1.1 - Symmetric expressions in roots | Cambridge A-Level Further Mathematics

The coefficients can reveal more than sums and products of roots. You will use identities and the original equation to find sums of squares, cubes, fourth powers and reciprocals without solving the equation, checking the conditions that make each expression meaningful.

Sums of squares

How can you add the squares of roots you have never found? Start with three familiar numbers: $2,3,5$ . Their squares add to $38$ , whereas their sum squared is $100$ . The difference is twice the pair-product sum: $2(6+10+15)=62$ .

Squaring a sum produces each square once and each pair product twice. For three roots,

(\alpha+\beta+\gamma)^2 =\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\alpha\gamma+\beta\gamma).

The same counting works for two or four roots. Let $S_1$ be the sum of all roots and $S_2$ the sum of all distinct pair products, counting root positions with multiplicity.

Sum of squared roots

For the complete root list of a quadratic, cubic or quartic,

\text{sum of squared roots}=S_1^2-2S_2.

For a quadratic, $S_2$ is simply the product of its two roots.

Not in the formula booklet: learn it.

The identity uses algebra, so the roots need not be distinct or real. When using coefficients, calculate $S_1$ and $S_2$ first to keep their signs visible.

Worked example 1: a cubic square sum

Question. The roots of $x^3-4x^2-x+4=0$ are $\alpha,\beta,\gamma$ . Find $\alpha^2+\beta^2+\gamma^2$ .

Plan. Read the root sum and pair-product sum, then remove the cross terms from the square of the sum.

\begin{aligned} S_1&=4,&S_2&=-1 \\ \alpha^2+\beta^2+\gamma^2&=4^2-2(-1)=18. \end{aligned}

Answer. The sum of squares is $18$ .

Check. Factorising by grouping gives $(x-4)(x^2-1)$ . The roots $4,1,-1$ have squares $16,1,1$ , whose sum is $18$ .

Now increase the degree to four. The identity is unchanged, but $S_2$ contains six pairs rather than three.

Worked example 2: a quartic square sum

Question. Find the sum of the squared roots of $2x^4-8x^3-14x^2+44x+48=0$ .

Plan. Divide the relevant coefficients by $2$ , then use the same identity.

\begin{aligned} S_1&=-\frac{-8}{2}=4 \\ S_2&=\frac{-14}{2}=-7 \\ \text{sum of squares}&=4^2-2(-7)=30. \end{aligned}

Answer. The sum is $30$ .

Check. The polynomial factorises as $2(x+2)(x+1)(x-3)(x-4)$ . Directly adding $4+1+9+16$ gives $30$ .

The first two coefficients after the leading coefficient determine the answer; the lower coefficients still matter to the individual roots.

Watch out

$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2$ omits the cross terms. Subtract $2S_2$ , including its sign: subtracting a negative increases the result.

Use the identity with different coefficients.

Sums of higher powers

Cubing the root sum creates mixed products as well as cubes. Those mixed products can be removed using the pair and triple sums you already know.

Write $P_2$ for the sum of squared roots and $P_3$ for the sum of cubed roots. Multiplying $S_1P_2$ gives $P_3$ plus all the terms such as $\alpha^2\beta$ , where one root is squared and a different root appears once. Call their total $T$ .

Multiplying $S_1S_2$ gives those same terms $T$ , plus three copies of each triple product. A particular triple $\alpha\beta\gamma$ appears by selecting the single root as $\alpha$ , $\beta$ or $\gamma$ . Thus

\begin{aligned} S_1P_2&=P_3+T \\ S_1S_2&=T+3S_3 \\ P_3&=S_1(P_2-S_2)+3S_3. \end{aligned}

Substitute $P_2=S_1^2-2S_2$ to obtain the useful form below.

Sum of cubed roots

For all roots of a cubic or quartic, counted with multiplicity,

P_3=S_1^3-3S_1S_2+3S_3,

where $S_3$ is the product of all roots for a cubic and the sum of four triple products for a quartic.

Not in the formula booklet: learn it.

The derivation explains both the coefficient $3$ and why a quartic requires its triple-product sum, not its four-root product.

For a quadratic there are no triple products. The same expansion therefore gives $P_3=S_1^3-3S_1S_2$ for its two roots.

Worked example 3: removing mixed cube terms

Question. Find the sum of the cubed roots of $x^3-2x^2-5x+6=0$ .

Plan. Read the three symmetric combinations and substitute them into the cube identity.

\begin{aligned} S_1&=2,&S_2&=-5,&S_3&=-6 \\ P_3&=2^3-3(2)(-5)+3(-6) \\ &=8+30-18=20. \end{aligned}

Answer. The cubed roots add to $20$ .

Check. The factorisation $(x+2)(x-1)(x-3)$ gives $(-2)^3+1^3+3^3=-8+1+27=20$ .

For higher powers, use the original equation itself. Every root $r$ in Worked example 3 satisfies $r^3=2r^2+5r-6$ . Multiplying by $r$ gives $r^4=2r^3+5r^2-6r$ . Add that equality for all three roots to express the fourth-power sum in terms of powers already known.

Worked example 4: increasing the requested power

Question. Find the sum of the fourth powers of the roots of $x^3-2x^2-5x+6=0$ .

Plan. Keep the same equation and known cube sum, calculate the square sum, then sum the equality for each root's fourth power.

\begin{aligned} P_2&=2^2-2(-5)=14 \\ P_4&=2P_3+5P_2-6S_1 \\ &=2(20)+5(14)-6(2)=98. \end{aligned}

Answer. The fourth powers add to $98$ .

