IGCSE Additional Mathematics: Indices and Surds Practice Questions
A surd is an irrational root left in exact form. Simplify by extracting square factors, and rationalise a denominator by multiplying by the conjugate. Index equations are solved by writing both sides to a common base.
Topic 4 of Cambridge IGCSE Additional Mathematics 0606 demands exact answers, so decimals lose marks even when numerically correct. Rationalising with a conjugate and matching bases are the two techniques that recur. The questions below drill both.
What you need to know for Indices and Surds
- Simplifying a surdExtract the largest square factor. The square root of 50 becomes the square root of 25 times 2, which is 5 times the square root of 2.
- Rationalising a simple denominatorMultiply top and bottom by the surd in the denominator.
- Rationalising with a conjugateFor a denominator of the form a + the square root of b, multiply top and bottom by a minus the square root of b. The difference of two squares clears the surd.
- Common baseWrite both sides of an index equation as powers of the same base, then equate the indices.
- Fractional and negative indicesThe denominator of a fractional index gives the root and the numerator the power. A negative index means the reciprocal.
IGCSE Additional Mathematics Indices and Surds questions and answers
4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.
Simplify the square root of 50 plus the square root of 18, giving your answer in the form a times the square root of b.
Show the worked answer
- Extract square factors: the square root of 50 = the square root of 25 x 2 = 5 times the square root of 2.
- Similarly the square root of 18 = the square root of 9 x 2 = 3 times the square root of 2.
- Both terms now contain the same surd, so they can be added: 5 + 3 = 8 times the square root of 2.
Rationalise the denominator of 4 divided by (2 + the square root of 3).
Show the worked answer
- Multiply the numerator and denominator by the conjugate, 2 minus the square root of 3.
- The denominator becomes (2 + root 3)(2 minus root 3) = 4 minus 3 = 1, using the difference of two squares.
- The numerator becomes 4(2 minus root 3) = 8 minus 4 times the square root of 3.
- Since the denominator is 1, the answer is 8 minus 4 times the square root of 3.
Solve 32x = 81.
Show the worked answer
- Write the right hand side as a power of the same base: 81 = 34.
- The equation becomes 32x = 34.
- With equal bases, the indices must be equal: 2x = 4, so x = 2.
Solve 2x+1 = 8x-1.
Show the worked answer
- Write 8 as a power of 2: 8 = 23, so 8x-1 = 23(x-1) = 23x-3.
- The equation becomes 2x+1 = 23x-3.
- Equate the indices: x + 1 = 3x minus 3.
- Rearranging gives 4 = 2x, so x = 2. Check: 23 = 8 and 81 = 8.
Common mistakes in this topic
- Adding surds without simplifying them to a common surd first.
- Using the wrong sign in a conjugate.
- Leaving a surd in a denominator when an exact answer is required.
- Giving a decimal when the question asks for surd form.
- Forgetting brackets when raising a power to a power.
Exam tips
- Look for the largest square factor when simplifying a surd. Missing it leaves the answer unsimplified.
- The conjugate always has the opposite sign in the middle.
- For index equations, ask which base both sides share. Usually 2, 3 or 5.
- Exact form means surds and fractions, never decimals.
- Check a rationalised answer by evaluating both the original and the result on a calculator.
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Indices and Surds FAQs
How do I simplify a surd?
Find the largest square factor of the number under the root and take its square root outside. The square root of 50 becomes the square root of 25 times 2, which simplifies to 5 times the square root of 2. Always check that no square factor remains.
How do I rationalise a denominator?
For a single surd, multiply the numerator and denominator by that surd. For a denominator of the form a plus a surd, multiply by its conjugate, a minus that surd, which uses the difference of two squares to clear the root.
How do I solve an equation with the unknown in the index?
Write both sides as powers of the same base, then equate the indices and solve the resulting equation. If the bases cannot be matched, take logarithms of both sides instead.
Why do I lose marks for giving a decimal?
Additional Mathematics frequently asks for exact answers, meaning surds, fractions or logarithms left unevaluated. A decimal is a rounded approximation, so it does not satisfy a request for an exact value however many figures are given.
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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.