctly. CAO: correct answer only.

Why the relations hold

Suppose the cubic $ax^3 + bx^2 + cx + d = 0$ has roots $\alpha$ , $\beta$ and $\gamma$ .

Then the cubic is $a(x - \alpha)(x - \beta)(x - \gamma)$ , and expanding gives

a\left[x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha\beta + \beta\gamma + \gamma\alpha)x - \alpha\beta\gamma\right].

Comparing coefficients with $ax^3 + bx^2 + cx + d$ gives the three relations below.

The same expansion works for any degree, and the signs always alternate: minus, plus, minus, plus.

Equation

Sum of roots

Sum of products in pairs

Sum of products in threes

Product of all roots

$ax^2 + bx + c = 0$

$\alpha + \beta = -\dfrac{b}{a}$

$\alpha\beta = \dfrac{c}{a}$

$ax^3 + bx^2 + cx + d = 0$

$\sum\alpha = -\dfrac{b}{a}$

$\sum\alpha\beta = \dfrac{c}{a}$

$\alpha\beta\gamma = -\dfrac{d}{a}$

$ax^4 + bx^3 + cx^2 + dx + e = 0$

$\sum\alpha = -\dfrac{b}{a}$

$\sum\alpha\beta = \dfrac{c}{a}$

$\sum\alpha\beta\gamma = -\dfrac{d}{a}$

$\alpha\beta\gamma\delta = \dfrac{e}{a}$

Here $\sum\alpha\beta$ means the sum over every pair of different roots: three terms for a cubic, six for a quartic.

The syllabus restricts this work to equations of degree 2, 3 or 4 only.

These relations are not in MF19, so learn them.

The relations hold whether the roots are real or not.

A polynomial with real coefficients has non-real roots in conjugate pairs, so a cubic with real coefficients always has at least one real root.

Symmetric functions

An expression in the roots is symmetric if swapping any two roots leaves it unchanged.

Every symmetric function can be written using the sums and products above.

The ones you will use most often, for a cubic, are:

$\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta$

$\sum\dfrac{1}{\alpha} = \dfrac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \dfrac{\sum\alpha\beta}{\alpha\beta\gamma}$

$\sum\dfrac{1}{\alpha\beta} = \dfrac{\sum\alpha}{\alpha\beta\gamma}$ , and so $\sum\dfrac{1}{\alpha^2} = \left(\sum\dfrac{1}{\alpha}\right)^2 - 2\sum\dfrac{1}{\alpha\beta}$

$\sum\alpha^2\beta$ (all six terms such as $\alpha^2\beta$ and $\alpha\beta^2$ ) $= \sum\alpha\sum\alpha\beta - 3\alpha\beta\gamma$

$(1 + \alpha)(1 + \beta)(1 + \gamma) = 1 + \sum\alpha + \sum\alpha\beta + \alpha\beta\gamma$

For a quartic, the first identity is the same, and $\sum\dfrac{1}{\alpha} = \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta}$ .

Check any new identity by expanding, not by memory.

Higher powers such as $\sum\alpha^3$ have a better route, which the next section teaches.

A useful consequence: if the coefficients are real and $\sum\alpha^2 < 0$ , the roots cannot all be real, because the squares of real numbers are never negative.

Key terms

Symmetric function of the roots: an expression in the roots that is unchanged when any two roots are swapped.

$\sum\alpha\beta$ : the sum of the products of the roots taken two at a time, over every pair of different roots.

Conjugate pair: two non-real roots $p + q\mathrm{i}$ and $p - q\mathrm{i}$ ; they occur together whenever the coefficients are real.

Worked example 1: symmetric functions with negative coefficients

The cubic equation $2x^3 - 3x^2 + 4x + 5 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ .

Find the values of $\sum\alpha^2$ , $\sum\dfrac{1}{\alpha}$ and $\sum\alpha^2\beta$ , and deduce that the equation has exactly one real root.

1. $\sum\alpha = -\dfrac{(-3)}{2} = \dfrac{3}{2}$ , $\sum\alpha\beta = \dfrac{4}{2} = 2$ , $\alpha\beta\gamma = -\dfrac{5}{2}$ . B1

2. $\sum\alpha^2 = \left(\dfrac{3}{2}\right)^2 - 2(2)$ M1

3. $= \dfrac{9}{4} - 4 = -\dfrac{7}{4}$ A1

4. $\sum\dfrac{1}{\alpha} = \dfrac{\sum\alpha\beta}{\alpha\beta\gamma} = \dfrac{2}{\left(-\frac{5}{2}\right)} = -\dfrac{4}{5}$ M1 A1

5. $\sum\alpha^2\beta = \left(\dfrac{3}{2}\right)(2) - 3\left(-\dfrac{5}{2}\right) = 3 + \dfrac{15}{2} = \dfrac{21}{2}$ M1 A1

6. The coefficients are real, so the cubic has at least one real root. If all three were real, $\sum\alpha^2$ would be at least 0. It is $-\dfrac{7}{4}$ , so two roots are a non-real conjugate pair. B1

Answer: $\sum\alpha^2 = -\dfrac{7}{4}$ , $\sum\dfrac{1}{\alpha} = -\dfrac{4}{5}$ , $\sum\alpha^2\beta = \dfrac{21}{2}$ ; exactly one real root.

