Algebraic Roots and Indices · Study Notes · Edexcel IGCSE 4MA1 Higher

Apply index laws to algebraic expressions. Negative and fractional indices in algebra. Solve exponential equations.

At a glance

Same-base laws apply to algebra too.
$(2x^{3})^{4} = 2^{4} \cdot x^{12} = 16x^{12}$ .
Negative index → reciprocal.
Fractional index = root + power.
Same-base equation: equate indices.
Always show positive indices in final answer.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.1 — Apply index laws to algebraic expressions.
2.1 — Use negative and fractional indices in algebraic contexts.
2.1 — Solve exponential equations by matching bases.

Applying the index laws to algebra

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Treat algebraic expressions like numerical bases. Same rules.

The seven index laws (review from Topic 1) all apply when bases are algebraic:

Law	In algebra
Product	$x^{a} \cdot x^{b} = x^{a+b}$
Quotient	$x^{a} / x^{b} = x^{a-b}$
Power of power	$(x^{a})^{b} = x^{ab}$
Power of product	$(xy)^{n} = x^{n} y^{n}$
Power of quotient	$(x/y)^{n} = x^{n}/y^{n}$
Zero	$x^{0} = 1$ ( $x \ne 0$ )
Negative	$x^{-n} = 1/x^{n}$

Multi-variable expressions: treat each variable separately.

Example: $\dfrac{6x^{5}y^{3}}{2x^{2}y}$

Numbers: $6/2 = 3$ .
$x$ : $x^{5-2} = x^{3}$ .
$y$ : $y^{3-1} = y^{2}$ .
Combine: $3x^{3}y^{2}$ .

Brackets with constants and variables:

$(2x^{3})^{4}$

Constant: $2^{4} = 16$ .
Variable: $(x^{3})^{4} = x^{12}$ .
Combine: $16x^{12}$ .

Worked qualitative. What's $(3xy^{2})^{3}$ ?

Constant: $3^{3} = 27$ .
$x$ : $x^{3}$ .
$y$ : $(y^{2})^{3} = y^{6}$ .
Combine: $27x^{3}y^{6}$ .

Edexcel tip. When a bracket contains constants, ALWAYS apply the power to them too. Forgetting is a classic error.

Index laws transfer to algebra.
Treat each variable separately.
Powers apply to constants in brackets too.
Multiply numbers, multiply indices for like bases.

Fractional and negative indices

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Negative = reciprocal. Fractional = root then power.

Negative indices in algebra:

$x^{-n} = \dfrac{1}{x^{n}}$ .

Examples:

$x^{-2} = \dfrac{1}{x^{2}}$ .
$\dfrac{2}{x^{-3}} = 2x^{3}$ (negative in denominator → positive in numerator).
$(2x)^{-3} = \dfrac{1}{(2x)^{3}} = \dfrac{1}{8x^{3}}$ .

Fractional indices in algebra:

$x^{m/n} = \sqrt[n]{x^{m}}$ .

Denominator = the root to take; numerator = the power to raise to.

Examples:

$x^{1/2} = \sqrt{x}$ .
$x^{2/3} = (x^{2})^{1/3} = \sqrt[3]{x^{2}}$ .
$\left(\dfrac{x^{6}}{16}\right)^{1/2} = \dfrac{x^{3}}{4}$ .

Combined manipulation:

$\dfrac{x^{5} \cdot x^{-2}}{x^{4}}$

Combine numerator: $x^{5-2} = x^{3}$ .
Divide: $x^{3-4} = x^{-1}$ .
Final form: $\dfrac{1}{x}$ .

Worked qualitative. Why prefer $\dfrac{1}{x}$ to $x^{-1}$ ?

Both are mathematically equivalent.
Edexcel mark schemes prefer the positive-index form.
Easier for students to visualise.

Edexcel tip. Always finish with positive indices unless the question explicitly asks otherwise. State the conversion: ' $x^{-1} = \dfrac{1}{x}$ '.

Negative: reciprocal.
Fractional: root + power.
Final answer: positive indices.
Show conversion explicitly.

Solving exponential equations

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Match bases on both sides; then equate indices.

