# Surds – Indices

#### Chapter 10

5 Steps - 3 Clicks

# Surds – Indices

### Introduction

Surds – Indices deals with problems on surds and indices involved with laws of surds and laws of indices respectively.

### Methods

Surds:

Surds are the  irrational numbers that contain the radical sign $$\sqrt{}$$. The set of irrational numbers contain numbers such as $$\sqrt{2}$$, $$\sqrt[3]{2}$$, $$\pi$$, etc.

• Surds can be added or subtracted only if they are like surds.
Example: $$\sqrt{2}$$ + 5$$\sqrt{2}$$ = 6$$\sqrt{2}$$

• Surds can be multiplied using $$\sqrt{x}$$ x $$\sqrt{x}$$= $$x$$ and $$\sqrt{x}$$ x $$\sqrt{y}$$ = $$\sqrt{xy}$$

• To rationalize the denominator of a fraction, we need to convert it into equivalent fraction.
Example: $$\frac{1}{\sqrt{7}}$$ = $$\frac{1}{\sqrt{7}}$$ x $$\frac{\sqrt{7}}{\sqrt{7}}$$ = $$\frac{\sqrt{7}}{7}$$

• A surd can be expressed in index form as a fractional index.
Example: $$\sqrt[n]{a}$$ = $$a^{(\frac{1}{n})}$$
here $$\sqrt[n]{a}$$ is in surd form and $$a^{(\frac{1}{n})}$$ is in index form.

Indices

Indices are a useful way expressing large numbers with the help of powers or also known as indices. Exponent is another commonly used term for a power. In the below example the power/exponent of 4 is 6.

Example: 4096 is obtained by multiplying 4 by itself for 6 times and so 4096 can be expresses as:$$4^6$$  where 4 x 4 x 4 x 4 x 4 x 4 = 4096

• Laws of indices can be applied only to the expressions having equal bases.
Example: $$2^6$$, $$2^4$$, $$2^8, etc$$

Example 1:
Show that for any positive real number p, the expression $$a^{-p}$$ is equivalent to $$\frac{1}{a^{p}}$$.

Solution:

We proceed with the following manipulation – $$a^{-p}$$ = $$a^{0 – p}$$

Using Law 2 i.e. $$\frac{a^{m}}{a^{n}}$$ = $$a^{(m – n)}$$, we can rewrite the above expression as –

$$\frac{a^{0}}{a^{p}}$$

= $$\frac{1}{a^{p}}$$, which is the required result.

Example 2:
Simplify and evaluate $$(\frac{16}{81})^{-(\frac{3}{4})}$$

Solution:

Using the laws of indices and some manipulation –

$$(\frac{16}{81})^{-(\frac{3}{4})}$$ = $$\frac{1}{(\frac{16}{81})^{\frac{3}{4}}}$$

= $$(\frac{81}{16})^{\frac{3}{4}}$$

= $$((\frac{81}{16})^{\frac{1}{4}})^{3}$$

= $$(\frac{81^{\frac{1}{4}}}{16^{\frac{1}{4}}})^{3}$$

= $$(\frac{3}{2})^{3}$$

= $$\frac{3^{3}}{2^{3}}$$

= $$\frac{27}{8}$$

Example 3:
Simplify the expression $$y = x^{a – b} \times x^{b – c} \times x^{c – a} \times x^{-a – b}$$

Solution:

Using the laws of indices:

$$y = x^{a – b} \times x^{b – c} \times x^{c – a} \times x^{-a – b}$$

$$y = x^{(a – b) + (b – c) + (c – a) + (-a – b)}$$

$$y = x^{-a – b}$$

$$y = \frac{1}{x^{a + b}}$$

Laws of Surds/ Rules of Surds:

Example:
Simplify $$\sqrt{18}$$

Solution:

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

$$\sqrt{18}$$ = $$\sqrt{3^{2} \times 2}$$

= $$\sqrt{3^{2}}$$ x $$\sqrt{2}$$ (Using the rule $$\sqrt{(a \times b)}$$ = $$\sqrt{a}$$ x $$\sqrt{b}$$)