The brackets round $-3$ and $-\frac{5}{2}$ are what stop the sign slips here.

Worked example 2: a quartic with a missing term

The quartic equation $x^4 - 3x^2 + 2x + 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ , $\delta$ .

Find $\sum\alpha^2$ and $\sum\dfrac{1}{\alpha^2}$ .

1. Write every coefficient, zeros included: $a = 1$ , $b = 0$ , $c = -3$ , $d = 2$ , $e = 6$ .

2. $\sum\alpha = 0$ , $\sum\alpha\beta = -3$ , $\sum\alpha\beta\gamma = -2$ , $\alpha\beta\gamma\delta = 6$ . B1

3. $\sum\alpha^2 = 0^2 - 2(-3) = 6$ M1 A1

4. $\sum\dfrac{1}{\alpha} = \dfrac{\sum\alpha\beta\gamma}{\alpha\beta\gamma\delta} = \dfrac{-2}{6} = -\dfrac{1}{3}$ and $\sum\dfrac{1}{\alpha\beta} = \dfrac{\sum\alpha\beta}{\alpha\beta\gamma\delta} = \dfrac{-3}{6} = -\dfrac{1}{2}$ M1

5. $\sum\dfrac{1}{\alpha^2} = \left(-\dfrac{1}{3}\right)^2 - 2\left(-\dfrac{1}{2}\right) = \dfrac{1}{9} + 1 = \dfrac{10}{9}$ A1

Answer: $\sum\alpha^2 = 6$ and $\sum\dfrac{1}{\alpha^2} = \dfrac{10}{9}$ .

Reading $c$ as the coefficient of the $x$ term (because the $x^3$ term is missing) is the slip this layout prevents.

Worked example 3: an unknown coefficient, ending at a given result

The equation $3x^3 - 10x^2 + kx - 6 = 0$ has roots $\alpha$ , $\dfrac{1}{\alpha}$ and $\beta$ , where $k$ is a constant.

(a) Show that $\beta = 2$ . [2]

(b) Find $k$ , and the exact values of $\alpha$ . [4]

1. (a) The product of the roots of $ax^3 + bx^2 + cx + d = 0$ is $-\dfrac{d}{a}$ , so $\alpha \times \dfrac{1}{\alpha} \times \beta = -\dfrac{(-6)}{3}$ M1

2. The left side simplifies to $\beta$ and the right side to $2$ , so $\beta = 2$ . A1 AG

3. (b) $x = 2$ is a root: $3(8) - 10(4) + 2k - 6 = 0$ , so $24 - 40 + 2k - 6 = 0$ M1

4. $2k = 22$ , $k = 11$ A1

5. $3x^3 - 10x^2 + 11x - 6 = (x - 2)(3x^2 - 4x + 3)$ , and $\alpha$ , $\dfrac{1}{\alpha}$ are the roots of $3x^2 - 4x + 3 = 0$ (their product is $\frac{3}{3} = 1$ , as it must be). M1

6. $\alpha = \dfrac{4 \pm \sqrt{16 - 36}}{6} = \dfrac{2 \pm \sqrt{5}\,\mathrm{i}}{3}$ A1

Answer: $k = 11$ ; $\alpha = \dfrac{2 + \sqrt{5}\,\mathrm{i}}{3}$ or $\dfrac{2 - \sqrt{5}\,\mathrm{i}}{3}$ (each is the reciprocal of the other).

Step 1 names the relation and shows the substituted coefficients: that line is what earns the AG mark.

Exam tip

Write the three (or four) relations as a separate first line, with brackets round every negative coefficient, before you use them: that line is often a B1 on its own.

When a result is printed, show the substituted line before it, and name the relation you used ("the product of the roots is $-\frac{d}{a}$ "), because a printed answer reached with a gap loses the A1.

Common mistakes

Treating a missing power as if it were not there. In $x^4 - 3x^2 + 2x + 6$ the $x^3$ coefficient is $0$ , so $\sum\alpha = 0$ ; it does not move $-3$ into that place.