Edexcel-style technique. When you see $a^{x} = b$ :

Try to write BOTH sides with the same base.
Once they have the same base, the indices must be equal.
Solve for $x$ .

Once both sides share a base, the indices must be equal.

Example. $2^{x+1} = 32$ .

$32 = 2^{5}$ .
So $2^{x+1} = 2^{5}$ .
Equate indices: $x + 1 = 5$ , so $x = 4$ .

Example. $9^{x} = 27$ .

$9 = 3^{2}$ , $27 = 3^{3}$ . So $(3^{2})^{x} = 3^{3}$ .
Simplify: $3^{2x} = 3^{3}$ .
Equate: $2x = 3$ , so $x = 1.5$ .

Example. $\left(\dfrac{1}{4}\right)^{x} = 16$ .

$\dfrac{1}{4} = 4^{-1}$ , $16 = 4^{2}$ . So $4^{-x} = 4^{2}$ .
Equate: $-x = 2$ , so $x = -2$ .

When same-base doesn't work, you'd need logs (A-Level material — not 4MA1 Higher). 4MA1 questions always allow same-base matching.

Worked qualitative. Solve $4^{x+1} \cdot 8 = 32^{x-1}$ .

Common base $2$ : $4 = 2^{2}$ , $8 = 2^{3}$ , $32 = 2^{5}$ .
Rewrite: $(2^{2})^{x+1} \cdot 2^{3} = (2^{5})^{x-1}$ .
Simplify: $2^{2(x+1) + 3} = 2^{5(x-1)}$ .
Indices: $2x + 2 + 3 = 5x - 5$ → $2x + 5 = 5x - 5$ → $10 = 3x$ → $x = 10/3$ .

Edexcel tip. Always show how you're matching the base. State 'rewriting with base $2$ ', then equate indices on the LAST line.

Match bases.
Equate indices.
Solve like a normal equation.
Common bases: $2, 3, 4, 5, 8, 9, 27$ .

Quick recap

Index laws apply to algebra.
Powers apply to constants in brackets.
Negative: reciprocal. Fractional: root + power.
Final answer with positive indices.
Same base → equate indices.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$x^{a} \cdot x^{b} = x^{a+b}$
$(xy)^{n} = x^{n} y^{n}$
$x^{-n} = 1/x^{n}$
$x^{m/n} = \sqrt[n]{x^{m}}$
Same base: equate indices.

How it’s examined

Algebraic indices appear Paper 1H + 2H (3-5 marks: simplify, often combined with surds or fractions). Exponential equations appear Paper 2H (4 marks). Examiner reports flag (1) forgetting to raise constants to the power in brackets, (2) leaving negative indices, (3) wrong base matching.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on algebraic roots and indices, written exactly the way a tutor would explain them at the board.

