B1, B2 - Indices and surds | AQA A-Level Mathematics

Use rational powers confidently, keeping the real-number conditions that make the index laws valid. Then simplify and combine surds and rationalise denominators, so exact answers stay exact even when individual numbers are irrational.

Rational powers and real roots

How can two irrational numbers multiply to exactly $1$ ? The pair $\sqrt5+2$ and $\sqrt5-2$ does exactly that. Before uncovering the cancellation, we need to be precise about roots and fractional powers.

A square of area $81\,\mathrm{cm}^2$ has side length $9\,\mathrm{cm}$ : a square root reverses squaring. Similarly, since $3^4=81$ , the fourth root of $81$ is $3$ . Thus $81^{3/4}$ means take that fourth root and cube it. Taking the root first keeps the arithmetic small. The integers $\mathbb Z$ include zero and negative whole numbers; the rational numbers $\mathbb Q$ are fractions of integers with nonzero denominators.

Rational power

Write $p/q$ in lowest terms, where $p$ is an integer and $q$ is a positive integer. For $a>0$ , $a^{p/q}=(\sqrt[q]{a})^p$ . For example, $81^{3/4}=3^3=27$ . A negative exponent means a reciprocal; it does not mean a negative answer.

The root symbol selects the non-negative root when the root has even order. In particular, $\sqrt9=3$ , although the equation $x^2=9$ has both $x=3$ and $x=-3$ as solutions.

Roots and reciprocals

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

a^{1/q}=\sqrt[q]{a},\qquad a^{-p/q}=\frac1{a^{p/q}}

Not in the formula booklet: learn it.

Start by interpreting the denominator of the exponent, then deal with its numerator.

Worked example 1: a positive rational power

Question. Evaluate $32^{2/5}$ exactly.

Plan. Take the fifth root first, then square it.

\begin{aligned} \sqrt[5]{32}&=2 && 2^5=32 \\ 32^{2/5}&=(\sqrt[5]{32})^2 && \text{interpret the exponent} \\ &=2^2=4 && \text{square the root} \end{aligned}

Why this step? The denominator $5$ specifies the root; it is not a divisor of $32$ .

Answer. $32^{2/5}=4$ .

Check. $4^5=1024=32^2$ .

The second example changes only the exponent's sign. The root calculation stays the same; the final operation becomes a reciprocal.

Worked example 2: a negative rational power

Question. Evaluate $32^{-2/5}$ exactly.

Plan. Evaluate the positive power and take its reciprocal.

\begin{aligned} 32^{-2/5}&=\frac1{32^{2/5}} && \text{negative exponent} \\ &=\frac1{(\sqrt[5]{32})^2} && \text{root before power} \\ &=\frac1{2^2}=\frac14 && \text{evaluate} \end{aligned}

Why this step? A reciprocal undoes multiplication by the positive power.

Answer. $32^{-2/5}=\frac14$ .

Check. $4\times\frac14=1$ , as required.

Zero to a positive rational power is zero. Negative powers of zero and $0^0$ are undefined here. For a negative base, a reduced rational exponent has a real value when its denominator is odd: $(-8)^{2/3}=(-2)^2=4$ . An even root of a negative number has no real value. Reduce the exponent first; this convention does not license unrestricted index laws for negative bases.

Watch out

Wrong: $32^{-2/5}=-4$ . The correct value is $1/4$ . Also keep brackets: $(-3)^2=9$ , whereas $-3^2=-(3^2)=-9$ .

Change the base now, keeping the same root-then-reciprocal decisions.

Index laws and their conditions

Index laws collect repeated multiplication into a shorter expression, but the operation and the domain both matter.

You already know

For a positive base, a rational exponent combines a root and a power. A negative exponent takes a reciprocal, so its base cannot be zero.

For instance, $2^3\times2^2$ contains five factors of $2$ , giving $2^5$ . Division cancels common factors instead, and a power of a power repeats every factor. The same laws extend to rational exponents for positive bases.

Index laws

For $a,b>0$ and rational $r,s$ ,

\begin{aligned} a^ra^s&=a^{r+s} & \frac{a^r}{a^s}&=a^{r-s} \\ (a^r)^s&=a^{rs} & (ab)^r&=a^rb^r \\ \left(\frac ab\right)^r&=\frac{a^r}{b^r} & a^0&=1 \end{aligned}

Not in the formula booklet: learn it.

Multiplication, division and nesting call for different operations on the exponents. Work with fractions so no approximation enters the algebra.

Worked example 3: collect powers of one base

Question. For $x>0$ , simplify $\dfrac{x^{3/2}x^{-1/4}}{x^{1/2}}$ as a single power.

