Why is "Substituting $x = ky$ instead of $x = \frac{y}{k}$ (or $x = y+k$ instead of $y-k$)" a common mistake in Substitutions to obtain an equation?

Why it happens: Substituting y = f(x) directly without first making x the subject. How to avoid it: Always rearrange to 'x = ' explicitly. New root y = 2x means x = y/2, NOT x = 2y. Source: 9231 Paper 1 Examiner Reports

Why is "Leaving fractional or surd coefficients in the final equation" a common mistake in Substitutions to obtain an equation?

Why it happens: Stopping at y^2/4 - 3y/2 + 5 = 0 without clearing denominators. How to avoid it: Multiply through by the LCM of the denominators to present integer coefficients, as the question requires. Source: 9231 Paper 1 mark schemes

Why is "Forgetting to relabel $y$ back to $x$" a common mistake in Substitutions to obtain an equation?

Why it happens: Treating y as the final variable; the question expects an equation 'in x'. How to avoid it: After simplifying, write the final line in x unless the question explicitly asks for an equation in y. Source: 9231 Paper 1 Examiner Reports

Why is "Errors expanding $(y - k)^3$ or $(y+3)^3$" a common mistake in Substitutions to obtain an equation?

Why it happens: Expanding from memory rather than using the binomial pattern 1, 3, 3, 1. How to avoid it: Use (y k)^3 = y^3 3ky^2 + 3k^2y k^3 carefully; expand one block at a time. Source: 9231 Paper 1 Examiner Reports

Why is "Using $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2$ for squared roots" a common mistake in Substitutions to obtain an equation?

Why it happens: Same dropped-cross-term error as with ^2 — forgetting the -2()() correction. How to avoid it: Memorise ^2^2 = ()^2 - 2()() as a companion to the ^2 identity. Source: 9231 Paper 1 — squared-roots questions

Why is "Substituting $x = \sqrt{y}$ and leaving surds in the equation" a common mistake in Substitutions to obtain an equation?

Why it happens: Trying to substitute directly for squared roots without isolating and squaring. How to avoid it: Isolate the single odd-power term, express x in terms of y, then SQUARE both sides — or use the Vieta assembly method to avoid surds entirely. Source: 9231 Paper 1 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

Substitutions to obtain an equation

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Detailed Study Notes

Detailed notes on Roots of Polynomial Equations for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Substitutions to Obtain an Equation with Related Roots — Cambridge 9231 Further Mathematics Paper 1 (2026-2027 syllabus)

Given a polynomial with roots $\alpha, \beta, \gamma$ , find a NEW polynomial whose roots are a function of them (e.g. $2\alpha$ , $\alpha+3$ , $\frac{1}{\alpha}$ , $\alpha^2$ ). Two reliable methods: direct substitution, and rebuilding with Vieta's formulae.

At a glance

Goal: build a new equation whose roots are $f(\alpha), f(\beta), \ldots$ — without solving the original.
Method 1 (substitution): set $y = f(x)$ , make $x$ the subject, substitute into the original equation.
Method 2 (Vieta): compute the new sum/product of roots, then write the new equation.
Reciprocal roots = reverse the coefficients (a beautiful shortcut).
Always replace $y$ with $x$ at the very end to present a clean equation.

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

1.2a — Use a substitution to obtain a polynomial equation whose roots are a given function of the roots of a given equation.
1.2b — Recognise standard transformations: $k\alpha$ , $\alpha + k$ , $\frac{1}{\alpha}$ , $\alpha^2$ , and combinations.
1.2c — Choose between the substitution method and the Vieta (symmetric-function) method efficiently.

The core idea — work backwards from the new root

If the new root is y = f(x), make x the subject and feed it back into the old equation.

The principle. Suppose the original equation $P(x) = 0$ has roots $\alpha, \beta, \gamma$ , and we want a new equation whose roots are $y = f(\alpha), f(\beta), f(\gamma)$ .

If $y = f(x)$ describes the relationship between an old root $x$ and a new root $y$ , then rearrange to make $x$ the subject: $x = g(y)$ . Now every old root satisfies $P(x) = 0$ , so substituting $x = g(y)$ gives: $P\big(g(y)\big) = 0$ This is an equation in $y$ whose roots are exactly the transformed values. Finally, relabel $y$ as $x$ to present the answer.