1Simplify using index laws
Foundation• indices
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Question
Simplify $\dfrac{6x^{5}y^{3}}{2x^{2}y}$ .
Step-by-step solution
1. Step 1
  Divide coefficients: $6 / 2 = 3$ .
2. Step 2
  Subtract indices for each variable: $x^{5-2} = x^{3}$ , $y^{3-1} = y^{2}$ .
3. Step 3
  Combine: $3x^{3}y^{2}$ .
Answer
$3x^{3}y^{2}$
Examiner tip
Treat each variable separately. Numbers, $x$ 's, and $y$ 's — divide them independently using their own rules.
2Power of a product
Foundation• indices, bracket
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Question
Simplify $(2x^{3}y^{2})^{4}$ .
Step-by-step solution
1. Step 1
  Power applies to EVERYTHING inside: $(ab)^{n} = a^{n}b^{n}$ .
2. Step 2
  $2^{4} = 16$ . $(x^{3})^{4} = x^{12}$ . $(y^{2})^{4} = y^{8}$ .
3. Step 3
  Combine: $16x^{12}y^{8}$ .
Answer
$16x^{12}y^{8}$
Examiner tip
DON'T forget to raise the constant to the power: $2^{4} = 16$ . Common error: writing $(2x^{3})^{4} = 2x^{12}$ .
3Negative indices in algebra
Higher• Adapted from 4MA1/1H Jan 2024 Q12• indices, negative
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Question
Simplify $\dfrac{x^{5} \times x^{-2}}{x^{4}}$ .
Step-by-step solution
1. Step 1
  Numerator: $x^{5} \times x^{-2} = x^{5-2} = x^{3}$ .
2. Step 2
  Divide: $x^{3} / x^{4} = x^{3-4} = x^{-1}$ .
3. Step 3
  Negative index → reciprocal: $x^{-1} = \dfrac{1}{x}$ .
Answer
$\dfrac{1}{x}$
Examiner tip
Always express the FINAL answer with positive indices unless told otherwise. Mark schemes prefer $1/x$ over $x^{-1}$ .
4Fractional indices in algebra
Higher• indices, fractional
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Question
Simplify $\left(\dfrac{x^{6}}{16}\right)^{1/2}$ .
Step-by-step solution
1. Step 1
  Square root applies to numerator and denominator separately.
  $\left(\frac{x^{6}}{16}\right)^{1/2} = \frac{(x^{6})^{1/2}}{16^{1/2}} = \frac{x^{3}}{4}$
Answer
$\dfrac{x^{3}}{4}$
Examiner tip
Square root of $x^{6}$ is $x^{3}$ (multiply indices: $6 \times 1/2 = 3$ ). Square root of $16$ is $4$ .
5Solve an equation using indices
Higher• indices, equation
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Question
Solve $2^{x+1} = 32$ .
Step-by-step solution
1. Step 1
  Express both sides with the SAME base.
2. Step 2
  $32 = 2^{5}$ . So $2^{x+1} = 2^{5}$ .
3. Step 3
  Same base means the indices are equal: $x + 1 = 5$ , so $x = 4$ .
Answer
$x = 4$
Examiner tip
Once both sides have the same base, equate the indices. This is the Higher-Tier exponential-equation technique.
6A zero power with a coefficient [1 mark]
Getting started• Paper 1H/2H style — 1 mark• zero index, indices
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Question
Simplify $5y^{0}$ , where $y \neq 0$ [1]
Step-by-step solution
1. Step 1
  B1 The power $0$ belongs only to $y$ , and any non-zero value to the power $0$ is $1$ . The $5$ is not raised to any power.
  $5y^{0} = 5 \times y^{0} = 5 \times 1 = 5$
Answer
$5$
Examiner tip
Examiner tip: An index only applies to the letter or bracket directly in front of it. Compare $5y^{0} = 5$ with $(5y)^{0} = 1$ .

Common mistake: Answering $1$ (raising the $5$ to the power $0$ as well) or $0$ (thinking the power $0$ wipes the term out).
7A negative power of an algebraic fraction [2 marks]
Getting started• Paper 1H/2H style — 2 marks• negative index, algebraic fraction
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Question
Simplify $\left(\dfrac{x^{2}}{3}\right)^{-2}$ [2]
Step-by-step solution
1. Step 1
  M1 The negative power turns the fraction upside down.
  $\left(\dfrac{x^{2}}{3}\right)^{-2} = \left(\dfrac{3}{x^{2}}\right)^{2}$
2. Step 2
  A1 Square the numerator and the denominator.
  $= \dfrac{3^{2}}{(x^{2})^{2}} = \dfrac{9}{x^{4}}$
Answer
$\dfrac{9}{x^{4}}$ (or $9x^{-4}$ )
Examiner tip
Examiner tip: Deal with the minus sign first by inverting; what is left is an ordinary square of a fraction.

Common mistake: Writing $\frac{3}{x^{4}}$ — squaring the power of $x$ but forgetting the $3$ . The power applies to the whole numerator and the whole denominator.
8Fractional and negative powers in one quotient [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• fractional index, negative index, index laws
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Question
Simplify $\dfrac{x^{\frac{3}{4}} \times x^{\frac{5}{4}}}{\sqrt{x^{-2}}}$ [3]
Step-by-step solution
1. Step 1
  M1 Numerator: add the indices.
  $x^{\frac{3}{4}} \times x^{\frac{5}{4}} = x^{\frac{8}{4}} = x^{2}$
2. Step 2
  M1 Denominator: a square root halves the index.
  $\sqrt{x^{-2}} = x^{-2 \times \frac{1}{2}} = x^{-1}$
3. Step 3
  A1 Divide by subtracting the indices.
  $\dfrac{x^{2}}{x^{-1}} = x^{2-(-1)} = x^{3}$
Answer
$x^{3}$
Examiner tip
Examiner tip: Simplify the numerator and the denominator separately, then divide. Three short lines earn three marks.