Plan. Add the numerator exponents and subtract the denominator exponent.

\begin{aligned} \frac{x^{3/2}x^{-1/4}}{x^{1/2}}&=x^{3/2-1/4-1/2} && \text{same base} \\ &=x^{6/4-1/4-2/4} && \text{common denominator} \\ &=x^{3/4} && \text{collect} \end{aligned}

Why this step? Division subtracts an exponent; the negative exponent already in the numerator keeps its own minus sign.

Answer. $x^{3/4}$ , for $x>0$ .

Check. At $x=16$ , the original is $64(1/2)/4=8=16^{3/4}$ .

Now change the denominator exponent from positive to negative. Subtracting a negative exponent increases the resulting power.

Worked example 4: divide by a negative power

Question. For $x>0$ , simplify $\dfrac{x^{3/2}x^{-1/4}}{x^{-1/2}}$ as a single power.

Plan. Subtract the entire denominator exponent, including its sign.

\begin{aligned} \frac{x^{3/2}x^{-1/4}}{x^{-1/2}}&=x^{3/2-1/4-(-1/2)} && \text{quotient law} \\ &=x^{6/4-1/4+2/4} && \text{subtract a negative} \\ &=x^{7/4} && \text{collect} \end{aligned}

Why this step? Dividing by $x^{-1/2}$ is multiplying by $x^{1/2}$ .

Answer. $x^{7/4}$ , for $x>0$ .

Check. At $x=16$ , the original is $64(1/2)/(1/4)=128=16^{7/4}$ .

Positive bases are a safe common domain for all these laws. Some extend to negative bases for integer powers, but the nested-power rule needs care: $\bigl((-1)^2\bigr)^{1/2}=1$ , whereas $(-1)^{2(1/2)}=-1$ . In general $\sqrt{x^2}=|x|$ for real $x$ , because the principal square root is non-negative. It equals $x$ only when $x\ge0$ . This absolute-value result is not in the formula booklet: learn it.

Watch out

Wrong: $x^2+x^3=x^5$ . Adding terms does not concatenate their factors. Right: $x^2x^3=x^5$ , while the sum factorises as $x^2(1+x)$ .

Index simplification will later turn roots and reciprocals into powers ready for calculus. For now, practise the sign in the quotient law.

Simplifying and combining surds

Removing square factors reveals which surd terms are alike, just as collecting algebraic terms requires matching variables.

The diagonal of a unit square is $\sqrt2$ , an exact length that no terminating decimal can give. A decimal is useful for size; the surd retains exact equality.

Surd

A surd is an irrational root kept in exact form, such as $\sqrt2$ or $\sqrt[3]5$ . The expression $\sqrt9=3$ is rational, so it is not a surd.

Split a non-negative radicand into a square factor and what remains. The square factor then comes outside the root.

Square-root arithmetic

For $a,b\ge0$ ,

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

For $a\ge0$ and $b>0$ ,

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

Not in the formula booklet: learn it.

The largest square factors save repeated simplification. Once every term is simplified, collect the coefficients of the matching surd.

Worked example 5: reveal like surds

Question. Simplify $\sqrt{72}+\sqrt{50}-\sqrt8$ .

Plan. Extract square factors, then collect multiples of $\sqrt2$ .

\begin{aligned} \sqrt{72}+\sqrt{50}-\sqrt8 &=\sqrt{36\times2}+\sqrt{25\times2}-\sqrt{4\times2} && \text{square factors} \\ &=6\sqrt2+5\sqrt2-2\sqrt2 && \text{take their roots} \\ &=9\sqrt2 && \text{collect coefficients} \end{aligned}

Why this step? The radicands look different, but all three roots are multiples of the same irrational number.

Answer. $9\sqrt2$ .

Check. Dividing the original by $\sqrt2$ gives $6+5-2=9$ .

The second example changes the common surd from $\sqrt2$ to $\sqrt3$ . The collection step is unchanged.

Worked example 6: a different common surd

Question. Simplify $\sqrt{108}+\sqrt{75}-\sqrt{12}$ .

Plan. Extract square factors and collect multiples of $\sqrt3$ .

\begin{aligned} \sqrt{108}+\sqrt{75}-\sqrt{12} &=\sqrt{36\times3}+\sqrt{25\times3}-\sqrt{4\times3} && \text{square factors} \\ &=6\sqrt3+5\sqrt3-2\sqrt3 && \text{simplify} \\ &=9\sqrt3 && \text{collect coefficients} \end{aligned}

Why this step? Only the common irrational factor changes; ordinary coefficient arithmetic still does the collecting.

Answer. $9\sqrt3$ .

Check. Dividing by $\sqrt3$ leaves $6+5-2=9$ .