= 3$$\sqrt{2}$$

Example:
Simplify $$\sqrt{\frac{12}{121}}$$

Solution:

$$\sqrt{\frac{12}{121}}$$ = $$\frac{\sqrt{12}}{\sqrt{121}}$$ (Using the rule $$\sqrt{\frac{a}{b}}$$ = $$\frac{\sqrt{a}}{\sqrt{b}}$$)

= $$\frac{\sqrt{2^{2} \times 3}}{11}$$ (Since 4 is the largest perfect square factor of 12)

= $$\frac{\sqrt{2^{2} \times \sqrt{3}}}{11}$$ (Using the rule $$\sqrt{(a \times b)}$$ = $$\sqrt{a}$$ x $$\sqrt{b}$$)

= $$\frac{2\sqrt{3}}{11}$$

Example:
Rationalise $$\frac{5}{\sqrt{7}}$$

Solution:

$$\frac{5}{\sqrt{7}}$$ = $$\frac{5}{\sqrt{7}}$$ x $$\frac{\sqrt{7}}{\sqrt{7}}$$ (Multiply both numerator and denominator by $$\sqrt{7}$$)

= $$5\frac{\sqrt{7}}{7}$$

Example:
Simplify $$5\sqrt{6} + 4\sqrt{6}$$

Solution:

$$5\sqrt{6} + 4\sqrt{6}$$ = (5 + 4)$$\sqrt{6}$$ (Using the rule $$a\sqrt{c} \pm b\sqrt{c} = (a \pm b)\sqrt{c}$$)

= $$9\sqrt{6}$$

Following this rule enables to rationalise the denominator.

Example:
Rationalise $$\frac{3}{2 + \sqrt{2}}$$

Solution:

$$\frac{3}{2 + \sqrt{2}}$$ = $$\frac{3}{2 + \sqrt{2}}$$ x $$\frac{2 – \sqrt{2}}{2 – \sqrt{2}}$$ (Multiply the numerator and denominator by $$2 – \sqrt{2}$$)

= $$\frac{6 – 3\sqrt{2}}{4 – 2}$$

= $$\frac{6 – 3\sqrt{2}}{2}$$

Following this rule enables to rationalise the denominator.

Example:
Rationalise $$\frac{3}{2 – \sqrt{2}}$$

Solution:

$$\frac{3}{2 – \sqrt{2}}$$ = $$\frac{3}{2 – \sqrt{2}}$$ x $$\frac{2 + \sqrt{2}}{2 + \sqrt{2}}$$ (Multiply the numerator and denominator by $$2 + \sqrt{2}$$)

= $$\frac{6 + 3\sqrt{2}}{4 – 2}$$

= $$\frac{6 + 3\sqrt{2}}{2}$$

### Formulae

1. Laws of Indices/ Laws of Exponents:

• $$a^m$$ x $$a^n$$ = $$a^{m + n}$$

• $$\frac{a^m}{a^n}$$ = $$a^{m – n}$$

• $$({a^m})^{n}$$ = $$a^{mn}$$

• $$(ab)^m$$ = $$a^m$$ x $$b^m$$

• $$(\frac{a}{b})^n$$ = $$\frac{a^n}{b^n}$$

• $$a^0$$ = 1

2. Laws of Surds:

• $$\sqrt[n]{a}$$ = $$a^{(\frac{1}{n})}$$

• $$\sqrt[n]{ab}$$ = $$\sqrt[n]{a}$$ x $$\sqrt[n]{b}$$

• $$\sqrt[n]{\frac{a}{b}}$$ = $$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$$

• $$(\sqrt[n]{a})^n$$ = a

• $$\sqrt[m]{\sqrt[n]{a}}$$ = $$\sqrt[mn]{a}$$

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

### Samples

1 Find $$\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}$$ = ?

Solution:

Assume $$\sqrt{{8}\sqrt{{8}{\sqrt{{8…..}}}}}$$ = $$x$$

Now, consider $$\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}$$

⇒$$\sqrt{8 x}$$ = $$x$$

⇒$$x^2$$ = 8$$x$$

⇒$$x$$ = 8

Therefore, the value of $$\sqrt{{8}\sqrt{{8}\sqrt{{8}\sqrt{8…..∝}}}}$$ = 8

2. Simplify the value of $$(125)^{(\frac{2}{3})}$$

Solution:

Consider$$(125)^{(\frac{2}{3})}$$

=$$(5^3)^{(\frac{2}{3})}$$

=$$5^2$$

=25

Therefore, the value of $$(125)^{(\frac{2}{3})}$$ = 25

3. Find the value of [$$(8)^{\frac{5}{3}}$$ + $$(8)^{\frac{-5}{3}}$$]?

Solution:

Consider [$$(8)^{\frac{5}{3}}$$ + $$(8)^{\frac{-5}{3}}$$]

= [$$(2^3)^{\frac{5}{3}}$$ + $$(2^3)^{\frac{-5}{3}}$$]

= [$$(2^5)$$ + $$(2^{-5})$$]

= $$(2^5)$$ + $$\frac{1}{(2^5)}$$

= $$\frac{2^5 + 1}{2^5}$$

= $$\frac{1024 + 1}{32}$$

= $$\frac{1025}{32}$$

Therefore, [$$(8)^{\frac{5}{3}}$$ + $$(8)^{\frac{-5}{3}}$$] = $$\frac{1025}{32}$$

4. What will be the quotient if $$(x^{-1}-1)$$ is divided by $$(x – 1)$$?

Solution:

Given that

Dividend = $$(x^{-1}-1)$$

Divisor = $$(x – 1)$$

Now, Consider $$\frac{(x^{-1}-1)}{(x – 1)}$$

= $$\frac{\frac{1}{x} -1}{(x – 1)}$$

= $$\frac{\frac{(1-x)}{x}}{(x – 1)}$$

= $$\frac{(1 – x)}{x}$$ x $$\frac{1}{(x – 1)}$$

= $$\frac{(1-x)}{(-x)(1-x)}$$

= –$$\frac{1}{x}$$

Therefore, the required quotient = –$$\frac{1}{x}$$

5. Which is larger among $$\sqrt[4]{6}$$, $$\sqrt[3]{4}$$, $$\sqrt[2]{5}$$ ?

Solution:

Given surds are $$\sqrt[4]{6}$$, $$\sqrt[3]{4}$$, $$\sqrt[2]{5}$$

Surds are in the order of 4, 3, 2 respectively.

L.C.M. of 4, 3, 2 is 12

Now, change the each given surd in the order of 12

$$\sqrt[4]{6}$$ = $${6}^{(\frac{1}{4})}$$ = $${6}^{(\frac{1}{4} * \frac{3}{3})}$$ = $${6}^{(\frac{3}{12})}$$ = $${(6^3)}^{(\frac{1}{12})}$$ = $${(216)}^{(\frac{1}{12})}$$

$$\sqrt[3]{4}$$ =$${4}^{(\frac{1}{3})}$$ = $${4}^{(\frac{1}{3} * \frac{4}{4})}$$ = $${4}^{(\frac{4}{12})}$$ = $${(4^4)}^{(\frac{1}{12})}$$ = $${(256)}^{(\frac{1}{12})}$$

$$\sqrt[2]{5}$$ =$${5}^{(\frac{1}{2})}$$ = $${5}^{(\frac{1}{2} * \frac{6}{6})}$$ = $${5}^{(\frac{6}{12})}$$ = $${(5^6)}^{(\frac{1}{12})}$$ = $${(3125)}^{(\frac{1}{12})}$$

Hence, $${(3125)}^{(\frac{1}{12})}$$ > $${(256)}^{(\frac{1}{12})}$$ > $${(216)}^{(\frac{1}{12})}$$

Therefore, $$\sqrt[2]{5}$$ > $$\sqrt[3]{4}$$ > $$\sqrt[4]{6}$$