Common mistake: Multiplying the fractional indices to get $x^{\frac{15}{16}}$ . When two powers of $x$ are multiplied, the indices are added.
9Splitting a quotient over a cube root into two powers of x [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• fractional index, negative index, algebraic fraction
▼
Question
Write $\dfrac{6x^{2} - 5}{2\sqrt[3]{x}}$ in the form $ax^{p} + bx^{q}$ , where $a$ , $b$ , $p$ and $q$ are constants. [4]
Step-by-step solution
1. Step 1
  M1 Write the cube root as a power of $x$ .
  $2\sqrt[3]{x} = 2x^{\frac{1}{3}}$
2. Step 2
  M1 Divide each term of the numerator by the whole denominator.
  $\dfrac{6x^{2}}{2x^{\frac{1}{3}}} - \dfrac{5}{2x^{\frac{1}{3}}}$
3. Step 3
  A1 First term: divide the numbers and subtract the indices, $2 - \frac{1}{3} = \frac{5}{3}$ .
  $3x^{\frac{5}{3}}$
4. Step 4
  A1 Second term: the power of $x$ moves up with its sign changed; the $2$ stays in the denominator.
  $-\dfrac{5}{2}x^{-\frac{1}{3}}$
Answer
$3x^{\frac{5}{3}} - \dfrac{5}{2}x^{-\frac{1}{3}}$ ( $a = 3$ , $p = \frac{5}{3}$ , $b = -\frac{5}{2}$ , $q = -\frac{1}{3}$ )
Examiner tip
Examiner tip: Split the fraction into two before touching any index. Both terms must be divided by all of $2x^{\frac{1}{3}}$ — the number and the power of $x$ .

Common mistake: Dividing only the first term, giving $3x^{\frac{5}{3}} - 5$ . The June 2024 examiners' report names this error — splitting a fraction and dividing only one term. A second slip is halving only the $6$ and writing $-5x^{-\frac{1}{3}}$ .
10Two unknown powers from two identities [5 marks]
Stretch• Paper 1H/2H style — 5 marks• index laws, identities, simultaneous equations
▼
Question
$x > 0$

$\dfrac{(2x^{a})^{3}}{x^{b}} = 8x^{7}$ and $x^{a} \times \sqrt{x^{b}} = x^{4}$

Find the value of $a$ and the value of $b$ . Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 First identity: $(2x^{a})^{3} = 8x^{3a}$ , and dividing by $x^{b}$ subtracts $b$ . The $8$ matches on both sides, so compare indices.
  $3a - b = 7$
2. Step 2
  M1 Second identity: $\sqrt{x^{b}} = x^{\frac{b}{2}}$ , and multiplying adds indices.
  $a + \dfrac{b}{2} = 4 \quad\Rightarrow\quad 2a + b = 8$
3. Step 3
  M1 Add the two equations to eliminate $b$ .
  $5a = 15$
4. Step 4
  A1 Solve for $a$ .
  $a = 3$
5. Step 5
  A1 Substitute back.
  $3(3) - b = 7 \quad\Rightarrow\quad b = 2$
Answer
$a = 3$ , $b = 2$
Examiner tip
Examiner tip: Check both identities with your values: $\frac{(2x^{3})^{3}}{x^{2}} = 8x^{7}$ and $x^{3} \times \sqrt{x^{2}} = x^{4}$ .

