9th Class Mathematics - Chapter 1

Exercise 1.2 Solved

Real and Complex Numbers

9th Class Mathematics - Chapter 1 - Solved Reference Guide

Question 1 Rationalization

Rationalize the denominator of the following:

(i)

13 4 + \sqrt 3

Solution

To rationalize the denominator, multiply and divide by the conjugate of the denominator, which is $4 - \sqrt 3$ :

13 4 + \sqrt 3 \times 4 - \sqrt 3 4 - \sqrt 3

Apply the algebraic identity $(a + b)(a - b) = a 2 - b 2$ to the denominator:

= 13(4 - \sqrt 3) (4) 2 - (\sqrt 3) 2

= 13(4 - \sqrt 3) 16 - 3

= 13(4 - \sqrt 3) 13

= 4 - \sqrt 3

(ii)

\sqrt 2 + \sqrt 5 \sqrt 3

Solution

Multiply and divide by the radical in the denominator, which is $\sqrt 3$ :

\sqrt 2 + \sqrt 5 \sqrt 3 \times \sqrt 3 \sqrt 3

= (\sqrt 2 + \sqrt 5) \times \sqrt 3 (\sqrt 3) 2

Distribute $\sqrt 3$ in the numerator: $\sqrt a \cdot \sqrt b = \sqrt ab$ :

= \sqrt 2 \times 3 + \sqrt 5 \times 3 3

= \sqrt 6 + \sqrt 15 3

(iii)

3 \sqrt 5

Solution

Multiply and divide by $\sqrt 5$ :

3 \sqrt 5 \times \sqrt 5 \sqrt 5

= 3 \times \sqrt 5 (\sqrt 5) 2

= 3 \sqrt 5 5

(iv)

6 - 4 \sqrt 2 6 + 4 \sqrt 2

Solution

Multiply and divide by the conjugate $6 - 4 \sqrt 2$ :

6 - 4 \sqrt 2 6 + 4 \sqrt 2 \times 6 - 4 \sqrt 2 6 - 4 \sqrt 2

Numerator becomes a perfect square: $(6 - 4 \sqrt 2) 2$ . Denominator simplifies via difference of squares:

= (6 - 4 \sqrt 2) 2 (6) 2 - (4 \sqrt 2) 2

Expand the numerator using $(a - b) 2 = a 2 + b 2 - 2ab$ :

= (6) 2 + (4 \sqrt 2) 2 - 2(6)(4 \sqrt 2) 36 - 16(2)

= 36 + 16(2) - 48 \sqrt 2 36 - 32

= 36 + 32 - 48 \sqrt 2 4

= 68 - 48 \sqrt 2 4

Factor out 4 from the numerator to simplify:

= 4(17 - 12 \sqrt 2) 4

= 17 - 12 \sqrt 2

(v)

\sqrt 3 - \sqrt 2 \sqrt 3 + \sqrt 2

Solution

Multiply and divide by the conjugate $\sqrt 3 - \sqrt 2$ :

\sqrt 3 - \sqrt 2 \sqrt 3 + \sqrt 2 \times \sqrt 3 - \sqrt 2 \sqrt 3 - \sqrt 2

= (\sqrt 3 - \sqrt 2) 2 (\sqrt 3) 2 - (\sqrt 2) 2

Expand the numerator: $(a - b) 2 = a 2 + b 2 - 2ab$ :

= (\sqrt 3) 2 + (\sqrt 2) 2 - 2(\sqrt 3)(\sqrt 2) 3 - 2

= 3 + 2 - 2 \sqrt 6 1

= 5 - 2 \sqrt 6

(vi)

4 \sqrt 3 \sqrt 7 + \sqrt 5

Solution

Multiply and divide by the conjugate $\sqrt 7 - \sqrt 5$ :

4 \sqrt 3 \sqrt 7 + \sqrt 5 \times \sqrt 7 - \sqrt 5 \sqrt 7 - \sqrt 5

= 4 \sqrt 3 (\sqrt 7 - \sqrt 5) (\sqrt 7) 2 - (\sqrt 5) 2

= 4 \sqrt 3 (\sqrt 7 - \sqrt 5) 7 - 5

= 4 \sqrt 3 (\sqrt 7 - \sqrt 5) 2

= 2 \sqrt 3 (\sqrt 7 - \sqrt 5)

