Square root, cube root, surds and indices MCQ - Practice Questions with Answers

Fractional indices are roots written in index form and that common-base conversion is usually quickest. Since 27 = 3^3, 27^(2/3) becomes (3^3)^(2/3) = 3^2 = 9. Since 32 = 2^5, 32^(3/5) becomes 2^3 = 8. Since 81 = 3^4, 81^(3/4) becomes 3^3 = 27. This gives the matching 27^(2/3) with 9, 32^(3/5) with 8, and 81^(3/4) with 27.

A^(m/n) can be read as the nth root of a^m, or as the mth power of the nth root, wherever the expression is defined. For 27^(2/3), first write 27 as 3^3. Then (3^3)^(2/3) becomes 3^2, which is 9. By contrast, 16^(1/2) is sqrt(16), not 16 divided by 2, and fractional indices should not be treated as ordinary division.

A perfect square can end only in 0, 1, 4, 5, 6 or 9. A number ending in 2, 3, 7 or 8 is rejected immediately by this filter. The number 378 ends in 8, so it cannot be a perfect square. The numbers 221, 625 and 784 are not rejected by the unit-digit rule; in fact, 625 and 784 are known squares given.

For a perfect cube, prime factors in groups of three is used. The factorisation of 5832 is given as three 2s and six 3s. One group of 2 contributes 2, and two groups of 3 contribute 3 and 3. Multiplying the outside factors gives 2 x 3 x 3 = 18. Therefore the factorisation and the stated cube root are both consistent with the cube-root rule.

28 squared is listed as 784, so sqrt(784) is 28. They also use 729 as 9 cubed, so cubert(729) is 9. By prime factorisation, sqrt(1296) as 36. For 27000, it splits as 27 times 1000; the cube roots are 3 and 10, giving 30. The matching is therefore sqrt(784)-28, cubert(729)-9, sqrt(1296)-36 and cubert(27000)-30.