Three-step recipe:

Write the connection $y = f(x)$ , then solve for $x$ in terms of $y$ .
Substitute that expression for $x$ into the original polynomial.
Expand, simplify to a polynomial $= 0$ , and replace $y$ with $x$ .

Cambridge tip. The examiner wants a polynomial equation with integer (or simplest) coefficients. Clear all fractions and surds before writing the final line.

New root $y = f(x)$ → make $x$ the subject → substitute.
Substituting into $P(x)=0$ gives the new equation in $y$ .
Relabel $y \to x$ and tidy to integer coefficients.

See the full worked example for substitutions to obtain an equation →

The standard transformations

Scaling, translating, reciprocating and squaring each have a clean substitution.

Learn these four building blocks — most exam questions are one of them or a combination.

Roots multiplied by $k$ (new roots $k\alpha$ ): set $y = kx \Rightarrow x = \dfrac{y}{k}$ .

Roots increased by $k$ (new roots $\alpha + k$ ): set $y = x + k \Rightarrow x = y - k$ .

Reciprocal roots (new roots $\dfrac{1}{\alpha}$ ): set $y = \dfrac{1}{x} \Rightarrow x = \dfrac{1}{y}$ , then multiply through to clear denominators.

Shortcut: the reciprocal-roots equation is the original with its coefficients reversed. For $ax^3 + bx^2 + cx + d = 0$ the reciprocal equation is $dx^3 + cx^2 + bx + a = 0$ .

Squared roots (new roots $\alpha^2$ ): set $y = x^2 \Rightarrow x = \sqrt{y}$ . To avoid surds, isolate the odd-power terms on one side and square both sides, or use the Vieta method instead.

A substitution slides, stretches or inverts every root at once. Here y = x + 2 shifts all roots two units right.

Cambridge tip. For " $\alpha + k$ " type questions the substitution $x = y - k$ is far less error-prone than expanding $(\,y-k\,)^3$ from a guessed binomial — write $x = y - k$ explicitly first.

$k\alpha$ : $x = \frac{y}{k}$ . $\alpha + k$ : $x = y - k$ .
$\frac{1}{\alpha}$ : reverse the coefficients.
$\alpha^2$ : $y = x^2$ , then square to remove surds (or use Vieta).

See the full worked example for substitutions to obtain an equation →

The Vieta (symmetric-function) method

Compute the new sum, pair-sum and product directly, then assemble the equation.

When the transformation is awkward to substitute (especially squared roots), it is often cleaner to compute the new symmetric functions and build the equation.

For a new cubic with roots $\alpha', \beta', \gamma'$ , the target equation is: $x^3 - \left(\sum\alpha'\right)x^2 + \left(\sum\alpha'\beta'\right)x - \alpha'\beta'\gamma' = 0$

(That is the monic form — note the alternating signs, the same Vieta pattern in reverse.)

Example — squared roots. If the original cubic has $\sum\alpha$ , $\sum\alpha\beta$ , $\alpha\beta\gamma$ known, then for new roots $\alpha^2, \beta^2, \gamma^2$ :

$\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$
$\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2(\alpha\beta\gamma)(\sum\alpha)$
$\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2$

Slot these three numbers into the assembly formula and you have the new equation — no surds, no squaring.

Choosing your method:

Transformation	Best method
$k\alpha$ , $\alpha + k$	Direct substitution
$\frac{1}{\alpha}$	Reverse coefficients (fastest)
$\alpha^2$	Vieta assembly (avoids surds)

Cambridge tip. When you assemble via Vieta, state the new $\sum, \sum\text{pair}, \text{product}$ explicitly — these are method marks even before you write the final equation.

Assemble: $x^3 - (\sum\alpha')x^2 + (\sum\alpha'\beta')x - \alpha'\beta'\gamma' = 0$ .
Squared roots: compute $\sum\alpha^2$ , $\sum\alpha^2\beta^2$ , $(\alpha\beta\gamma)^2$ .
Reciprocals: reverse the coefficients.