Getting the signs wrong: the product of the roots of a cubic is $-\dfrac{d}{a}$ , but of a quartic it is $+\dfrac{e}{a}$ .

Forgetting to divide by $a$ when the leading coefficient is not 1.

Writing $\sum\alpha^2 = \left(\sum\alpha\right)^2$ . The cross terms $2\sum\alpha\beta$ must be taken away.

Jumping to a given answer such as $\beta = 2$ without the line that shows which coefficients were used.

Check your understanding

1. The cubic equation $x^3 + 4x^2 - 2x + 7 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Find the value of $\alpha^2 + \beta^2 + \gamma^2$ . [2]

2. The quartic equation $2x^4 - x^3 + 3x - 4 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ , $\delta$ .
(a) Find the value of $\sum\dfrac{1}{\alpha}$ . [2]
(b) Find the value of $\sum\alpha^2$ . [2]

3. The cubic equation $x^3 - 2x^2 + 5x + 3 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ .
(a) Show that $\alpha^2 + \beta^2 + \gamma^2 = -6$ . [2]
(b) Deduce the number of real roots of the equation. [1]

4. The cubic equation $x^3 + 3x^2 - x + 2 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ . Find the value of $(1 + \alpha)(1 + \beta)(1 + \gamma)$ . [3]

5. The equation $2x^3 - 9x^2 + kx + 6 = 0$ has roots $\alpha$ , $\beta$ , $\gamma$ , where $k$ is a constant and $\alpha + \beta = 3$ .
(a) Find $\gamma$ and $k$ . [3]
(b) Find the exact values of $\alpha$ and $\beta$ . [2]

6. The quadratic equation $3x^2 - 5x + 1 = 0$ has roots $\alpha$ and $\beta$ . Find the exact value of $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha}$ . [3]

Answers

1. With $\sum\alpha = -4$ and $\sum\alpha\beta = -2$ : $\sum\alpha^2 = (-4)^2 - 2(-2)$ M1 $= 20$ A1

2. (a) $\sum\alpha\beta\gamma = -\dfrac{3}{2}$ , $\alpha\beta\gamma\delta = -\dfrac{4}{2} = -2$ B1; $\sum\dfrac{1}{\alpha} = \dfrac{-\frac{3}{2}}{-2} = \dfrac{3}{4}$ A1
(b) $\sum\alpha = \dfrac{1}{2}$ and, since the $x^2$ term is missing, $\sum\alpha\beta = 0$ B1; $\sum\alpha^2 = \dfrac{1}{4} - 0 = \dfrac{1}{4}$ A1

3. (a) $\sum\alpha = 2$ , $\sum\alpha\beta = 5$ B1; $\sum\alpha^2 = 2^2 - 2(5) = 4 - 10 = -6$ A1 AG
(b) Real coefficients, so at least one real root; if all three were real then $\sum\alpha^2 \geqslant 0$ , which is false. Exactly one real root. B1

4. $\sum\alpha = -3$ , $\sum\alpha\beta = -1$ , $\alpha\beta\gamma = -2$ B1; $(1 + \alpha)(1 + \beta)(1 + \gamma) = 1 + \sum\alpha + \sum\alpha\beta + \alpha\beta\gamma$ M1 $= 1 - 3 - 1 - 2 = -5$ A1. (Check: $x^3 + 3x^2 - x + 2 = (x - \alpha)(x - \beta)(x - \gamma)$ , and at $x = -1$ the left side is $5$ , which equals $-(1 + \alpha)(1 + \beta)(1 + \gamma)$ .)

5. (a) $\sum\alpha = \dfrac{9}{2}$ , so $\gamma = \dfrac{9}{2} - 3 = \dfrac{3}{2}$ B1; substitute $x = \dfrac{3}{2}$ : $2\left(\dfrac{27}{8}\right) - 9\left(\dfrac{9}{4}\right) + \dfrac{3k}{2} + 6 = 0$ M1; $k = 5$ A1
(b) $2x^3 - 9x^2 + 5x + 6 = (2x - 3)(x^2 - 3x - 2)$ M1; $\alpha, \beta = \dfrac{3 \pm \sqrt{17}}{2}$ A1 (their sum is 3, as given).

6. $\alpha + \beta = \dfrac{5}{3}$ , $\alpha\beta = \dfrac{1}{3}$ B1; $\dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{\alpha^2 + \beta^2}{\alpha\beta} = \dfrac{\frac{25}{9} - \frac{2}{3}}{\frac{1}{3}}$ M1 $= \dfrac{19}{3}$ A1

Roots and coefficients: symmetric functions

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