Common mistake: Writing $2x^{3a}$ for $(2x^{a})^{3}$ in the first identity and then being unable to match the $8$ . Cube the coefficient as well as the power of $x$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
Write $\sqrt[4]{x^{3}}$ as a power of $x$ [1]
Model answer
$\sqrt[4]{x^{3}} = \left(x^{3}\right)^{\frac{1}{4}} = x^{\frac{3}{4}}$

Answer: $x^{\frac{3}{4}}$
Why this scores
Mark scheme:
- B1 for $x^{\frac{3}{4}}$ oe, e.g. $x^{0.75}$
Why this scores full marks: The candidate puts the root as the denominator and the power as the numerator. The usual wrong answer, $x^{\frac{4}{3}}$ , has them the wrong way round.
Question 2
Paper 1H/2H style — 2 marks2 marks
Simplify $3a^{-2} \times 4a^{\frac{1}{2}}$

Give your answer in the form $ka^{n}$ [2]
Model answer
$3a^{-2} \times 4a^{\frac{1}{2}} = 12a^{-2 + \frac{1}{2}} = 12a^{-\frac{3}{2}}$

Answer: $12a^{-\frac{3}{2}}$
Why this scores
Mark scheme:
- B1 for $12$ as the coefficient
- B1 for $a^{-\frac{3}{2}}$ oe
Why this scores full marks: The numbers are multiplied and the indices added, as two separate jobs. Scripts adding the coefficients to get $7$ lose the first mark; $-2 + \frac{1}{2}$ is $-\frac{3}{2}$ , not $-\frac{5}{2}$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
$y^{n} = \dfrac{\sqrt{y}}{y^{3}}$

Find the value of $n$ [2]
Model answer
$\dfrac{\sqrt{y}}{y^{3}} = \dfrac{y^{\frac{1}{2}}}{y^{3}} = y^{\frac{1}{2} - 3} = y^{-\frac{5}{2}}$

Answer: $n = -\dfrac{5}{2}$
Why this scores
Mark scheme:
- M1 for $\sqrt{y} = y^{\frac{1}{2}}$ used, or $\frac{1}{2} - 3$
- A1 for $-\dfrac{5}{2}$ oe
Why this scores full marks: Writing the root as a power first turns the question into a single subtraction of indices. Scripts that divide $\frac{1}{2}$ by $3$ and give $n = \frac{1}{6}$ score $0$ .
Question 4
Paper 1H/2H style — 2 marks2 marks
Sam simplifies $\dfrac{1}{(4x)^{-2}}$ and writes

$\dfrac{1}{(4x)^{-2}} = 4x^{2}$

(a) Explain the mistake Sam has made. [1]

(b) Write down the correct simplification. [1]
Model answer
(a) Sam has squared only the $x$ . The power applies to everything in the bracket, so the $4$ must be squared too.

(b) $\dfrac{1}{(4x)^{-2}} = (4x)^{2} = 16x^{2}$

Answer: (a) the $4$ has not been squared (b) $16x^{2}$
Why this scores
Mark scheme:
- B1 for an explanation that the $4$ should also be squared oe
- B1 for $16x^{2}$ cao
Why this scores full marks: The explanation names the specific error — the coefficient in the bracket was not raised to the power — rather than just saying "wrong index". Sam did handle the reciprocal correctly, and the candidate does not criticise that.
Question 5
Paper 1H/2H style — 3 marks3 marks
$x > 0$ and $y > 0$

Simplify fully $\sqrt{\dfrac{9x^{2}}{y^{6}}} \times \dfrac{2y^{4}}{x^{3}}$ [3]
Model answer
$\sqrt{\dfrac{9x^{2}}{y^{6}}} = \dfrac{3x}{y^{3}}$

$\dfrac{3x}{y^{3}} \times \dfrac{2y^{4}}{x^{3}} = \dfrac{6xy^{4}}{x^{3}y^{3}} = \dfrac{6y}{x^{2}}$

Answer: $\dfrac{6y}{x^{2}}$
Why this scores
Mark scheme:
- M1 for $\dfrac{3x}{y^{3}}$ oe
- M1 for multiplying to a single fraction, e.g. $\dfrac{6xy^{4}}{x^{3}y^{3}}$
- A1 for $\dfrac{6y}{x^{2}}$ oe, e.g. $6x^{-2}y$
Why this scores full marks: The square root is applied to all three factors — $9$ , $x^{2}$ and $y^{6}$ — before anything else. Weaker scripts root the number but halve only one index, the two-out-of-three error the examiners describe.
Question 6
Paper 1H/2H style — 3 marks3 marks
$x > 0$

(a) Factorise fully $x^{\frac{5}{2}} - 9x^{\frac{1}{2}}$ [2]