See the full worked example for substitutions to obtain an equation →

Quick recap

New roots $y = f(x)$ : make $x$ the subject, substitute into $P(x)=0$ , relabel $y\to x$ .
$k\alpha$ : $x = \frac{y}{k}$ ; $\alpha+k$ : $x = y-k$ .
Reciprocal roots: reverse the coefficients.
Squared roots: use the Vieta assembly to dodge surds.
Always present a polynomial with integer / simplest coefficients.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

The three-step substitution recipe ( $y=f(x)\to x=g(y)\to$ substitute).
Standard substitutions for $k\alpha$ , $\alpha+k$ , $\frac1\alpha$ , $\alpha^2$ .
Reciprocal = reverse coefficients shortcut.
Vieta assembly formula: $x^3 - (\sum)x^2 + (\sum\text{pair})x - \text{product}=0$ .
Squared-roots Vieta identities for the new sum, pair-sum and product.

How it’s examined

Topic 1.2 appears every series, almost always as part (b) or (c) of the Topic 1 question (combined 6-10 marks with 1.1). Cambridge frequently asks for an equation with roots $\alpha^2$ or $2\alpha - 1$ . The syllabus note that 'the substitution will not be given for the simplest cases' means you must know $x = y-k$ , $x = y/k$ etc. by heart. Marks are lost through algebraic slips when expanding, and by leaving fractional/surd coefficients in the final equation.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 1); 9231/11, 9231/13 Further Pure Mathematics 1 — past papers and mark schemes. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — Substitutions to obtain an equation

Step-by-step solutions to past-paper-style questions on substitutions to obtain an equation, written exactly the way a tutor would explain them at the board.

Question type:

1Roots multiplied by a constant
Getting startedDirect calculation• scaling, substitution
Question
The equation $x^2 - 3x + 5 = 0$ has roots $\alpha, \beta$ . Find an equation with roots $2\alpha, 2\beta$ .
Step-by-step solution
1. Step 1
  New root $y = 2x$ , so make $x$ the subject: $x = \dfrac{y}{2}$ .
2. Step 2
  Substitute $x = \frac{y}{2}$ into the original equation.
  $\left(\frac{y}{2}\right)^2 - 3\left(\frac{y}{2}\right) + 5 = 0$
3. Step 3
  Simplify and multiply through by $4$ to clear fractions.
  $\frac{y^2}{4} - \frac{3y}{2} + 5 = 0 \;\Rightarrow\; y^2 - 6y + 20 = 0$
4. Step 4
  Relabel $y \to x$ .
Answer
$x^2 - 6x + 20 = 0$ .
Examiner tip
Check with Vieta: original $\alpha+\beta=3$ , so new sum $=6$ ✓; original $\alpha\beta=5$ , new product $=4(5)=20$ ✓.
2Roots shifted by a constant
Getting startedDirect calculation• translation, substitution
Question
The equation $x^2 + 4x - 1 = 0$ has roots $\alpha, \beta$ . Find an equation with roots $\alpha - 2, \beta - 2$ .
Step-by-step solution
1. Step 1
  New root $y = x - 2$ , so $x = y + 2$ .
2. Step 2
  Substitute $x = y + 2$ .
  $(y+2)^2 + 4(y+2) - 1 = 0$
3. Step 3
  Expand and collect.
  $y^2 + 4y + 4 + 4y + 8 - 1 = 0 \;\Rightarrow\; y^2 + 8y + 11 = 0$
Answer
$x^2 + 8x + 11 = 0$ .
Examiner tip
Subtracting 2 from each root shifts left, so the substitution is $x = y + 2$ — the sign is the opposite of what students expect.
3Reciprocal roots — the reverse-coefficients shortcut
Building confidenceDirect calculation• reciprocal, shortcut
Question
The equation $2x^3 - 5x^2 + x + 4 = 0$ has roots $\alpha, \beta, \gamma$ . Find an equation with roots $\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}$ .
Step-by-step solution
1. Step 1
  Method (full): $y = \frac{1}{x} \Rightarrow x = \frac{1}{y}$ . Substitute:
  $2\left(\frac{1}{y}\right)^3 - 5\left(\frac{1}{y}\right)^2 + \frac{1}{y} + 4 = 0$
2. Step 2
  Multiply through by $y^3$ .
  $2 - 5y + y^2 + 4y^3 = 0$
3. Step 3
  Tidy to descending powers: $4y^3 + y^2 - 5y + 2 = 0$ . Notice this is the original with coefficients reversed ( $2,-5,1,4 \to 4,1,-5,2$ ).
Answer
$4x^3 + x^2 - 5x + 2 = 0$ .
Examiner tip
The reverse-coefficients shortcut gets you straight to the answer; the substitution shown is the justification examiners want to see at least once.
4A combined transformation
Building confidenceMulti-step problem• combined, substitution
Question
The equation $x^3 - 2x^2 + 4 = 0$ has roots $\alpha, \beta, \gamma$ . Find an equation with roots $2\alpha + 1, 2\beta + 1, 2\gamma + 1$ .
Step-by-step solution
1. Step 1
  New root $y = 2x + 1 \Rightarrow x = \dfrac{y - 1}{2}$ .
2. Step 2
  Substitute into $x^3 - 2x^2 + 4 = 0$ .
  $\left(\frac{y-1}{2}\right)^3 - 2\left(\frac{y-1}{2}\right)^2 + 4 = 0$
3. Step 3
  Expand: $\frac{(y-1)^3}{8} - \frac{2(y-1)^2}{4} + 4 = 0$ . Multiply by $8$ .
  $(y-1)^3 - 4(y-1)^2 + 32 = 0$
4. Step 4
  Expand fully: $(y^3 - 3y^2 + 3y - 1) - 4(y^2 - 2y + 1) + 32 = 0$ .
  $y^3 - 7y^2 + 11y + 27 = 0$
Answer
$x^3 - 7x^2 + 11x + 27 = 0$ .
Examiner tip
Do the algebra on $(y-1)$ as a single block — expanding $(2x+1)$ in the other direction multiplies the chances of a slip.
5Squared roots via the Vieta assembly method
StretchMulti-step problem• squared-roots, vieta
Question
The equation $x^3 + 3x - 2 = 0$ has roots $\alpha, \beta, \gamma$ . Find an equation with roots $\alpha^2, \beta^2, \gamma^2$ .
Step-by-step solution
1. Step 1
  Original Vieta: $\sum\alpha = 0$ , $\sum\alpha\beta = 3$ , $\alpha\beta\gamma = 2$ .
2. Step 2
  New sum $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta$ .
  $\sum\alpha^2 = 0 - 2(3) = -6$
3. Step 3
  New pair-sum $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2(\alpha\beta\gamma)(\sum\alpha)$ .
  $\sum\alpha^2\beta^2 = 3^2 - 2(2)(0) = 9$
4. Step 4
  New product $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 4$ .
5. Step 5
  Assemble $x^3 - (\sum)x^2 + (\sum\text{pair})x - \text{product} = 0$ .
  $x^3 - (-6)x^2 + 9x - 4 = 0$
Answer
$x^3 + 6x^2 + 9x - 4 = 0$ .
Examiner tip
Vieta assembly dodges the surds that direct substitution ( $x=\sqrt{y}$ ) would create. State all three new symmetric values — each is worth a mark.
6Squared roots by squaring the substitution
StretchShow that / prove• squared-roots, substitution
Question
The equation $x^3 - x^2 + 2 = 0$ has roots $\alpha, \beta, \gamma$ . By using $y = x^2$ , find an equation with roots $\alpha^2, \beta^2, \gamma^2$ .
Step-by-step solution
1. Step 1
  Separate even and odd powers so the square root appears only once. $x^3 - x^2 + 2 = 0 \Rightarrow x^3 = x^2 - 2$ .
2. Step 2
  Write in terms of $y = x^2$ : $x\cdot x^2 = x^2 - 2$ , i.e. $x\,y = y - 2$ , so $x = \dfrac{y-2}{y}$ . (Here $x = \sqrt{y}$ .)
3. Step 3
  Square to remove $x$ : $x^2 = y$ gives $\left(\dfrac{y-2}{y}\right)^2 = y$ .
  $(y-2)^2 = y^3$
4. Step 4
  Expand and rearrange.
  $y^3 - y^2 + 4y - 4 = 0$
Answer
$x^3 - x^2 + 4x - 4 = 0$ .
Examiner tip
Isolating the single odd-power term then squaring is the standard way to remove the surd. Cross-check via Vieta: $\sum\alpha^2 = 1^2 - 0 = 1$ ✓ (coefficient of $x^2$ is $-1$ ).