Distribute $2 \sqrt 3$ into the terms:

= 2 \sqrt 3 \times 7 - 2 \sqrt 3 \times 5

= 2 \sqrt 21 - 2 \sqrt 15

Question 2 Laws of Exponents

Simplify the following:

(i)

(8116) -3/4

Solution

Use the property $(a b) -m = (b a) m$ to make the exponent positive:

= (1681) 3/4

Represent 16 and 81 as powers of 2 and 3:

= (2434) 3/4

= [ (23) 4] 3/4

Multiply the powers: $(a m) n = a m \cdot n$ :

= (23) 4 \times 3/4

= (23) 3

= 2 \times 2 \times 2 3 \times 3 \times 3 = 827

(ii)

(34) -2 \div (49) 3 \times 1627

Solution

Convert exponents to positive and write bases in prime factors ( $4 = 2 2$ , $9 = 3 2$ , $16 = 2 4$ , $27 = 3 3$ ):

= (43) 2 \div (49) 3 \times 1627

= (223) 2 \div (2232) 3 \times 2433

= 2432 \div 2636 \times 2433

Change division to multiplication by taking the reciprocal:

= 2432 \times 3626 \times 2433

Group numerator and denominator terms: $a m \cdot a n = a m+n$ :

= 2 4 + 4 \times 3 6 26 \times 3 2 + 3

= 28 \times 3 6 26 \times 3 5

Apply subtraction of powers for division: $a m / a n = a m-n$ :

= 2 8 - 6 \times 3 6 - 5

= 22 \times 3 1

= 4 \times 3 = 12

(iii)

(0.027) -1/3

Solution

Convert decimal to fraction:

= (271000) -1/3

Invert fraction to make the exponent positive:

= (100027) 1/3

Express 1000 and 27 as cubes ( $103$ and $33$ ):

= (10333) 1/3

= [ (103) 3] 1/3

= (103) 3 \times 1/3

= 103 = 313

(iv)

7 \sqrt x 14 \cdot y 21 \cdot z 35 y 14 \cdot z 7

Solution

Simplify variables inside the radical by subtracting denominator powers:

= 7 \sqrt x 14 \cdot y 21 - 14 \cdot z 35 - 7

= 7 \sqrt x 14 \cdot y 7 \cdot z 28

Convert the 7th root to fractional exponent $1/7$ :

= (x 14 \cdot y 7 \cdot z 28) 1/7

Distribute the exponent to each variable:

= x 14 \times 1/7 \cdot y 7 \times 1/7 \cdot z 28 \times 1/7

= x 2 y z 4

(v)

5 \cdot (25) n+1 - 25 \cdot (5) 2n 5 \cdot (5) 2n+3 - (25) n+1

Solution

Express 25 as base 5 ( $25 = 5 2$ ):

= 5 \cdot (5 2) n+1 - 5 2 \cdot 5 2n 5 \cdot 5 2n+3 - (5 2) n+1

= 5 \cdot 5 2n+2 - 5 2n+2 5 2n+4 - 5 2n+2

Factor out the smallest common term, which is $5 2n+2$ . Rewrite $5 2n+4 = 5 2n+2 \cdot 5 2$ :

= 5 2n+2 (5 - 1) 5 2n+2 \cdot 5 2 - 5 2n+2 \cdot 1

= 5 2n+2 (5 - 1) 5 2n+2 (5 2 - 1)

Cancel out $5 2n+2$ from numerator and denominator:

= 5 - 1 25 - 1

= 424 = 16

(vi)

16 x+1 + 20(4 2x) 2 x-3 \times 8 x+2

Solution

Express all bases in prime base 2 ( $16 = 2 4$ , $20 = 5 \cdot 2 2$ , $4 = 2 2$ , $8 = 2 3$ ):

