Maths › Algebra and functions › Indices and surds
Indices and surds
Three laws govern every power there is, and once fractional and negative exponents join in, roots stop being a separate subject. Then surds: the art of leaving answers exact, tidying them, and never letting a root squat in a denominator.
IN THIS TOPIC
- Use the laws of indices for all rational exponents, including negative and fractional powers.
- Simplify surds using the multiplication rule and exact arithmetic.
- Rationalise denominators, including with the conjugate.
WHAT YOU PROBABLY THINK
√(a + b) = √a + √b.
The laws of indices
Everything about powers follows from three laws, and all three are on the must-learn list:
The course extends them to all rational exponents, and the extensions are forced, not chosen. Dividing a3 by a3 gives a0 by the second law and 1 by common sense, so a0 = 1. Push one step further and a−n = 1/an. And since (a1/2)2 = a1 by the third law, a1/2 must be the square root of a: in general am/n means the nth root of am, and the two notations are interchangeable.
WORKED EXAMPLE
A negative fractional power, unpacked
Evaluate 27−2/3 without a calculator.
Work outside-in: the minus sign means reciprocal, the third means cube root, the 2 means square.
27−2/3 = 1/272/3 = 1/(∛27)2 = 1/32 = 1/9.
Taking the root before the square keeps the numbers small; (27²) first would mean cube-rooting 729. Same answer, worse afternoon.
Surds: exact by choice
A surd is a root left unevaluated, √2 or 5√3, because writing 1.414… would throw information away. Exact answers are the house currency of A-level maths, and surds obey rules inherited from indices, chief among them:
along with (√x)2 = x. The first rule, read right to left, is the simplifying move: pull the largest square factor out of the root. And note what is not on the list: the lie above. Roots do not distribute over addition, and one counter example settles it in the manner of the proof unit: √(9 + 16) = √25 = 5, while √9 + √16 = 7.
WORKED EXAMPLE
Simplifying a surd
Write √48 in the form k√3, and hence simplify √75 − √27.
√48 = √(16 × 3) = √16 × √3 = 4√3.
√75 = √(25 × 3) = 5√3 and √27 = √(9 × 3) = 3√3, so √75 − √27 = 2√3.
Like terms in √3 collect the same way terms in x would: once each surd is fully simplified, surd arithmetic is ordinary algebra.
Rationalising the denominator
Convention, and most mark schemes, want denominators free of surds. For a lone surd, multiply top and bottom by it: 1/√2 = √2/2. When the denominator is a sum or difference involving a surd, multiply by its conjugate, the same expression with the middle sign flipped, because the difference-of-squares identity (√x + √y)(√x − √y) = x − y wipes the roots out.
WORKED EXAMPLE
The conjugate at work
Express 1/(3 − √2) with a rational denominator.
Multiply top and bottom by the conjugate 3 + √2: the denominator becomes (3 − √2)(3 + √2) = 9 − 2 = 7.
So 1/(3 − √2) = (3 + √2)/7.
The conjugate is chosen precisely so the cross terms cancel; nothing else about the fraction changes, because multiplying by (3 + √2)/(3 + √2) is multiplying by 1.
YOUR TURN
A fuller fraction
Express (2 + √5)/(3 − √5) in the form (a + b√5)/c, before opening the working.
Show the working
Multiply by (3 + √5)/(3 + √5). Denominator: 9 − 5 = 4.
Numerator: (2 + √5)(3 + √5) = 6 + 2√5 + 3√5 + 5 = 11 + 5√5. So the answer is (11 + 5√5)/4.
Expand the numerator like any pair of brackets; the only special step in the whole method is the choice of conjugate.
TRY IT UNSEEN
The quadratic you nearly missed
Solve, using algebra and showing each stage of your working, the equation x − 6√x + 4 = 0.
Show the working
Substitute u = √x (so u ≥ 0): the equation becomes u2 − 6u + 4 = 0.
By the quadratic formula, u = (6 ± √20)/2 = 3 ± √5, and both values are positive, so both are allowed.
Then x = u2 = (3 ± √5)2 = 9 ± 6√5 + 5 = 14 + 6√5 or 14 − 6√5.
This equation appears verbatim in the specification. Index laws to see the hidden quadratic, surd arithmetic to finish it, and a check that each root of u survives the u ≥ 0 condition.
THE EXAM BIT
- The three index laws are on the must-learn list and are quoted, not derived, in working. Everything else, a0 = 1, negative and fractional powers, follows from them if pressed.
- Evaluate fractional powers root-first: am/n as (nth root of a)m keeps the arithmetic small and calculator-free, which is how these questions are phrased.
- Simplify every surd fully before collecting: pull out the largest square factor, then treat k√3 terms exactly like terms in a variable.
- "Show each stage of your working" signals a no-calculator method mark scheme: name the substitution, show the conjugate multiplication, keep every line exact.
- Never leave a surd in a denominator in a final answer; rationalise with the surd itself, or with the conjugate when the denominator has two terms.
CHECK YOURSELF
Evaluate 323/5 without a calculator, and express 6/(√7 − 1) in the form a + √b.
Show a hint
Fifth root first; then the conjugate of √7 − 1 is √7 + 1.
Show the answer
323/5 = (⁵√32)3 = 23 = 8.
Multiply 6/(√7 − 1) by (√7 + 1)/(√7 + 1): denominator 7 − 1 = 6, so the fraction is 6(√7 + 1)/6 = √7 + 1, which is 1 + √7 in the requested form with a = 1, b = 7.
Both answers are exact, and neither needed a decimal at any stage: that is the standard this course holds answers to.
Three index laws rule every power; fractions in the exponent are roots.
Simplify surds by their largest square factor, and rationalise with the conjugate.
CHECK YOUR PROGRESS
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- Use the laws of indices for all rational exponents, including negative and fractional powers.
- Simplify surds using the multiplication rule and exact arithmetic.
- Rationalise denominators, including with the conjugate.
No animated video for this topic yet; these notes stand alone.