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Aptitude Surds and Indices Theory

Surds and Indices - Important Formulas :

1.Laws of Indices:

(i) am x an = am + n

(ii) $\dfrac{a^{m}}{a^{n}}= a^{m - n}$

(iii) $(a^{m})^{n} = a^{mn}$

(iv) $(ab)^{n} = a^{n}b^{n}$

(v) $\left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}}$

(vi) $a^{0} = 1$

2.Surds:

Let a be rational number and n be a positive integer such that$ a^{(1/n)} = \sqrt[n]{a}$
Then, $\sqrt[n]{a}$ is called a surd of order n.

3. Laws of Surds:

(i) $\sqrt[n]{a}=a^{(1/n)}$

(ii) $\sqrt[n]{ab}=\sqrt[n]{a} \times \sqrt[n]{b}$

(iii) $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$

(iv) $(\sqrt[n]{a}){^n}=a$

(v) $\sqrt[m]{\sqrt[n]{a}}=\sqrt[mn]{a}$

(vi) $(\sqrt[n]{a})^{m}=\sqrt[n]{a^{m}}$

Surds and Indices - Theory

Surds

Introduction

Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers.

There are certain rules that we follow to simplify an expression involving surds. Rationalising the denominator is one way to simplify these expressions. It is done by eliminating the surd in the denominator. This is shown in Rules 3, 5 and 6.

It can often be necessary to find the largest perfect square factor in order to simplify surds. The largest perfect square factor is found by looking at any possible factors of the number that is being square rooted. Lets say that you are looking at the square root of 242. Can you simplify this? Well, 2 x 121 is 242 and we can take the square root of 121 without leaving a surd (because we get 11). Since we cannot take the square root of a larger number that can be multiplied by another to give 242 then we say that 121 is the largest perfect square factor.

Six Rules of Surds


Rule 1:

$\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$

Example:

Simplify $\sqrt{18}$:

solution
Since ,18=9 x 2 = 32 x 2, as 9 is the largest perfect square factor of 18.
$\therefore \sqrt{18} = \sqrt{3^{2} \times 2}$
=$\sqrt{3^{2}} \times \sqrt{2}$
=$3\sqrt{2}$

Exercise :

44449.simplify $\sqrt{125x^{3}}$
5x$\sqrt{5x}$
25x$\sqrt{5x}$
5x$\sqrt{25x}$
15x$\sqrt{5x}$
Explanation:

$\sqrt{125x^{3}}=\left(\sqrt{(25x^{2})} \times \sqrt{(5x)}\right)$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]

=5x$\sqrt{5x}$

Rule 2:

$\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$

Example:

Simplify $\sqrt{\dfrac{12}{121}}$

solution
$\sqrt{\dfrac{12}{121}}=\dfrac{\sqrt{12}}{\sqrt{121}}$
Since 4 is the largest perfect square factor of 12
$\dfrac{\sqrt{2^{2} \times 3}}{11}$
$\dfrac{\sqrt{2^{2}} \times \sqrt{3}}{11}$
$\dfrac{2\sqrt{3}}{11}$

Exercise :

44450.Write $\sqrt{\dfrac{32}{144}}$ in the simplified form $a\sqrt{n}$
$\dfrac{\sqrt{32}}{3}$
$\dfrac{\sqrt{2}}{13}$
$\dfrac{\sqrt{2}}{3}$
$\dfrac{\sqrt{3}}{2}$
Explanation:

$\sqrt{\dfrac{32}{144}}=\dfrac{\sqrt{32}}{\sqrt{144}}$[using the rule $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$]

=$\dfrac{(\sqrt{16}\times \sqrt{2})}{12}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]

=$\dfrac{4 \sqrt{2}}{12}$ [simplify $\dfrac{4}{12}$]

=$\dfrac{\sqrt{2}}{3}$

Rule 3:

$\dfrac{b}{\sqrt{a}}=\dfrac{b}{\sqrt{a}} \times \dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{b \sqrt{a}}{a}$
By multiplying both the numberator and denominator by the denominator you can rationalise the denominator.

Example:

rationalise $\dfrac{5}{\sqrt{7}}$ :

solution :
$\dfrac{5}{\sqrt{7}}=\dfrac{5}{\sqrt{7}}\times \dfrac{\sqrt{7}}{\sqrt{7}}$ (multiply both numerator and denominator by $\sqrt{7}$)
=$\dfrac{5 \sqrt{7}}{7}$

Exercise :

44451.simplify $\dfrac{10\sqrt{3}}{\sqrt{5}}$
4$\sqrt{15}$
2$\sqrt{15}$
2$\sqrt{5}$
4$\sqrt{5}$
Explanation:

$\dfrac{10\sqrt{3}}{\sqrt{5}}=\dfrac{10\sqrt{3}}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}}$ [using the rule $\dfrac{b}{\sqrt{a}}=\dfrac{b}{\sqrt{a}} \times \dfrac{\sqrt{a}}{\sqrt{a}}=\dfrac{b \sqrt{a}}{a}$]