(b) Hence simplify $\dfrac{x^{\frac{5}{2}} - 9x^{\frac{1}{2}}}{x^{\frac{3}{2}} + 3x^{\frac{1}{2}}}$ [1]
Model answer
(a) $x^{\frac{5}{2}} - 9x^{\frac{1}{2}} = x^{\frac{1}{2}}\left(x^{2} - 9\right) = x^{\frac{1}{2}}(x - 3)(x + 3)$

(b) $\dfrac{x^{\frac{1}{2}}(x - 3)(x + 3)}{x^{\frac{1}{2}}(x + 3)} = x - 3$

Answer: (a) $x^{\frac{1}{2}}(x - 3)(x + 3)$ (b) $x - 3$
Why this scores
Mark scheme:
- M1 for $x^{\frac{1}{2}}\left(x^{2} - 9\right)$ , or for a correct factorisation of $x^{2} - 9$ in a partly factorised expression
- A1 for $x^{\frac{1}{2}}(x - 3)(x + 3)$ oe, e.g. $\sqrt{x}(x - 3)(x + 3)$
- B1 for $x - 3$ cao
Why this scores full marks: The common factor is the lowest power, $x^{\frac{1}{2}}$ ; dividing it out uses $\frac{5}{2} - \frac{1}{2} = 2$ . The candidate then spots the difference of two squares, so the answer is fully factorised. In (b), the denominator is factorised the same way, $x^{\frac{1}{2}}(x + 3)$ , and only whole factors are cancelled.
Question 7
Paper 1H/2H style — 4 marks4 marks
(a) Write $\dfrac{1}{\sqrt[3]{x}}$ as a power of $x$ [1]

(b) Hence solve $\dfrac{128}{\sqrt[3]{x}} = x^{2}$ , where $x > 0$

Show clear algebraic working. [3]
Model answer
(a) $\dfrac{1}{\sqrt[3]{x}} = x^{-\frac{1}{3}}$

(b) $128x^{-\frac{1}{3}} = x^{2}$

$128 = x^{2} \times x^{\frac{1}{3}} = x^{\frac{7}{3}}$

$x = 128^{\frac{3}{7}} = \left(\sqrt[7]{128}\right)^{3} = 2^{3} = 8$

Answer: (a) $x^{-\frac{1}{3}}$ (b) $x = 8$
Why this scores
Mark scheme:
- B1 for $x^{-\frac{1}{3}}$
- M1 for $x^{\frac{7}{3}} = 128$ oe
- M1 for $x = 128^{\frac{3}{7}}$ or $\left(\sqrt[7]{128}\right)^{3}$
- A1 for $x = 8$ cao
Why this scores full marks: "Hence" is used: (a) turns the left-hand side into $128x^{-\frac{1}{3}}$ , and collecting the powers of $x$ gives a single power, $x^{2 + \frac{1}{3}} = x^{\frac{7}{3}}$ , equal to a number. A check, $\frac{128}{\sqrt[3]{8}} = \frac{128}{2} = 64 = 8^{2}$ , confirms it.
Question 8
Paper 1H/2H style — 4 marks4 marks
$x = 9y^{-4}$ , where $y > 0$

Write each of the following in terms of $y$ . Give each answer in its simplest form.

(a) $\sqrt{x}$ [1]

(b) $x^{-\frac{3}{2}}$ [2]

(c) $\dfrac{x^{2}}{3}$ [1]
Model answer
(a) $\sqrt{x} = (9y^{-4})^{\frac{1}{2}} = 3y^{-2}$

(b) $x^{-\frac{3}{2}} = \left(3y^{-2}\right)^{-3} = 3^{-3}y^{6} = \dfrac{y^{6}}{27}$