Model Answers — Substitutions to obtain an equation

High-scoring sample answers for substitutions to obtain an equation on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
4 marks
The roots of $x^2 - 6x + 4 = 0$ are $\alpha$ and $\beta$ . Find a quadratic equation with integer coefficients whose roots are $3\alpha$ and $3\beta$ .
Model answer
Let $y = 3x$ , so $x = \dfrac{y}{3}$ . Substitute into the original: $\left(\frac{y}{3}\right)^2 - 6\left(\frac{y}{3}\right) + 4 = 0$

$\frac{y^2}{9} - 2y + 4 = 0$

Multiply through by $9$ : $y^2 - 18y + 36 = 0$

Relabelling $y \to x$ : $\;x^2 - 18x + 36 = 0$ .
Why this scores
M1 for $x = y/3$ and substitution, M1 for clearing fractions, A1 A1 for correct integer coefficients. Vieta check: new sum $= 3(6) = 18$ , new product $= 9(4) = 36$ .
Question 2
5 marks
The roots of $x^3 - 3x^2 + 2x - 5 = 0$ are $\alpha, \beta, \gamma$ . Find a cubic equation whose roots are $\alpha + 1, \beta + 1, \gamma + 1$ .
Model answer
Let $y = x + 1$ , so $x = y - 1$ . Substitute: $(y-1)^3 - 3(y-1)^2 + 2(y-1) - 5 = 0$

Expand each block: $(y^3 - 3y^2 + 3y - 1) - 3(y^2 - 2y + 1) + (2y - 2) - 5 = 0$

$y^3 - 3y^2 + 3y - 1 - 3y^2 + 6y - 3 + 2y - 2 - 5 = 0$

Collect like terms: $y^3 - 6y^2 + 11y - 11 = 0$

Relabelling: $\;x^3 - 6x^2 + 11x - 11 = 0$ .
Why this scores
M1 for $x = y - 1$ , M1 for substitution, M1 for full expansion, A1 A1 for the simplified cubic. Keeping $(y-1)$ as a block and expanding once each minimises sign errors.
Question 3
4 marks
The roots of $3x^3 + 2x^2 - x + 5 = 0$ are $\alpha, \beta, \gamma$ . Find an equation, with integer coefficients, whose roots are $\dfrac{1}{\alpha}, \dfrac{1}{\beta}, \dfrac{1}{\gamma}$ .
Model answer
Let $y = \dfrac{1}{x}$ , so $x = \dfrac{1}{y}$ . Substitute: $3\left(\frac{1}{y}\right)^3 + 2\left(\frac{1}{y}\right)^2 - \frac{1}{y} + 5 = 0$

Multiply through by $y^3$ : $3 + 2y - y^2 + 5y^3 = 0$

Arranging in descending powers and relabelling $y \to x$ : $5x^3 - x^2 + 2x + 3 = 0$

(This is the original equation $3,2,-1,5$ with coefficients reversed to $5,-1,2,3$ .)
Why this scores
M1 for $x = 1/y$ , M1 for multiplying by $y^3$ , A1 A1 for the correct equation. Either the substitution or an explicit statement of the reverse-coefficient property earns the method marks.
Question 4
6 marks
The roots of $x^3 + 4x^2 - 2x + 1 = 0$ are $\alpha, \beta, \gamma$ . Find a cubic equation whose roots are $2\alpha - 3, 2\beta - 3, 2\gamma - 3$ .
Model answer
Let $y = 2x - 3$ , so $x = \dfrac{y + 3}{2}$ . Substitute into the cubic: $\left(\frac{y+3}{2}\right)^3 + 4\left(\frac{y+3}{2}\right)^2 - 2\left(\frac{y+3}{2}\right) + 1 = 0$

$\frac{(y+3)^3}{8} + \frac{4(y+3)^2}{4} - (y+3) + 1 = 0$

Multiply through by $8$ : $(y+3)^3 + 8(y+3)^2 - 8(y+3) + 8 = 0$

Expand: $(y^3 + 9y^2 + 27y + 27) + 8(y^2 + 6y + 9) - 8y - 24 + 8 = 0$ $y^3 + 17y^2 + 67y + 83 = 0$

Relabelling: $\;x^3 + 17x^2 + 67x + 83 = 0$ .
Why this scores
M1 for $x = \frac{y+3}{2}$ , M1 for clearing the denominator, M1 for full expansion, A1 A1 A1 for each correct coefficient. Treating $(y+3)$ as one block is essential to control the algebra.
Question 5
7 marks
The roots of $x^3 - 2x^2 + 5x - 3 = 0$ are $\alpha, \beta, \gamma$ . Find a cubic equation whose roots are $\alpha^2, \beta^2, \gamma^2$ .
Model answer
Original Vieta: $\sum\alpha = 2$ , $\sum\alpha\beta = 5$ , $\alpha\beta\gamma = 3$ .

