Algebraic Roots & Indices (Edexcel IGCSE Maths): Revision notes

Section 1

What do fractional and negative indices mean?

The laws of indices you use for numbers apply exactly the same way when the base is an algebraic term such as $x$ , $2x$ or $x^2y$ .

A negative index means reciprocal: $x^{-n} = \dfrac{1}{x^n}$
A fractional index with denominator $n$ means an $n$ th root: $x^{\frac{1}{n}} = \sqrt[n]{x}$
A general fractional index combines root and power: $x^{\frac{m}{n}} = \left(\sqrt[n]{x}\right)^m = \sqrt[n]{x^m}$

Example: $x^{\frac{2}{3}} = \left(\sqrt[3]{x}\right)^2$ . Always take the root first (smaller numbers), then raise to the power.

Key termsindex (exponent)reciprocalnth root

Exam tip

Memorise the order: denominator = root, numerator = power. 'Root first, power second' keeps the numbers small and avoids arithmetic mistakes.

Example

$8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$

Section 2

How do you simplify algebraic expressions with indices?

The three core laws still apply when the base is a letter or algebraic term:

Multiplying: $x^a \times x^b = x^{a+b}$
Dividing: $x^a \div x^b = x^{a-b}$
Power of a power: $\left(x^a\right)^b = x^{ab}$

These also work when coefficients and multiple letters are involved — deal with numbers and each letter separately.

Example: $\dfrac{6x^5y^3}{2x^2y} = 3x^{5-2}y^{3-1} = 3x^3y^2$

Any non-zero base to the power 0 equals 1, including algebraic terms: $\left(3x^2y\right)^0 = 1$ (as long as $x \neq 0$ and $y \neq 0$ ).

Key termscoefficientzero index

Common mistake

Do not add or subtract the coefficients using the index laws — only the powers of the letters follow the addition/subtraction rule. $6x^5 \div 2x^2 = 3x^3$ , not $4x^3$ and not $3x^{2.5}$ .

Exam tip

Split compound terms into 'numbers' and 'each letter' before applying a law, then recombine at the end.

Section 3

How do you simplify algebraic surds?

A surd is an irrational root such as $\sqrt{2}$ or $\sqrt{5x}$ that cannot be written as an exact fraction. To simplify:

Product rule: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$
Quotient rule: $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$
Extract square factors: $\sqrt{18x^2} = \sqrt{9x^2 \times 2} = 3x\sqrt{2}$

Always look for the largest perfect-square factor (4, 9, 16, 25...) inside the surd, including any squared variable, and pull it outside the root sign.

Example: $\sqrt{50x^3} = \sqrt{25x^2 \times 2x} = 5x\sqrt{2x}$

Key termssurdperfect square

Think of it like this

Think of a surd like an unopened box: factorising is checking what's inside for a 'square' item you can pull straight out, leaving only the awkward part under the root.

Common mistake

$\sqrt{a} + \sqrt{b} \neq \sqrt{a+b}$ . Surds can only be combined this way through multiplication and division, never addition or subtraction (unless the surd parts already match).

Section 4

How do you rationalise a denominator?

A fraction should never be left with a surd in the denominator. To rationalise:

For $\dfrac{a}{\sqrt{b}}$ , multiply top and bottom by $\sqrt{b}$ : $\dfrac{a}{\sqrt{b}} = \dfrac{a\sqrt{b}}{b}$
For $\dfrac{a}{p + \sqrt{q}}$ , multiply top and bottom by the conjugate $p - \sqrt{q}$ , using the difference of two squares to remove the surd from the bottom.

Example: $\dfrac{5}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}$

Example with a conjugate: $\dfrac{4}{2+\sqrt{3}} = \dfrac{4(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \dfrac{4(2-\sqrt{3})}{4-3} = 4(2-\sqrt{3})$

Key termsrationaliseconjugate

Exam tip

Multiplying a surd expression by its conjugate always removes the root, because $(p+\sqrt{q})(p-\sqrt{q}) = p^2 - q$ .

Must Know

$x^{-n} = \dfrac{1}{x^n}$ and $x^0 = 1$ (for non-zero $x$ )
$x^{\frac{1}{n}} = \sqrt[n]{x}$ ; $x^{\frac{m}{n}} = \left(\sqrt[n]{x}\right)^m$ — root first, power second
$x^a \times x^b = x^{a+b}$ , $x^a \div x^b = x^{a-b}$ , $(x^a)^b = x^{ab}$ — apply to letters and numbers separately
Simplify surds by extracting the largest perfect-square factor: $\sqrt{50} = 5\sqrt{2}$
Rationalise a single surd denominator by multiplying top and bottom by that surd
Rationalise a binomial surd denominator by multiplying by the conjugate

That's the notes covered.

Carry on to the next subtopic.