The laws of indices including negative and fractional powers, and simplifying, multiplying and rationalising surds at Higher tier.

What this dot point is asking

AQA wants you to apply the laws of indices, including zero, negative and fractional powers, and at Higher tier to simplify, multiply, add and rationalise surds. Index laws appear in standard form, algebra and growth problems; surds are a Higher-tier non-calculator skill that produces exact answers, especially in Pythagoras and trigonometry questions.

The laws of indices

When multiplying powers of the same base you add the indices: $x^3 \times x^4 = x^7$. When dividing you subtract: $\dfrac{x^7}{x^2} = x^5$. A power of a power multiplies: $(x^2)^3 = x^6$. A negative index means a reciprocal: $5^{-2} = \tfrac{1}{25}$. A zero index always gives $1$: $7^0 = 1$.

A fractional power combines a root and a power. The denominator gives the root and the numerator gives the power, so $27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$. Negative fractional powers do both: $8^{-\frac{1}{3}} = \dfrac{1}{\sqrt[3]{8}} = \dfrac{1}{2}$.

Working with surds at Higher tier

A surd is a root that is irrational, such as $\sqrt{2}$ or $\sqrt{5}$, left in exact form rather than rounded. Simplify a surd by pulling out the largest square factor: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. You can add or subtract like surds: $3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}$, but $\sqrt{2} + \sqrt{3}$ cannot be combined. Multiplying uses $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$, so $\sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3}$.

Rationalising the denominator

A surd in a denominator is usually removed (rationalised) to leave a whole-number denominator. Multiply the top and bottom by the surd in the denominator.

For $\dfrac{10}{\sqrt{5}}$, multiply by $\dfrac{\sqrt{5}}{\sqrt{5}}$ to get $\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$. When the denominator is a sum like $\dfrac{1}{3 + \sqrt{2}}$, multiply by the conjugate $3 - \sqrt{2}$, using the difference of two squares: $\dfrac{3 - \sqrt{2}}{(3 + \sqrt{2})(3 - \sqrt{2})} = \dfrac{3 - \sqrt{2}}{9 - 2} = \dfrac{3 - \sqrt{2}}{7}$.

Negative indices and reciprocals in algebra

The index laws extend straight into algebra, where they are tested heavily. A term like $\dfrac{1}{x^3}$ is written $x^{-3}$, which makes differentiating and simplifying expressions cleaner at A-level later. Simplifying $\dfrac{6x^5}{2x^2}$ uses the division law: divide the coefficients ($6 \div 2 = 3$) and subtract the indices ($x^{5-2} = x^3$), giving $3x^3$. A bracket raised to a power applies the power to every factor inside: $(2x^3)^4 = 2^4 x^{12} = 16x^{12}$, a step often half-done by forgetting to raise the coefficient.

Why surds give exact answers

Surds matter because they preserve exact values that decimals only approximate. In Pythagoras and trigonometry, a side might come out as $\sqrt{50}$, and writing it as $5\sqrt{2}$ keeps the answer exact rather than rounding to $7.07$. Many non-calculator questions deliberately produce surd answers so they can be left exact. The exact trigonometric values, such as $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$, are surds, which is why surd manipulation and right-angled triangle work are so closely linked in the specification.