New sum of roots: $\sum\alpha^2 = (\sum\alpha)^2 - 2\sum\alpha\beta = 2^2 - 2(5) = 4 - 10 = -6$

New sum of products in pairs: $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2(\alpha\beta\gamma)(\sum\alpha) = 5^2 - 2(3)(2) = 25 - 12 = 13$

New product: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = 3^2 = 9$

Assemble using $x^3 - (\sum)x^2 + (\sum\text{pair})x - \text{product} = 0$ : $x^3 - (-6)x^2 + 13x - 9 = 0 \;\Rightarrow\; x^3 + 6x^2 + 13x - 9 = 0$
Why this scores
M1 A1 for each of the three new symmetric functions, M1 for the assembly, A1 for the final equation. The pair-sum identity $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2(\alpha\beta\gamma)(\sum\alpha)$ is the part most often missed.
Question 6
7 marks
The roots of $x^3 + 2x^2 - 4 = 0$ are $\alpha, \beta, \gamma$ . By using the substitution $y = x^2$ , or otherwise, find a cubic equation in $y$ whose roots are $\alpha^2, \beta^2, \gamma^2$ .
Model answer
Separate the even and odd powers of $x$ . From $x^3 + 2x^2 - 4 = 0$ : $x^3 = -2x^2 + 4$

The left side has the single odd power. Write $x^3 = x \cdot x^2 = x\,y$ and $x^2 = y$ : $x\,y = -2y + 4 \;\Rightarrow\; x = \frac{-2y + 4}{y} = \frac{4 - 2y}{y}$

Now use $x^2 = y$ , i.e. square both sides: $\left(\frac{4 - 2y}{y}\right)^2 = y$

$(4 - 2y)^2 = y^3$

$16 - 16y + 4y^2 = y^3$

$y^3 - 4y^2 + 16y - 16 = 0$
Why this scores
M1 for isolating the odd-power term, M1 for expressing $x$ via $y$ , M1 for squaring to eliminate $x$ , A1 A1 A1 for the correct cubic. Cross-check: $\sum\alpha^2 = 2^2 - 0 = 4$ , matching the $-4y^2$ coefficient.

Key Formulae — Substitutions to obtain an equation

The formulae you need to memorise for substitutions to obtain an equation on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Roots multiplied by $k$
$y = kx \;\Rightarrow\; x = \frac{y}{k}$
$k$
the scaling factor
When to use
New roots are $k\alpha$ . Substitute $x = y/k$ and clear fractions.
Roots increased by $k$
$y = x + k \;\Rightarrow\; x = y - k$
$k$
the shift added to each root
When to use
New roots are $\alpha + k$ . Substitute $x = y - k$ (note the sign flip).
Reciprocal roots (reverse coefficients)
$ax^n + \cdots + k \;\to\; kx^n + \cdots + a$
When to use
New roots are $\frac{1}{\alpha}$ . The new equation is the original with its coefficients written in reverse order.
Assembling a monic cubic from its roots
$x^3 - \left(\sum\alpha'\right)x^2 + \left(\sum\alpha'\beta'\right)x - \alpha'\beta'\gamma' = 0$
$\alpha', \beta', \gamma'$
the new (transformed) roots
When to use
Vieta method — once you know the new sum, pair-sum and product, slot them straight in.
Pair-sum of squared roots
$\sum\alpha^2\beta^2 = \left(\sum\alpha\beta\right)^2 - 2(\alpha\beta\gamma)\left(\sum\alpha\right)$
When to use
Building the squared-roots cubic by the Vieta method — this is the trickiest of the three new symmetric functions.

Key Definitions and Keywords — Substitutions to obtain an equation

Definitions to memorise and the exact keywords mark schemes credit for substitutions to obtain an equation answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Related-roots equation
Examiner keyword
A new polynomial equation whose roots are a specified function $f(\alpha)$ of the roots of a given equation.
Substitution method
Examiner keyword
Set $y = f(x)$ , solve for $x = g(y)$ , substitute into $P(x) = 0$ , then relabel $y$ as $x$ to obtain the new equation.
Reverse-coefficient property
Examiner keyword
The equation with reciprocal roots is the original equation read with its coefficients in reverse order.
Integer / simplest coefficients
Examiner keyword
Examiners require the final equation to have no fractions or surds; multiply through by the lowest common denominator at the end.
Block expansion
Keeping a transformed term such as $(y - k)$ as a single block before expanding, to reduce algebraic errors.

