Indices and Surds
Senior 2 · Mathematics
Indices and Surds
Lesson 5 of 5
25 min
Which is easier to work out by hand: 1 ÷ 1.414, or 1.414 ÷ 2? The second, clearly. Both are approximations of 1/√2, rewritten in two ways. Moving a surd out of the denominator is called rationalising. It makes answers easier to compare, and national exams expect every final answer to have a rational denominator.
You also need to find square roots without a calculator: by factorising, by estimating, and by the digit-by-digit general method.
Multiply the top and bottom by the surd in the denominator. This is the same as multiplying by 1, so the value does not change, and √a × √a = a removes the root.
Worked example 1
6/√3 = (6 × √3)/(√3 × √3) = 6√33 = 2√3.
Worked example 2
5/(2√5) = 5√5/(2 × 5) = 5√510 = √52.
When the denominator is a sum or difference such as 3 − √5, multiply by its conjugate: the same two terms with the sign between them changed (3 + √5). Then (a − √b)(a + √b) = a2 − b, which has no root.
Worked example 3
Rationalise 4/(3 − √5).
Multiply top and bottom by 3 + √5. Bottom: 32 − (√5)2 = 9 − 5 = 4. Top: 4(3 + √5). So the fraction is 4(3 + √5)/4 = 3 + √5.
Write the number as a product of primes, then halve every index.
Worked example 4
1764 = 22 × 32 × 72, so √1764 = 2 × 3 × 7 = 42. This works only when every prime appears an even number of times, which means the number is a perfect square.
Worked example 5
Estimate √50.
50 lies between the perfect squares 49 = 72 and 64 = 82, and it is much closer to 49. So √50 is just above 7. Try 7.1: 7.12 = 50.41, which is slightly too big. So √50 ≈ 7.07 (check: 7.072 = 49.98).
This method finds a square root one digit at a time, like long division.
Worked example 6
Find √2209.
Check: 47 × 47 = 2209 ✓. For a number that is not a perfect square, add pairs of zeros after the decimal point and continue; for example, the method gives √20 = 4.47… to 2 decimal places.
Tip
Exam technique. To rationalise, multiply by exactly what removes the root: the surd itself (√a) for a single surd, or the conjugate (flip the middle sign) for a two-term denominator. Then simplify the fraction — many students stop at 6√33 and lose the mark for not writing 2√3.
Rationalise 10/√5.
Solution: 10√55 = 2√5.
Rationalise 3/(2√3).
Solution: 3√3/(2 × 3) = 3√36 = √32.
Rationalise 2/(√3 + 1).
Solution: Multiply by (√3 − 1). Bottom: 3 − 1 = 2. So 2(√3 − 1)/2 = √3 − 1.
Find √1296 using prime factors.
Solution: 1296 = 24 × 34. Halve the indices: 22 × 32 = 4 × 9 = 36.
Use the general method to find √3136.
Solution: Pairs 31 | 36. Largest square not above 31 is 25, so the first digit is 5; 31 − 25 = 6. Bring down 36 to get 636. Double 5 to get 10; try (100 + 6) × 6 = 636. The next digit is 6 and the remainder is 0, so √3136 = 56.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
