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Study Guide: Surds & Indices
Source: https://www.fatskills.com/quantitative-aptitude-and-numerical-ability-for-competitive-examinations/chapter/surds-indices

Surds & Indices

By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.

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Indices
The process of multiplying any quantity by itself any number of times is known as involution and the product thus obtained is called power of that quantity. The number which expresses the power is known as index of power or exponent of power. This number is written above a quantity to its right which is known as Base. Index or exponent indicates how many times the base is to be multiplied by itself. For example if a quantity a is multiplied by itself m times, then we write it as follows:
a . a . a . a .……m times = am. Here 'a' is the base, 'm' is the index or exponent, am is the mth power of 'a'. We read it as a to the power m.

Thus, it is obvious that, (a) × (a) = a2 (a) × (a) × (a) = a2 (a) × (a) × (a) × (a) = a4 etc.
The plural of index is indices.
span class="body-bold-italic'>If no index is written on any base, then, the index is regarded as 1. For example, a = a1, 4 = 41,

35 = 351 etc.

Laws of Indices
(i) am× an = a m + n
(ii) am÷ an = a m - n
(iii) (am)n = amn
(iv) a0 = 1 where a ? 0
(v) a - m =
(vi)
(vii) (ab)m = ambm
(viii) = a
(ix) n = a1/n
(x) = (a1/q)p
(xi) am = an m = n
(xii) am = bm a = b

For solution of Indices problems always remember:
Reduce all bigger numbers to the lowest shape of factors
Try to compare the basis or the powers
Generally avoid dealing with - ve powers
Generally the answer is 1 where cycle is complete.

Formulae to remember:
a(b + c) = ab + ac (distributive law)
(a + b)2 = a2 + 2ab + b2
(a ? b)2 = a2 ? 2ab + b2
(a + b)2 + (a ? b)2 = 2(a2 + b2)
(a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + 3ab(a + b) + b3
(a ? b)3 = a3 ? 3a2b + 3ab2 ? b3 = a3 ? 3ab(a ? b) ? b3
a2 ? b2 = (a ? b)(a + b)
a3 + b3 = (a + b)(a2 ? ab + b2)
a3 ? b3 = (a ? b)(a2 + ab + b2)
an ? bn = (a ? b)(an - 1 + an - 2 b + an - 3 b2 + ··· + bn - 1)
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b)(a + c) = a2 + (b + c)a + bc
an = a.a.a ... (n times)
am.an...ap = am + n + p
am.an = am + n
am/an = a m - n
(am)n = amn = (an)m
(ab)n = anbn
(ab)n = anbn
a ? n = 1/an
an = 1/a?n
(x + a)2 = x2 + 2ax + a2
(x ? a)2 = x2 ? 2ax + a2
(x + a)(x + b) = x2 + (a + b)x + ab
x2 ? a2 = (x ? a)(x + a)
(xn + 1) is completely divisible by (x + 1) when n is odd
(xn + an) is completely divisible by (x + a) when n is odd
(xn ? an) is completely divisible by (x ? a) for every natural number n
(xn ? an) is completely divisible by (x + a) when n is even
a0 = 1 where a?R,a ? 0
if am = an where a ? 0 and a ? ±1, then m = n




Example: Simplify.



=
= a a0 = 1.

Example: Find the square root of 11 + 6

Let = + Squaring both sides, we get
11 + 6 = x + y + 2 comparing rational and irrational parts we get
x + y = 11 & 2 = 6 say = 3 or xy = 18
by applying the formula (x - y)2 = (x + y)x - 4 xy
we get x - y = 7 solving further x = 9, y = 2
writing in the proper shape = + = + = 3 + Ans

Example: Simplify

= - + = 5 - 3 + 2 = 4


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