# Concept: Surds & Indices

CONTENTS

INDICES

Indices Rules

Note: a, b, m, n ≠0, ±1

1. 1m = 1

2. 0m = 0

3. a1 = a

4. a0 = 1

5. am × an = am+n

6. am ÷ an = am-n

7. am × bm = (a × b)m

8. am ÷ bm = (a ÷ b)m

9. ${\mathrm{a}}^{1/\mathrm{m}}$ = $\sqrt[\mathrm{m}]{\mathrm{a}}$

10. ${\mathrm{a}}^{-\mathrm{m}}$ = $\frac{1}{{\mathrm{a}}^{\mathrm{m}}}$

11. ${\left({\mathrm{a}}^{\mathrm{m}}\right)}^{n}$ = ${\mathrm{a}}^{\mathrm{m}×\mathrm{n}}$

12. ${\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\frac{1}{n}}$ = ${\mathrm{a}}^{\frac{m}{n}}$

13. If am = an, then m = n

14. If am = bm, then

• a = b, if m is odd
• a = ± b, if m is even

SURDS

A surd is an irrational number which is represented as the nth root of a rational number. It is written as $\sqrt[\mathrm{n}]{\mathrm{a}}$.

Therefore √5,∛6,∜7 are surds.

Note: ∛27 is not a surd as it can be simplified as 3 which is not an irrational number.

$\sqrt[\mathrm{n}]{\mathrm{a}}$ is read as nth root of a, where n is the index of the surd.

∛5 is read as the third root (or the cube root) of 5 and ∜7 is read as the fourth root of 7. If the index of the radical is not given, it is assumed to be 2.

Remember
• A surd is an irrational number
• A surd is of the type $\sqrt[\mathrm{n}]{\mathrm{a}}$, where a is a positive rational number and n is a natural number greater than 1.

Hence, rational numbers like √9 or ∛125 are not surds, because these are not unresolved. √9 is equal to +3 and ∛125 is equal to 5.

Again, √2 × √8 is equal to √16 or +4, hence it is not a surd.

Pure Surds and Mixed Surds

In a surd of the form b$\sqrt[\mathrm{n}]{\mathrm{a}}$, b is called the coefficient of the surd.

Surds with coefficient equal to 1 are called pure surds while surds with coefficients other than unity are known as mixed surds.

For example: √3,∛5,∜11 are pure surds whereas 2√5,3∛11 and 5∜7 are mixed surds.

The following operations can be performed with surds:

• $\sqrt[\mathrm{n}]{\mathrm{a}}$ × $\sqrt[\mathrm{n}]{\mathrm{b}}$ =  $\sqrt[\mathrm{n}]{\mathrm{a}×\mathrm{b}}$
• $\frac{\sqrt[\mathrm{n}]{\mathrm{a}}}{\sqrt[\mathrm{n}]{\mathrm{b}}}$ =  $\sqrt[\mathrm{n}]{\frac{\mathrm{a}}{\mathrm{b}}}$
• $x×\sqrt[\mathrm{n}]{\mathrm{a}}$ × $y×\sqrt[\mathrm{n}]{\mathrm{a}}$ = $\left(x+y\right)×\sqrt[\mathrm{n}]{\mathrm{a}}$

Rationalization of Surds

Surd is an irrational number. The process of converting a surd to a rational number is called rationalization. To rationalize, we multiply the surd with a rationalizing factor. When the rationalizing factor is multiplied with the surd, we get a rational number.

For example, to rationalize √3, we multiply it with √3, (the rationalizing factor) to get √3 × √3 = 3 and to rationalize √8 we multiply it with √2 to get √8 × √2 = √16 = 4.

It is to be noted that there are always multiple rationalizing factors available. In the above example, we could have also multiplied √3 with √27 to rationalize it, since √3 × √27 = √81 = 9

• Rationalizing factor of a sum of surds

Consider the term √a + √b which is the sum of two surds. Conjugate of (√a + √b) is (√a - √b).

Similarly, the conjugate of (√a - √b) is (√a + √b).

The conjugate is the rationalizing factor (RF) for a sum of surds of the form (√a + √b).

When we multiply these two terms we get, (√a + √b) × (√a - √b) = (a – b) which is a rational number.

Note: Conjugate of (a + √b) is (a - √b) and that of (a - √b) is (a + √b)

• Use of Rationalization

Rationalization is mostly used to rationalize the denominator of an expression when the denominator is sum of surds.