= (2 4) x+1 + (5 \cdot 2 2)(2 2) 2x 2 x-3 \times (2 3) x+2

= 2 4x+4 + 5 \cdot 2 2 \cdot 2 4x 2 x-3 \cdot 2 3x+6

= 2 4x+4 + 5 \cdot 2 4x+2 2 x-3 + 3x + 6

= 2 4x+4 + 5 \cdot 2 4x+2 2 4x+3

Factor out $2 4x+2$ from both the numerator and the denominator:

= 2 4x+2 (2 2) + 5 \cdot 2 4x+2 2 4x+2 \cdot 2 1

= 2 4x+2 (2 2 + 5) 2 4x+2 (2)

Cancel out $2 4x+2$ :

= 22 + 5 2

= 4 + 5 2 = 92

(vii)

64 -2/3 \div 9 -3/2

Solution

Write as a fraction and make exponents positive using $a -m b -n = b n a m$ :

= 64 -2/3 9 -3/2 = 9 3/2 64 2/3

Write bases in prime power form ( $9 = 3 2$ , $64 = 2 6$ ):

= (3 2) 3/2 (2 6) 2/3

= 3 2 \times 3/2 2 6 \times 2/3

= 3324

= 2716

(viii)

3 n \times 9 n+1 3 n-1 \times 9 n-1

Solution

Write $9 = 3 2$ to get a single base 3:

= 3 n \times (3 2) n+1 3 n-1 \times (3 2) n-1

= 3 n \times 3 2n+2 3 n-1 \times 3 2n-2

Sum exponents in numerator and denominator: $3 a \cdot 3 b = 3 a+b$ :

= 3 n + 2n + 2 3 n - 1 + 2n - 2

= 3 3n + 2 3 3n - 3

Subtract denominator exponent from numerator exponent: $3 a / 3 b = 3 a-b$ :

= 3 (3n + 2) - (3n - 3)

= 3 3n + 2 - 3n + 3

= 35

= 3 \times 3 \times 3 \times 3 \times 3 = 243

(ix)

5 n+3 - 6 \cdot 5 n+1 9 \cdot 5 n - 4 \cdot 5 n

Solution

Factor out the base power term $5 n$ :

= 5 n \cdot 5 3 - 6 \cdot 5 n \cdot 5 1 5 n (9 - 4)

= 5 n (5 3 - 6 \cdot 5 1) 5 n (5)

Cancel out $5 n$ :

= 53 - 6(5) 5

= 125 - 30 5

= 955 = 19

Question 3 Evaluations

x = 3 + \sqrt 8

, then find the value of:
(i)

x + 1 x

(ii)

x - 1 x

(iii)

x 2 + 1 x 2

(iv)

x 2 - 1 x 2

(v)

x 4 + 1 x 4

(vi)

(x - 1 x) 2

Solution

First, find the reciprocal of $x$ , which is $1 x$ , and rationalize it:

Given: x = 3 + \sqrt 8 — Eq. (1)

1 x = 1 3 + \sqrt 8

Rationalize the denominator by multiplying by the conjugate $3 - \sqrt 8$ :

1 x = 1 3 + \sqrt 8 \times 3 - \sqrt 8 3 - \sqrt 8

= 3 - \sqrt 8 (3) 2 - (\sqrt 8) 2

= 3 - \sqrt 8 9 - 8 = 3 - \sqrt 8 1

1 x = 3 - \sqrt 8 — Eq. (2)

(i) Finding $x + 1 x$ :

Add Eq. (1) and Eq. (2):

x + 1 x = (3 + \sqrt 8) + (3 - \sqrt 8)

= 3 + 3 + \sqrt 8 - \sqrt 8

x + 1 x = 6 — Eq. (3)

(ii) Finding $x - 1 x$ :

Subtract Eq. (2) from Eq. (1):

x - 1 x = (3 + \sqrt 8) - (3 - \sqrt 8)

= 3 + \sqrt 8 - 3 + \sqrt 8

x - 1 x = 2 \sqrt 8 — Eq. (4)

(iii) Finding $x 2 + 1 x 2$ :

Take the square of Eq. (3):

(x + 1 x) 2 = (6) 2

x 2 + 1 x 2 + 2(x)(1 x) = 36

x 2 + 1 x 2 + 2 = 36

x 2 + 1 x 2 = 34 — Eq. (5)