Products use ordinary expansion as well as the root rules. For the opening pair,

\begin{aligned} (\sqrt5+2)(\sqrt5-2)&=5-2\sqrt5+2\sqrt5-4 \\ &=1 \end{aligned}

Both factors are irrational, yet the irrational terms cancel exactly. The difference-of-squares identity $(a+b)(a-b)=a^2-b^2$ is not in the formula booklet: learn it. In contrast, $(\sqrt5+2)^2=9+4\sqrt5$ : squaring does not make the middle terms vanish.

Watch out

Wrong: $\sqrt{9+16}=\sqrt9+\sqrt{16}=7$ . In fact $\sqrt{25}=5$ . Roots split over products under the stated conditions, not over sums. Also $2\sqrt3+4\sqrt5$ cannot be collected into one like term.

Use different square factors now, then check the signs of the coefficients.

Rationalising a denominator

Rationalising uses an equivalent fraction with a rational denominator.

The value is preserved by multiplying by $1$ . For a denominator $\sqrt6$ , use $\sqrt6/\sqrt6$ ; for two terms, choose a multiplier that cancels the middle terms in the denominator's expansion.

Conjugate of a two-term surd expression

The conjugate changes the sign between the terms: $a+\sqrt b$ and $a-\sqrt b$ are conjugates. Their product is $a^2-b$ , for real $a$ and $b\ge0$ .

The conjugate product is the difference of two squares from the previous part. First see the simpler case where the denominator has only one term.

Worked example 7: a single surd denominator

Question. Rationalise and simplify $\dfrac4{\sqrt7}$ .

Plan. Multiply top and bottom by $\sqrt7$ to make the denominator rational.

\begin{aligned} \frac4{\sqrt7}&=\frac4{\sqrt7}\frac{\sqrt7}{\sqrt7} && \text{multiply by }1 \\ &=\frac{4\sqrt7}{7} && (\sqrt7)^2=7 \end{aligned}

Why this step? Multiplying by the denominator turns its square root into a rational number.

Answer. $4\sqrt7/7$ .

Check. Multiplying by $\sqrt7$ gives $4(7)/7=4$ , the original numerator.

Now add just one term, $-\sqrt3$ , to the denominator. Multiplying by itself would leave a middle surd term; changing the sign makes that term cancel.

Worked example 8: a two-term denominator

Question. Rationalise and simplify $\dfrac4{\sqrt7-\sqrt3}$ , showing the rationalising step.

Plan. Multiply top and bottom by $\sqrt7+\sqrt3$ .

\begin{aligned} \frac4{\sqrt7-\sqrt3}&=\frac{4(\sqrt7+\sqrt3)}{(\sqrt7-\sqrt3)(\sqrt7+\sqrt3)} && \text{multiply by }1 \\ &=\frac{4(\sqrt7+\sqrt3)}{7-3} && \text{middle terms cancel} \\ &=\sqrt7+\sqrt3 && \text{divide by }4 \end{aligned}

Why this step? The conjugate produces $7-3$ , whereas squaring the denominator would produce $10-2\sqrt{21}$ .

Answer. $\sqrt7+\sqrt3$ .

Check. $(\sqrt7+\sqrt3)(\sqrt7-\sqrt3)=4$ , recovering the numerator.

The multiplier must be defined and nonzero. For example, $1/(2-\sqrt t)$ requires $t\ge0$ and $t\ne4$ . Its conjugate $2+\sqrt t$ is positive, so rationalising gives $(2+\sqrt t)/(4-t)$ on that same domain. A zero denominator gives no fraction to rationalise. These identities are not in the formula booklet: learn the process.

Exam tip

When AQA asks you to show a surd result, display the conjugate multiplication and simplification of the roots. A rounded calculator answer, or a jump straight to the printed answer, does not show that chain.

The opening pair now has another interpretation: $\sqrt5-2=1/(\sqrt5+2)$ . Two irrational numbers can be exact reciprocals.

Watch out

Wrong: $3/\sqrt6=3\sqrt6$ , multiplying only the numerator. Right: multiply both numerator and denominator by $\sqrt6$ , giving $3\sqrt6/6=\sqrt6/2$ . Only a multiplier equal to $1$ preserves the fraction's value.

Keep that equal multiplier visible in the next two-term denominator.

Quick Recap

A reduced rational exponent specifies a root and a power; a negative exponent takes a reciprocal.
Apply index laws on a domain where the powers and the laws are valid; positive bases are the safe common case.
Extract square factors before collecting matching surd terms. Roots do not split over sums.
Rationalise with a suitable nonzero multiplier, using the conjugate for a two-term denominator.

The opening pair multiplies to $1$ because $(\sqrt5+2)(\sqrt5-2)=5-4=1$ ; the irrational middle terms cancel.

Last Accessed

Pure Mathematics

Mechanics

Statistics

B1, B2 - Indices and surds

Rational powers and real roots

Index laws and their conditions

Simplifying and combining surds

Rationalising a denominator

Quick Recap

Learn faster with Nova