N7, N8, A4 - Fractional indices and exact surd calculations

Use fractional indices to connect powers and roots, then calculate exactly with surds. Simplify, collect and expand surd expressions, and rewrite fractions with rational denominators without changing their values.

Fractional indices as roots and powers (Higher)

How can $\frac4{3+\sqrt2}$ be rewritten with no root in its denominator, while keeping exactly the same value? First we need to understand how roots behave in products and powers.

You already know

For a non-zero base, a negative integer index means reciprocal. A square-root symbol chooses the non-negative root. Index laws combine equal bases by adding, subtracting or multiplying indices as appropriate.

For a positive base, $(a^{1/2})^2 = a^1 = a$ . Therefore $a^{1/2} = \sqrt a$ . Similarly, $a^{1/3} = \sqrt[3]{a}$ . The denominator of a fractional index names the root; its numerator names the power.

Rational indices

For $a > 0$ , integer $p$ and positive integer $q$ ,

a^{p/q} = \left(\sqrt[q]{a}\right)^p = \sqrt[q]{a^p} \qquad a^{-p/q} = \frac1{a^{p/q}}

Not on the formula sheet: learn it.

Choose the easier order. For $27^{2/3}$ , the cube root first gives $3^2 = 9$ , avoiding the larger intermediate number $27^2 = 729$ .

Worked example 1: a root followed by a power

Question. Work out $16^{3/4}$ .

Plan. Take the fourth root first, then cube the result.

\begin{aligned} 16^{3/4} &= \left(\sqrt[4]{16}\right)^3 && \text{denominator gives the root} \\ &= 2^3 && \text{since }2^4 = 16 \\ &= 8 && \text{numerator gives the power} \end{aligned}

Why take the root first? It turns the input into a small whole number before the power is applied.

Answer. $8$ .

Check. $8^4 = 4096 = 16^3$ , so the two valid orders agree.

Change only the sign of the index. It now asks for the reciprocal of the previous value.

Worked example 2: a negative fractional index

Question. Work out $16^{-3/4}$ . Give an exact answer.

Plan. Find the positive-index value, then invert it.

\begin{aligned} 16^{-3/4} &= \frac1{16^{3/4}} && \text{negative means reciprocal} \\ &= \frac1{(\sqrt[4]{16})^3} && \text{interpret the positive index} \\ &= \frac1{2^3} = \frac18 && \text{evaluate exactly} \end{aligned}

Why is the answer not $-8$ ? The minus sign belongs to the index; the positive base still produces a positive value.

Answer. $\frac18$ .

Check. $16^{3/4} \times 16^{-3/4} = 8 \times \frac18 = 1$ .

The familiar laws still work for positive bases and rational indices. For instance, $8^{1/3} \times 8^{2/3} = 8^1 = 8$ , agreeing with $2 \times 4$ . Also $9^{3/2} \div 9^{1/2} = 9^1 = 9$ .

Watch out

The positive-base condition avoids ambiguity about even roots of negative numbers. $\sqrt[3]{-8} = -2$ is real, but $\sqrt{-8}$ has no real value. Do not apply a fractional-power rule to an arbitrary negative base without considering the root.

Keep the index from the second example and change only the base.

Surds and square factors (Higher)

A root can remain exact even when its decimal expansion never ends or repeats.

Surd

A surd is an irrational root expressed exactly. For example, $\sqrt2$ is a surd; $\sqrt9 = 3$ is not, because it is rational. The decimal $1.414$ approximates $\sqrt2$ ; it is not equal to it.

For non-negative inputs, $\sqrt{ab} = \sqrt a\sqrt b$ . A perfect-square factor can therefore leave the root as a whole-number factor. Aim for a final radicand, the number inside the root, with no square factor greater than one.

Square-root product and quotient rules

For $a, b \ge 0$ ,

\sqrt{ab} = \sqrt a\sqrt b

For $a \ge 0$ and $b > 0$ ,

\sqrt{\frac ab} = \frac{\sqrt a}{\sqrt b}

Not on the formula sheet: learn it.

Use the largest square factor you can recognise. A smaller one works too, but may leave more simplifying to do.

Worked example 3: extracting a square factor

Question. Simplify $\sqrt{72}$ .

Plan. Split $72$ into its largest square factor times the remaining factor.

\begin{aligned} \sqrt{72} &= \sqrt{36 \times 2} && \text{identify a square factor} \\ &= \sqrt{36}\sqrt2 && \text{use the product rule} \\ &= 6\sqrt2 && \text{evaluate the square root} \end{aligned}

Why choose $36$ ? Its square root is whole, and the remaining $2$ has no square factor to extract.

Answer. $6\sqrt2$ .

Check. $(6\sqrt2)^2 = 36 \times 2 = 72$ , and the proposed root is non-negative.