(c) $\dfrac{x^{2}}{3} = \dfrac{81y^{-8}}{3} = 27y^{-8}$

Answer: (a) $3y^{-2}$ (b) $\dfrac{y^{6}}{27}$ (c) $27y^{-8}$
Why this scores
Mark scheme:
- B1 for $3y^{-2}$ or $\dfrac{3}{y^{2}}$
- M1 for $\left(3y^{-2}\right)^{-3}$ or $27y^{-6}$ or $\dfrac{1}{27}$ seen
- A1 for $\dfrac{y^{6}}{27}$ oe
- B1 for $27y^{-8}$ or $\dfrac{27}{y^{8}}$
Why this scores full marks: Part (b) reuses the answer to (a): $x^{-\frac{3}{2}}$ is the square root cubed and inverted. Each factor — the $9$ and the $y^{-4}$ — gets the power, and substituting keeps $9y^{-4}$ in a bracket before squaring in (c).
Question 9
Paper 1H/2H style — 4 marks4 marks
Show that $\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)^{2} = \dfrac{(x + 1)^{2}}{x}$ for $x > 0$ [4]
Model answer
$\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)^{2} = x^{\frac{1}{2}}x^{\frac{1}{2}} + x^{\frac{1}{2}}x^{-\frac{1}{2}} + x^{-\frac{1}{2}}x^{\frac{1}{2}} + x^{-\frac{1}{2}}x^{-\frac{1}{2}}$

$= x + 1 + 1 + x^{-1} = x + 2 + \dfrac{1}{x}$

$= \dfrac{x^{2}}{x} + \dfrac{2x}{x} + \dfrac{1}{x} = \dfrac{x^{2} + 2x + 1}{x} = \dfrac{(x + 1)^{2}}{x}$
Why this scores
Mark scheme:
- M1 for expanding to four terms, at least three correct
- M1 for $x + 2 + x^{-1}$ oe, using $x^{\frac{1}{2}} \times x^{-\frac{1}{2}} = x^{0} = 1$
- M1 for writing over the common denominator $x$ , e.g. $\dfrac{x^{2} + 2x + 1}{x}$
- A1 for a complete and correct argument reaching $\dfrac{(x + 1)^{2}}{x}$
Why this scores full marks: The bracket is written out as four products, so the two middle terms are not lost. Squaring each term separately to get $x + x^{-1}$ is the most common error and scores $0$ . Every step to the printed result is shown, as a show-that requires.
Question 10
Paper 1H/2H style — 5 marks5 marks
The area of a rectangle is $12x^{\frac{7}{2}}$ cm². The length of the rectangle is $3\sqrt{x}$ cm.

(a) Find an expression for the width of the rectangle. Give your answer in the form $kx^{n}$ [2]

(b) The width of the rectangle is $32$ cm. Work out the area of the rectangle. Give your answer correct to 3 significant figures. [3]
Model answer
(a) $\text{width} = \dfrac{12x^{\frac{7}{2}}}{3x^{\frac{1}{2}}} = 4x^{\frac{7}{2} - \frac{1}{2}} = 4x^{3}$

(b) $4x^{3} = 32 \quad\Rightarrow\quad x^{3} = 8 \quad\Rightarrow\quad x = 2$

$\text{area} = 12 \times 2^{\frac{7}{2}} = 135.764\ldots$

Answer: (a) $4x^{3}$ cm (b) $136$ cm²
Why this scores
Mark scheme:
- M1 for $\dfrac{12x^{\frac{7}{2}}}{3\sqrt{x}}$ with $\sqrt{x} = x^{\frac{1}{2}}$ used
- A1 for $4x^{3}$ cao
- M1 for $4x^{3} = 32$ leading to $x = 2$ (ft their width)
- M1 for substituting their $x$ into $12x^{\frac{7}{2}}$
- A1 for awrt $136$
Why this scores full marks: The width is area ÷ length, and the root is converted to $x^{\frac{1}{2}}$ before the indices are subtracted. In (b) the candidate answers the question asked — the area, not $x$ — and keeps the full calculator value until the final rounding.

4MA1 Algebraic Roots and Indices

At a glance

What you’ll learn

Applying the index laws to algebra

Fractional and negative indices

Solving exponential equations

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Simplify using index laws

2Power of a product

3Negative indices in algebra

4Fractional indices in algebra

5Solve an equation using indices

6A zero power with a coefficient [1 mark]

7A negative power of an algebraic fraction [2 marks]

8Fractional and negative powers in one quotient [3 marks]

9Splitting a quotient over a cube root into two powers of x [4 marks]

10Two unknown powers from two identities [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10