=$\dfrac{10(\sqrt{15})}{5}$ [using the rule $\sqrt{a} \times \sqrt{b}=\sqrt{(a \times b)}$]

=2$\sqrt{15}$ [since $\dfrac{10}{5}=2$]

Rule 4:

$a \sqrt{c}\pm b\sqrt{c}=(a \pm b) \sqrt{c}$

Example:

Simplify $5 \sqrt{6}+ 4 \sqrt{6}$ :
solution
$5 \sqrt{6}+ 4 \sqrt{6}=(5+4) \sqrt{6}$
=9$ \sqrt{6}$

Exercise :

44452.Simplify $\dfrac{2\sqrt{3}}{5}+\sqrt{108}$
$\dfrac{3 \sqrt{3}}{5}$
$\dfrac{32 \sqrt{3}}{5}$
$\dfrac{32 \sqrt{2}}{5}$
$\dfrac{32 \sqrt{3}}{15}$
Explanation:

$\dfrac{2\sqrt{3}}{5}+\sqrt{108}=\dfrac{2\sqrt{3}+5\sqrt{108}}{5}$ [find the LCD to add]

=$\dfrac{2\sqrt{3}+5(\sqrt{36}\times \sqrt{3})}{5}$ [using the rule $\sqrt{(a \times b)}=\sqrt{a} \times \sqrt{b}$]

=$\dfrac{2\sqrt{3}+30 \sqrt{3}}{5}$ [Evaluate \sqrt{36} and multiply to 5]

=$\dfrac{(2+30) \sqrt{3}}{5}$ [using the rule $a \sqrt{c}\pm b\sqrt{c}=(a \pm b) \sqrt{c}$]

=$\dfrac{32 \sqrt{3}}{5}$

Rule 5:

$\dfrac{c}{a+b \sqrt{n}}$ multiply top and bottom by $a-b \sqrt{n}$
Following this rule enables you to rationalise the denominator.

Example:

Rationalise : $\dfrac{3}{2+ \sqrt{2}}$

solution
$\dfrac{3}{2+ \sqrt{2}}=\dfrac{3}{2+ \sqrt{2}} \times \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}}$ [multiply the numerator and denominator by $2 - \sqrt{2}$]
=$\dfrac{6-3\sqrt{2}}{4-2}$
=$\dfrac{6-3\sqrt{2}}{2}$

Exercise :

44453.Rationalise the denominaor in $\dfrac{7}{\sqrt{3}+2}$
$-7\sqrt{3}+14$
$7\sqrt{3}+14$
$-7\sqrt{3}-14$
$7\sqrt{3}-14$
Explanation:
$\dfrac{7}{\sqrt{3}+2}=\dfrac{7}{\sqrt{3}+2} \times \dfrac{\sqrt{3}-2}{\sqrt{3}-2}$ [multiply the numerator and denominator by $\sqrt{3}-2$]

=$\dfrac{7\sqrt{3}-14}{3-4}$

=$\dfrac{7\sqrt{3}-14}{-1}$

=$-7\sqrt{3}+14$

Rule 6:

$\dfrac{c}{a-b\sqrt{n}}$ multiply top and bottom by $a+b\sqrt{n}$
Following this rule enables you to rationalise the denominator.

Example:

Rationalise : $\dfrac{3}{2-\sqrt{2}}$

solution
$\dfrac{3}{2-\sqrt{2}}=\dfrac{3}{2-\sqrt{2}} \times \dfrac{2 + \sqrt{2}}{2+\sqrt{2}}$[multiply the numerator and denominator by $2 + \sqrt{2}$]
=$\dfrac{6 + 3\sqrt{2}}{4-2}$
=$\dfrac{6 + 3\sqrt{2}}{2}$

Exercise :

44454.Rationalise the denominator in $\dfrac{2}{1-\sqrt{2}}$
$2+2\sqrt{2}$
$2-2\sqrt{2}$
$-2+2\sqrt{2}$
$-2-2\sqrt{2}$
Explanation:

$\dfrac{2}{1-\sqrt{2}} = \dfrac{2}{1-\sqrt{2}} \times \dfrac{1+\sqrt{2}}{1+\sqrt{2}}$ [multiply the numerator and denominator by $1+\sqrt{2}$]

=$\dfrac{2+2\sqrt{2}}{1-2}$

=$\dfrac{2+2\sqrt{2}}{-1}$

=$-2-2\sqrt{2}$

Indices & the Law of Indices


Introduction

Indices are a useful way of more simply expressing large numbers. They also present us with many useful properties for manipulating them using what are called the Law of Indices.

What are Indices?

The expression 25 is defined as follows:
25=2 x 2 x 2 x 2 x 2
We call "2" the base and "5" the index.