Change only the input by multiplying it by four, a perfect square. Its square root should double.

Worked example 4: scaling the radicand

Question. Simplify $\sqrt{288}$ .

Plan. Use $288 = 144 \times 2$ , then extract the square root of $144$ .

\begin{aligned} \sqrt{288} &= \sqrt{144 \times 2} && \text{identify the largest square factor} \\ &= \sqrt{144}\sqrt2 && \text{separate the product} \\ &= 12\sqrt2 && \text{evaluate the square root} \end{aligned}

Why does the coefficient double? Multiplying the input by $4$ multiplies its non-negative square root by $\sqrt4 = 2$ .

Answer. $12\sqrt2$ .

Check. $(12\sqrt2)^2 = 144 \times 2 = 288$ , and this result is twice $6\sqrt2$ .

Using a smaller factor in the first example gives $\sqrt{72} = 2\sqrt{18} = 2 \times 3\sqrt2 = 6\sqrt2$ , the same answer. Products can simplify too: $\sqrt6 \times \sqrt{15} = \sqrt{90} = 3\sqrt{10}$ , while $\sqrt7 \times \sqrt7 = 7$ . A quotient example is $\sqrt{25/9} = \frac53$ .

Always keep the non-negative-root convention. $\sqrt{(-5)^2} = 5$ , and $\sqrt{x^2}$ is the non-negative magnitude of $x$ ; it is not always $x$ .

Watch out

Wrong: $\sqrt{9+16} = \sqrt9 + \sqrt{16}$ . The left side is $5$ and the right side is $7$ . The product rule does not distribute a root over a sum.

Change only the radicand in the second example, and again look for its largest square factor.

Collecting and expanding surd expressions (Higher)

Surds follow the ordinary distributive laws, but only matching surd terms can be collected.

Three copies of $\sqrt5$ plus two copies give $3\sqrt5 + 2\sqrt5 = 5\sqrt5$ . In contrast, $\sqrt2 + \sqrt3$ has no matching surd terms to collect. Simplify first when needed: $\sqrt{50} + \sqrt8 = 5\sqrt2 + 2\sqrt2 = 7\sqrt2$ , which is not $\sqrt{58}$ .

Multiply coefficients and roots separately. Thus $(3\sqrt2)(2\sqrt6) = 6\sqrt{12} = 12\sqrt3$ . Multiplying over a bracket gives $2\sqrt3\,(\sqrt{12}+5) = 12 + 10\sqrt3$ , because the outside factor multiplies both terms. Conversely, $6\sqrt2 + 9\sqrt2 = 3\sqrt2\,(2+3)$ . Unlike surds can still share a numerical factor: $4\sqrt2 + 6\sqrt3 = 2(2\sqrt2 + 3\sqrt3)$ .

Worked example 5: expanding a surd product

Question. Expand and simplify $(2+\sqrt3)(4-\sqrt3)$ .

Plan. Form all four products, then collect rational terms and matching surd terms.

\begin{aligned} (2+\sqrt3)(4-\sqrt3) &= 8 - 2\sqrt3 + 4\sqrt3 - 3 && \text{multiply every pair of terms} \\ &= (8-3) + (-2+4)\sqrt3 && \text{group like terms} \\ &= 5 + 2\sqrt3 && \text{collect} \end{aligned}

Why is the last product $-3$ ? $\sqrt3 \times (-\sqrt3) = -(\sqrt3)^2 = -3$ .

Answer. $5 + 2\sqrt3$ .

Check. Using $\sqrt3 \approx 1.732$ , the original product is approximately $3.732 \times 2.268 \approx 8.464$ , matching $5 + 2(1.732)$ .

Change only the minus sign in the second bracket to plus. Track which two products change sign.

Worked example 6: changing a bracket sign

Question. Expand and simplify $(2+\sqrt3)(4+\sqrt3)$ .

Plan. Expand all four products again, then combine matching terms.

\begin{aligned} (2+\sqrt3)(4+\sqrt3) &= 8 + 2\sqrt3 + 4\sqrt3 + 3 && \text{multiply every pair of terms} \\ &= (8+3) + (2+4)\sqrt3 && \text{group like terms} \\ &= 11 + 6\sqrt3 && \text{collect} \end{aligned}

Why recheck both changed terms? The positive surd in the second bracket multiplies both $2$ and $\sqrt3$ in the first bracket.

Answer. $11 + 6\sqrt3$ .

Check. The second bracket increased by $2\sqrt3$ , so the product increased by $(2+\sqrt3)2\sqrt3 = 6 + 4\sqrt3$ . This equals the difference between the two final answers.

Squaring a bracket includes two cross-products: $(1+\sqrt2)^2 = 1 + 2\sqrt2 + 2 = 3 + 2\sqrt2$ . Opposite-sign brackets behave differently: $(a+b)(a-b) = a^2-b^2$ , because their cross-products cancel. For example, $(3+\sqrt2)(3-\sqrt2) = 9-2 = 7$ .