Example: Rationalize $\frac{4}{\sqrt{7}+\sqrt{3}}$

Solution:
$\frac{4}{\sqrt{7}+\sqrt{3}}$ = $\frac{4}{\sqrt{7}+\sqrt{3}}$ × $\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}$

$\frac{4}{\sqrt{7}+\sqrt{3}}$ = $\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$

$\frac{4}{\sqrt{7}+\sqrt{3}}$ = $\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}$

$\frac{4}{\sqrt{7}+\sqrt{3}}$ = √7 - √3

Example: Rationalize $\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$

Solution:
Given, $\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$

We first multiply and divide with (√2 + √4) - √5

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ × $\frac{\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}}{\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]}{{\left(\sqrt{2}+\sqrt{4}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]}{1+4\sqrt{2}}$

Now, the rationalizing factor will be 1 - 4√2

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]}{1+4\sqrt{2}}$ × $\frac{1-4\sqrt{2}}{1-4\sqrt{2}}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]×\left(1-4\sqrt{2}\right)}{{1}^{2}-{\left(4\sqrt{2}\right)}^{2}}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]×\left(1-4\sqrt{2}\right)}{-31}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = $\frac{31\left[\left(\sqrt{2}+\sqrt{4}\right)-\sqrt{5}\right]×\left(1-4\sqrt{2}\right)}{-31}$

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = [√2 + √4 - √5](4√2 - 1)

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = [√2 + √4 - √5](4√2 - 1)

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = 8 + 4√8 - 4√10 - √2 - √4 + √5

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = 8 + 8√2 - 4√10 - √2 - 2 + √5

$\frac{31}{\sqrt{2}+\sqrt{4}+\sqrt{5}}$ = 6 + 7√2 + √5 - 4√10

Comparison of Surds

It is difficult to compare two surds of different indices. The strategy we follow is to change both surds to the same order. We can then compare them by the value of their radicands.

The new index of the two surds is the LCM of the original indices of the surds. The following example should make it clear.

Example: Compare √2 and ∛3

Solution:
The order of the surds is 2 and 3, hence we convert both the surds to surds of order 6 (which is the LCM of 2 and 3).

√2 = ${2}^{1/2}$${\left({2}^{3}\right)}^{1/6}$${8}^{1/6}$

and, ∛3 = ${3}^{1/3}$${\left({3}^{2}\right)}^{1/6}$${9}^{1/6}$

Now since, 8 < 9, we can say that ${8}^{1/6}$ < ${9}^{1/6}$

Hence, √2 < ∛3

Calculating Square Root of Surds

If there exists a square root of a surd of the type a + √b , then it will be of the form √x + √y . We can equate the square of √x + √y to a + √b and then solve for x and y.

Note: When there is an equation with rational and irrational terms, the rational part on the left hand side is equal to the rational part on the right hand side and, the irrational part on the left hand side is equal to the irrational part on the right hand side of the equation.

Example: Calculate square root of 5 + 2√6

Solution:
Square root of 5 + 2√6 will be of the form √x + √y.

∴ (√x + √y)2 = 5 + 2√6

⇒ x + y + 2√xy = 5 + 2√6

Now, rational part of RHS = rational part on LHS
⇒ x + y = 5   …(1)

Also, irrational part of RHS = irrational part on LHS
⇒ 2√xy = 2√6
⇒ xy = 6   …(2)

Solving (1) and (2), we get
x = 3 and y = 2 or x = 2 and y = 3.

In both cases square root of 5 + 2√6 = √2 + √3

Example: Calculate positive square root of 5 - 2√6

Solution:
Square root of 5 - 2√6 will be of the form √x - √y.

∴ (√x - √y)2 = 5 - 2√6

⇒ x + y - 2√xy = 5 - 2√6

Now, rational part of RHS = rational part on LHS
⇒ x + y = 5   …(1)

Also, irrational part of RHS = irrational part on LHS
⇒ - 2√xy = - 2√6
⇒ xy = 6   …(2)

Solving (1) and (2), we get
x = 3 and y = 2 or x = 2 and y = 3.

The square root of 5 - 2√6 will be either √3 - √2 or √2 - √3

We will reject √2 - √3 as it is negative

∴ Square root of 5 - 2√6 is √3 - √2

If there exists a square root of a surd of the type a + √b + √c + √d , then it will be of the form √x + √y + √z. [x, y, z ≥ 0]

Example: Calculate square root of 6 + 2√2 + 2√3 + 2√6

Solution:
Square root of 6 + 2√2 + 2√3 + 2√6  will be of the form √x + √y + √z.

∴ (√x + √y + √z)2 = 6 + 2√2 + 2√3 + 2√6

⇒ (x + y + z) + 2√xy + 2√xz + 2√yz = 6 + 2√2 + 2√3 + 2√6

⇒ x + y + z = 6   …(1)
⇒ xy = 2   …(2)
⇒ xz = 3   …(3)
⇒ yz = 6   …(4)

⇒ x = 1, y = 2 and z = 3

∴ Square root of 6 + 2√2 + 2√3 + 2√6  will be √1 + √2 + √3 = 1 + √2 + √3

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