What is forming a new equation?

Solving the equation when you should not. Questions usually say "without solving" or have awkward roots on purpose. Use the symmetric functions and identities rather than the quadratic formula.

What is wrong substitution direction?

For new roots y = α + 1, you substitute x = y - 1 (because the old root x = α corresponds to the new root). Substituting x = y + 1 shifts the wrong way.

The roots of x^2 + 3x - 10 = 0 are α and β. Find α + β and αβ. [2 marks]

Using those values, find α^2 + β^2. [2 marks]

The roots of x^3 + 4x^2 - x + 6 = 0 are α, β, γ. Write down αβγ. [1 mark]

CCEA CCEA AS 2021-style practice: The roots of x^2 - 6x + 4 = 0 are α and β. Without solving the equation, find the values of α^2 + β^2 and 1α + 1β.

From the equation, the sum and product of the roots are α + β = 6 and αβ = 4. For the sum of squares, use the identity α^2 + β^2 = (α + β)^2 - 2αβ: α^2 + β^2 = 6^2 - 2(4) = 36 - 8 = 28. For the sum of reciprocals, combine into a single fraction: 1α + 1β = α + βαβ = 64 = 32. Markers reward the correct sum and product, the identity for the sum of squares, and the reciprocal manipulation, all without finding α and β explicitly.

CCEA CCEA AS 2019-style practice: The roots of x^3 - 2x^2 + 5x - 1 = 0 are α, β and γ. Write down α + β + γ, αβ + βγ + γα and αβγ, and find the equation whose roots are α + 1, β + 1 and γ + 1.

For x^3 - 2x^2 + 5x - 1 = 0 the symmetric functions are sumα = 2, sumαβ = 5 and αβγ = 1 (signs alternate: -(b)/(a), (c)/(a), -(d)/(a)). To shift each root up by 1, substitute x = y - 1 (so a root y = α + 1 corresponds to x = α): (y - 1)^3 - 2(y - 1)^2 + 5(y - 1) - 1 = 0. Expanding: (y^3 - 3y^2 + 3y - 1) - 2(y^2 - 2y + 1) + 5y - 5 - 1 = 0, which gives y^3 - 5y^2 + 12y - 9 = 0. Markers reward the three symmetric functions with correct signs and the substitution x = y - 1 leading to the correct new cubic.

Northern IrelandFurther MathsSyllabus dot point

How are the roots of a polynomial related to its coefficients?

The relationships between the roots and coefficients of quadratic, cubic and quartic equations, and forming new equations whose roots are functions of the original roots.

A CCEA AS Further Maths answer on the relationships between roots and coefficients for quadratics, cubics and quartics, the symmetric functions of the roots, and how to form a new polynomial whose roots are functions of the original roots.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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What this dot point is asking

CCEA wants you to relate the roots of a polynomial to its coefficients without solving the equation. You need the sum and product (and intermediate symmetric functions) of the roots for quadratics, cubics and quartics, the standard identities that build expressions such as $\alpha^2 + \beta^2$ , and the substitution method for forming a new equation whose roots are functions of the originals.

The answer

Quadratics: sum and product of roots

Cubics and quartics

The pattern continues with alternating signs of the coefficient ratios.

Standard symmetric identities

To evaluate expressions in the roots, rewrite them in terms of the sums and products you know.

Forming a new equation

Forming an equation with reciprocal roots

The roots of $2x^2 - 5x + 1 = 0$ are $\alpha$ and $\beta$ . Find the equation whose roots are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ .

step Find the sum and product of the original roots

\alpha + \beta = -\frac{-5}{2} = \frac{5}{2}, \qquad \alpha\beta = \frac{1}{2}.

step Find the sum and product of the new roots

\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/2}{1/2} = 5, \qquad \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/2} = 2.

step Build the new quadratic

Using $y^2 - (\text{sum})y + (\text{product}) = 0$ :

y^2 - 5y + 2 = 0.

This matches the substitution

y = \frac{1}{x}

, that is

x = \frac{1}{y}

, into

2x^2 - 5x + 1 = 0

Examples in context

Example 1. Designing a quadratic to order. An engineer who needs a control system with two decay rates $\alpha$ and $\beta$ writes the characteristic quadratic straight from the required sum and product, $x^2 - (\alpha + \beta)x + \alpha\beta = 0$ , rather than guessing coefficients. The roots-and-coefficients link runs both ways.

Example 2. Checking a numerical solver. If software returns three cubic roots, summing them and comparing with $-\frac{b}{a}$ is a fast sanity check. A mismatch reveals a numerical error before it propagates further.

Try this

Q1. The roots of $x^2 + 3x - 10 = 0$ are $\alpha$ and $\beta$ . Find $\alpha + \beta$ and $\alpha\beta$ . [2 marks]

Cue. $\alpha + \beta = -3$ , $\alpha\beta = -10$ .

Q2. Using those values, find $\alpha^2 + \beta^2$ . [2 marks]

Cue. $(\alpha + \beta)^2 - 2\alpha\beta = 9 - 2(-10) = 29$ .

Q3. The roots of $x^3 + 4x^2 - x + 6 = 0$ are $\alpha, \beta, \gamma$ . Write down $\alpha\beta\gamma$ . [1 mark]

Cue. $\alpha\beta\gamma = -\frac{d}{a} = -6$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA AS 20216 marksThe roots of

x^2 - 6x + 4 = 0

are

\alpha

and

\beta

. Without solving the equation, find the values of

\alpha^2 + \beta^2

and

\dfrac{1}{\alpha} + \dfrac{1}{\beta}

Show worked answer →

From the equation, the sum and product of the roots are $\alpha + \beta = 6$ and $\alpha\beta = 4$ .

For the sum of squares, use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ :

$\alpha^2 + \beta^2 = 6^2 - 2(4) = 36 - 8 = 28.$

For the sum of reciprocals, combine into a single fraction:

$\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{6}{4} = \dfrac{3}{2}.$

Markers reward the correct sum and product, the identity for the sum of squares, and the reciprocal manipulation, all without finding $\alpha$ and $\beta$ explicitly.

CCEA AS 20197 marksThe roots of

x^3 - 2x^2 + 5x - 1 = 0

are

\alpha

\beta

and

\gamma

. Write down

\alpha + \beta + \gamma

\alpha\beta + \beta\gamma + \gamma\alpha

and

\alpha\beta\gamma

, and find the equation whose roots are

\alpha + 1

\beta + 1

and

\gamma + 1

Show worked answer →

For $x^3 - 2x^2 + 5x - 1 = 0$ the symmetric functions are $\sum\alpha = 2$ , $\sum\alpha\beta = 5$ and $\alpha\beta\gamma = 1$ (signs alternate: $-\frac{b}{a}, \frac{c}{a}, -\frac{d}{a}$ ).

To shift each root up by $1$ , substitute $x = y - 1$ (so a root $y = \alpha + 1$ corresponds to $x = \alpha$ ):

$(y - 1)^3 - 2(y - 1)^2 + 5(y - 1) - 1 = 0.$

Expanding: $(y^3 - 3y^2 + 3y - 1) - 2(y^2 - 2y + 1) + 5y - 5 - 1 = 0$ , which gives

$y^3 - 5y^2 + 12y - 9 = 0.$

Markers reward the three symmetric functions with correct signs and the substitution $x = y - 1$ leading to the correct new cubic.

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)