Check. The factorisation from Worked example 3 gives roots $-2,1,3$ . Their fourth powers add to $16+1+81=98$ , independently confirming the recurrence calculation.

The deliberate change was the requested power, not the roots. For the cube sum itself, adding $r^3=2r^2+5r-6$ gives $P_3=2P_2+5S_1-18$ : the constant $-6$ is counted three times. After multiplying by $r$ , the last term becomes $-6S_1$ , not $-18$ .

For a quartic $ar^4+br^3+cr^2+dr+e=0$ , summing directly over its four roots gives $aP_4+bP_3+cP_2+dS_1+4e=0$ . The factor $4$ counts four copies of the constant. This is the same principle applied at the equation's own degree; no new formula is needed.

Problem-solving

Use a symmetric identity when it matches the expression. When a power matches the degree of the original equation, rearranging that equation for each root offers another route.

Choose the identity for cubes, then the original equation for fourth powers in this check.

Reciprocal sums

A common denominator turns a sum of reciprocals into a ratio of symmetric combinations. The first decision is whether every root is nonzero: if a root is zero, its reciprocal is undefined.

For three nonzero roots,

\frac1\alpha+\frac1\beta+\frac1\gamma =\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma} =\frac{S_2}{S_3}.

For four nonzero roots, each numerator contains the other three roots, so the corresponding answer is $S_3/S_4$ .

Sums for reciprocal roots

For a cubic with nonzero roots, let $T_1$ be the reciprocal sum and $T_2$ the sum of pair products of reciprocals. Then

T_1=\frac{S_2}{S_3},\qquad T_2=\frac{S_1}{S_3}.

For a quartic with nonzero roots,

T_1=\frac{S_3}{S_4},\qquad T_2=\frac{S_2}{S_4}.

In either case the sum of squared reciprocals is $T_1^2-2T_2$ .

Not in the formula booklet: learn it.

For example, in a quartic, $1/(\alpha\beta)=\gamma\delta/S_4$ ; the six reciprocal pairs produce the six complementary pairs. This explains the second quartic relation instead of treating it as a new rule to memorise. For a quadratic the same denominator argument gives $T_1=S_1/S_2$ and $T_2=1/S_2$ , provided $S_2\ne0$ .

Worked example 5: reciprocal sums for a cubic

Question. For the roots of $x^3-4x^2-x+4=0$ , find the sum of their reciprocals and the sum of their squared reciprocals.

Plan. Confirm the product is nonzero, find $T_1,T_2$ , then use the square-sum identity on the reciprocals.

\begin{aligned} S_1&=4,&S_2&=-1,&S_3&=-4\ne0 \\ T_1&=\frac{-1}{-4}=\frac14 \\ T_2&=\frac4{-4}=-1 \\ T_1^2-2T_2&=\frac1{16}+2=\frac{33}{16}. \end{aligned}

Answer. The reciprocal sum is $\tfrac14$ ; the squared reciprocals add to $\tfrac{33}{16}$ .

Check. The roots are $-1,1,4$ . Their reciprocals $-1,1,\tfrac14$ add to $\tfrac14$ , and their squares add to $1+1+\tfrac1{16}=\tfrac{33}{16}$ .

Now change the degree to four. The denominator becomes the four-root product and the numerators are complementary triples or pairs.

Worked example 6: reciprocal sums for a quartic

Question. Find the sum of the reciprocals and the sum of the squared reciprocals of the roots of $x^4-12x^3+47x^2-72x+36=0$ .

Plan. Use the quartic relations, checking $S_4\ne0$ before dividing.

\begin{aligned} (S_1,S_2,S_3,S_4)&=(12,47,72,36) \\ T_1&=72/36=2 \\ T_2&=47/36 \\ T_1^2-2T_2&=4-\frac{94}{36}=\frac{25}{18}. \end{aligned}

Answer. The reciprocal sum is $2$ ; the squared reciprocals add to $\tfrac{25}{18}$ .

Check. The polynomial factorises as $(x-1)(x-2)(x-3)(x-6)$ . Directly, $1+\tfrac12+\tfrac13+\tfrac16=2$ and $1+\tfrac14+\tfrac19+\tfrac1{36}=\tfrac{25}{18}$ .

For $x^3-3x^2+2x=0$ , the constant is zero, so at least one root is zero. A request for the sum of all three reciprocals would therefore have no defined value; do not try to rescue it by cancelling a zero denominator. Our opening question about unknown roots has a useful answer, but only when the requested expression itself exists.

Watch out

The reciprocal of a sum is not the sum of reciprocals. For roots $1,2$ , the two values are $1/3$ and $3/2$ . Use a common denominator, and check it is nonzero.

Calculate the reciprocal quantities for a different quartic.

Quick Recap

Squaring a root sum adds cross terms: remove $2S_2$ .
Cubed roots add to $S_1^3-3S_1S_2+3S_3$ .
Use each root's original equation to reduce a fourth-power sum to known lower powers.
A common denominator expresses reciprocal sums using complementary root products.
Check that every root is nonzero before taking its reciprocal.

You can add powers of roots you have not found because the coefficients already supply the symmetric combinations needed by these identities.

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1.1 - Symmetric expressions in roots

Sums of squares

Sums of higher powers

Reciprocal sums

Quick Recap

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