Watch out

Wrong: $(1+\sqrt2)^2 = 1+2 = 3$ . Right: include the two cross-products to get $3+2\sqrt2$ . A squared bracket is a product of two whole brackets.

Change only the rational term in the second bracket from the second example.

Removing surds from denominators (Higher)

Rationalising rewrites a fraction with a rational denominator while preserving its exact value.

Multiplying numerator and denominator by the same non-zero expression multiplies the fraction by one. For a simple surd denominator, use the surd itself:

\frac3{\sqrt5} = \frac{3\sqrt5}{\sqrt5\sqrt5} = \frac{3\sqrt5}{5}

If there is a coefficient, keep it: $\frac5{2\sqrt3} = \frac{5\sqrt3}{6}$ . Both original and rationalised expressions are exact.

Conjugate

For a binomial surd denominator, its conjugate changes the sign between the two terms. The conjugate of $3+\sqrt2$ is $3-\sqrt2$ .

Their product is $(3+\sqrt2)(3-\sqrt2) = 3^2-(\sqrt2)^2 = 7$ , a rational number.

The difference-of-squares identity explains the choice: the irrational cross-products cancel. Check that both the original denominator and the multiplier used are non-zero.

Worked example 7: a binomial denominator

Question. Rationalise the denominator of $\frac4{3+\sqrt2}$ .

Plan. Multiply numerator and denominator by the conjugate $3-\sqrt2$ .

\begin{aligned} \frac4{3+\sqrt2} &= \frac{4(3-\sqrt2)}{(3+\sqrt2)(3-\sqrt2)} && \text{multiply by one} \\ &= \frac{12-4\sqrt2}{9-2} && \text{expand numerator and denominator} \\ &= \frac{12-4\sqrt2}{7} && \text{simplify the denominator} \end{aligned}

Why use the conjugate rather than the same bracket? Opposite signs cancel the cross-products; squaring the original bracket would retain a surd term.

Answer. $\frac{12-4\sqrt2}{7}$ .

Check. Multiplying this answer by $3+\sqrt2$ gives $\frac{4(9-2)}7 = 4$ , recovering the original numerator.

Change only the sign in the original denominator. Its conjugate changes sign too.

Worked example 8: the opposite denominator sign

Question. Rationalise the denominator of $\frac4{3-\sqrt2}$ .

Plan. Multiply numerator and denominator by $3+\sqrt2$ .

\begin{aligned} \frac4{3-\sqrt2} &= \frac{4(3+\sqrt2)}{(3-\sqrt2)(3+\sqrt2)} && \text{multiply by one} \\ &= \frac{12+4\sqrt2}{9-2} && \text{use difference of squares} \\ &= \frac{12+4\sqrt2}{7} && \text{simplify} \end{aligned}

Why is the denominator still $7$ ? Reversing the order of the same two factors does not change their product.

Answer. $\frac{12+4\sqrt2}{7}$ .

Check. Multiplication by $3-\sqrt2$ gives $\frac{4(9-2)}7 = 4$ again.

Two irrational values can even multiply to one: $(\sqrt2+1)(\sqrt2-1) = 2-1 = 1$ . Each is therefore the other's reciprocal. For a denominator with two roots, $\frac1{\sqrt5-\sqrt2}$ , the same reasoning gives $\frac{\sqrt5+\sqrt2}{5-2} = \frac{\sqrt5+\sqrt2}{3}$ .

Simplify roots first where useful. Cancel a numerical factor only when it divides the whole numerator: $\frac{6-2\sqrt3}{2} = 3-\sqrt3$ , since both terms divide by two.

Watch out

Wrong: $\frac3{\sqrt5} = \frac35$ , changing only the denominator. Right: $\frac3{\sqrt5} \times \frac{\sqrt5}{\sqrt5} = \frac{3\sqrt5}{5}$ . Preserve the value by multiplying both parts by the same non-zero expression.

Change only the numerator of the second example.

Quick Recap

A fractional index gives a root and a power; a negative index adds a reciprocal.
Extract square factors to simplify surds, and collect only matching surd terms.
Expand all products, including the cross-products in a squared bracket.
Rationalise by multiplying both parts of a fraction by the same suitable non-zero expression.

The opening fraction is exactly $\frac{12-4\sqrt2}{7}$ : multiplying by the conjugate removes the root from its denominator.

Last Accessed

Papers 1, 2 and 3

N7, N8, A4 - Fractional indices and exact surd calculations

Fractional indices as roots and powers (Higher)

Surds and square factors (Higher)

Collecting and expanding surd expressions (Higher)

Removing surds from denominators (Higher)

Quick Recap

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