Common Mistakes and Misconceptions — Substitutions to obtain an equation

The traps other students keep falling into on substitutions to obtain an equation questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Substituting $x = ky$ instead of $x = \frac{y}{k}$ (or $x = y+k$ instead of $y-k$ )
9231 Paper 1 Examiner Reports
Why it happens
Substituting $y = f(x)$ directly without first making $x$ the subject.
How to avoid it
Always rearrange to ' $x = \ldots$ ' explicitly. New root $y = 2x$ means $x = \frac{y}{2}$ , NOT $x = 2y$ .
✕Leaving fractional or surd coefficients in the final equation
9231 Paper 1 mark schemes
Why it happens
Stopping at $\frac{y^2}{4} - \frac{3y}{2} + 5 = 0$ without clearing denominators.
How to avoid it
Multiply through by the LCM of the denominators to present integer coefficients, as the question requires.
✕Forgetting to relabel $y$ back to $x$
9231 Paper 1 Examiner Reports
Why it happens
Treating $y$ as the final variable; the question expects an equation 'in $x$ '.
How to avoid it
After simplifying, write the final line in $x$ unless the question explicitly asks for an equation in $y$ .
✕Errors expanding $(y - k)^3$ or $(y+3)^3$
9231 Paper 1 Examiner Reports
Why it happens
Expanding from memory rather than using the binomial pattern $1, 3, 3, 1$ .
How to avoid it
Use $(y \pm k)^3 = y^3 \pm 3ky^2 + 3k^2y \pm k^3$ carefully; expand one block at a time.
✕Using $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2$ for squared roots
9231 Paper 1 — squared-roots questions
Why it happens
Same dropped-cross-term error as with $\sum\alpha^2$ — forgetting the $-2(\alpha\beta\gamma)(\sum\alpha)$ correction.
How to avoid it
Memorise $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2 - 2(\alpha\beta\gamma)(\sum\alpha)$ as a companion to the $\sum\alpha^2$ identity.
✕Substituting $x = \sqrt{y}$ and leaving surds in the equation
9231 Paper 1 Examiner Reports
Why it happens
Trying to substitute directly for squared roots without isolating and squaring.
How to avoid it
Isolate the single odd-power term, express $x$ in terms of $y$ , then SQUARE both sides — or use the Vieta assembly method to avoid surds entirely.

Substitutions to obtain an equation

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Roots multiplied by a constant

2Roots shifted by a constant

3Reciprocal roots — the reverse-coefficients shortcut

4A combined transformation

5Squared roots via the Vieta assembly method

6Squared roots by squaring the substitution

Roots multiplied by kkk

Roots increased by kkk

Reciprocal roots (reverse coefficients)

Assembling a monic cubic from its roots

Pair-sum of squared roots

Related-roots equation

Substitution method

Reverse-coefficient property

Integer / simplest coefficients

Block expansion

✕Substituting x=kyx = kyx=ky instead of x=ykx = \frac{y}{k}x=ky​ (or x=y+kx = y+kx=y+k instead of y−ky-ky−k)

✕Leaving fractional or surd coefficients in the final equation

✕Forgetting to relabel yyy back to xxx

✕Errors expanding (y−k)3(y - k)^3(y−k)3 or (y+3)3(y+3)^3(y+3)3

✕Using ∑α2β2=(∑αβ)2\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2∑α2β2=(∑αβ)2 for squared roots

✕Substituting x=yx = \sqrt{y}x=y​ and leaving surds in the equation

Roots multiplied by $k$

Roots increased by $k$

✕Substituting $x = ky$ instead of $x = \frac{y}{k}$ (or $x = y+k$ instead of $y-k$ )

✕Forgetting to relabel $y$ back to $x$

✕Errors expanding $(y - k)^3$ or $(y+3)^3$

✕Using $\sum\alpha^2\beta^2 = (\sum\alpha\beta)^2$ for squared roots

✕Substituting $x = \sqrt{y}$ and leaving surds in the equation