Law of Indices

To manipulate expressions, we can consider using the Law of Indices. These laws only apply to expressions with the same base, for example, 34 and 32 can be manipulated using the Law of Indices, but we cannot use the Law of Indices to manipulate the expressions 35 and 57 as their base differs (their bases are 3 and 5, respectively).

Six rules of the Law of Indices


Rule 1:

a0=1
Any number, except 0, whose index is 0 is always equal to 1, regardless of the value of the base.

Example:

solution

Simplify 20

solution
20=1

Exercise :

44455.Simplify 3m8n3 $\div$ (3m8n3)0
3m8n3
0
1
None of these
Explanation:

3m8n3 $\div$ (3m8n3)0=3m8n3 $\div$ 1 [using the rule a0=1]

=3m8n3

Rule 2:

a-m =$\dfrac{1}{a^{m}}$

Example:

Simplify: 2-2

solution
2-2=$\dfrac{1}{2^{2}}$ [using a-m =$\dfrac{1}{a^{m}}$ ]
=$\dfrac{1}{4}$

Exercise :

44456.Simplify : (3a)-2
$\dfrac{1}{9a^{-2}}$
$\dfrac{1}{3a^{2}}$
$\dfrac{1}{9a^{2}}$
$\dfrac{-1}{9a^{2}}$
Explanation:

(3a)-2=$\dfrac{1}{(3a)^{2}}$ [Using Rule a-m =$\dfrac{1}{a^{m}}$]

=$\dfrac{1}{3^{2}a^{2}}$ [Using Rule (am)n=amn]

=$\dfrac{1}{9a^{2}}$ [Evaluate 32]

Rule 3:

am x an = am + n
To multiply expressions with the same base, copy the base and add the indices.

Example:

Simplify : 5 x 53 : (note: 5=51)

Solution :
51 x 53 =51+3 [using am x an = am + n ]
=54
=5x5x5x5
=625

Exercise :

44457.Simplify : a4b2 x a2b2
a4b4
a6b4
a4b2
a2b4
Explanation:

a4b2 x a2b2=a4+2b2+2 [ Using Rule am x an = am + n ]

= a6b4

Rule 4:

$a^{m}\div a^{n}= a^{m - n}$
To divide expressions with the same base, copy the base and subtract the indices.

Example:

Simplify : 5(y9$\div$y5)

Solution :
5(y9$\div$y5) =5(y9-5) [using $\dfrac{a^{m}}{a^{n}}= a^{m - n}$]
=5y4

Exercise :

44458.Simplify : 5(8x4 $\div$ 2x6)
$\dfrac{8}{x^{2}}$
$\dfrac{2}{x^{2}}$
$\dfrac{20}{x^{2}}$
$\dfrac{20}{x^{4}}$
Explanation:

5(8x4 $\div$ 2x6) [since a$\div$b=$\dfrac{a}{b}$ , split the equation to make it easier to solve]

=5(8 $\div$ 2)(x4 $\div$ x6) [Get the quotient of 8 and 2.]

=5(4)(x4-6) [Using Rule $a^{m}\div a^{n}= a^{m - n}$]

=20(x-2) [Get the product of 5 and 4]

=$\dfrac{20}{x^{2}}$ [Using Rule $a^{-m}=\dfrac{1}{a^{m}}$]

Rule 5:

$(a^{m})^{n} = a^{mn}$
To raise an expression to the nth index, copy the base and multiply the indices.

Example:

Simplify : $(y^{2})^{6} $

Solution :
$(y^{2})^{6} = y^{2\times6}$ [using $(a^{m})^{n} = a^{mn}$]
=y12

Exercise :

44459.Simplify : $\left( \dfrac{5a}{b^{2}}\right)^{2}$
$\dfrac{5a^{2}}{b^{4}}$
$\dfrac{25a^{2}}{b^{2}}$
$\dfrac{25a^{2}}{b^{4}}$
$\dfrac{5a^{2}}{b^{2}}$
Explanation:

$\left( \dfrac{5a}{b^{2}}\right)^{2} =\dfrac{ 5^{2}a^{2}}{b^{2 \times 2}}$ [Using Rule $(a^{m})^{n} = a^{mn}$]

=$\dfrac{25a^{2}}{b^{4}}$ [Evaluate 52]

Rule 6:

am/n=$\sqrt[n]{a^{m}}=(\sqrt[n]{a})^{m}$

Example:

Simplify : 1252/3

Solution :
1252/3=$(\sqrt[3]{125})^{2}$ [Using am/n=$\sqrt[n]{a^{m}}=(\sqrt[n]{a})^{m}$]
=52 [Recognize cube root of 125 is 5]
=25

Exercise :

44460.Simplify : 5*82/3
5
8
40
20
Explanation:

5*82/3

= 5*$(\sqrt[3]{8})^{2}$ [using Rule am/n=$\sqrt[n]{a^{m}}=(\sqrt[n]{a})^{m}$]

=5*22 [Get the value of $\sqrt[3]{8}$]

=5*4 [Get the value of 22]

=20 [Get the product of 5 and 4]

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