(iv) Finding $x 2 - 1 x 2$ :

Use the difference of squares identity: $a 2 - b 2 = (a + b)(a - b)$ :

x 2 - 1 x 2 = (x + 1 x)(x - 1 x)

Substitute values from Eq. (3) and Eq. (4):

= (6)(2 \sqrt 8)

x 2 - 1 x 2 = 12 \sqrt 8

(v) Finding $x 4 + 1 x 4$ :

Take the square of Eq. (5):

(x 2 + 1 x 2) 2 = (34) 2

(x 2) 2 + (1 x 2) 2 + 2(x 2)(1 x 2) = 1156

x 4 + 1 x 4 + 2 = 1156

x 4 + 1 x 4 = 1156 - 2

x 4 + 1 x 4 = 1154

(vi) Finding $(x - 1 x) 2$ :

Take the square of Eq. (4):

(x - 1 x) 2 = (2 \sqrt 8) 2

= (2) 2 \cdot (\sqrt 8) 2

= 4(8)

(x - 1 x) 2 = 32

Question 4 Comparing Coefficients

Find rational numbers p and q such that:

8 - 3 \sqrt 2 4 + 3 \sqrt 2 = p + q \sqrt 2

Solution

Start by rationalizing the denominator of the LHS expression by multiplying by conjugate $4 - 3 \sqrt 2$ :

LHS = 8 - 3 \sqrt 2 4 + 3 \sqrt 2 \times 4 - 3 \sqrt 2 4 - 3 \sqrt 2

= (8 - 3 \sqrt 2)(4 - 3 \sqrt 2) (4) 2 - (3 \sqrt 2) 2

Expand the numerator:

= 8(4) - 8(3 \sqrt 2) - 3 \sqrt 2 (4) + 3 \sqrt 2 (3 \sqrt 2) 16 - 9(2)

= 32 - 24 \sqrt 2 - 12 \sqrt 2 + 9(2) 16 - 18

= 32 - 36 \sqrt 2 + 18 -2

= 50 - 36 \sqrt 2 -2

Divide each term in the numerator by the denominator:

= 50 -2 - 36 \sqrt 2 -2

= -25 - (-18 \sqrt 2)

= -25 + 18 \sqrt 2

Equate to LHS:

p + q \sqrt 2 = -25 + 18 \sqrt 2

By comparing rational and irrational coefficients on both sides, we get:

p = -25 and q = 18

Question 5 Advanced Simplifications

Simplify the following:

(i)

(25) 3/2 \times (243) 3/5 (16) 5/4 \times (8) 4/3

Solution

Write bases in prime factored power forms ( $25 = 5 2$ , $243 = 3 5$ , $16 = 2 4$ , $8 = 2 3$ ):

= (5 2) 3/2 \times (3 5) 3/5 (2 4) 5/4 \times (2 3) 4/3

Multiply powers: $(a m) n = a m \cdot n$ :

= 5 2 \times 3/2 \times 3 5 \times 3/5 2 4 \times 5/4 \times 2 3 \times 4/3

= 53 \times 3 3 25 \times 2 4

In the numerator, group base products under a single power since exponents are same: $a m \cdot b m = (ab) m$ . In the denominator, sum powers: $a p \cdot a q = a p+q$ :

= (5 \times 3) 3 2 5 + 4

= 15329

= 15 \times 15 \times 15 512 = 3375512

(ii)

54 \times 3 \sqrt (27) 2x 9 x+1 + 216(3 2x-1)

Solution

Write numbers in base 3 forms ( $54 = 2 \cdot 27 = 2 \cdot 3 3$ , $27 = 3 3$ , $9 = 3 2$ , $216 = 8 \cdot 27 = 2 3 \cdot 3 3$ ):

= (2 \times 3 3) \times 3 \sqrt (3 3) 2x (3 2) x+1 + 2 3 \cdot 3 3 \cdot 3 2x-1

= (2 \times 3 3) \times 3 \sqrt 3 6x 3 2x+2 + 2 3 \cdot 3 3 + 2x - 1

Simplify radical: $3 \sqrt 3 6x = (3 6x) 1/3 = 3 2x$ :

= (2 \times 3 3) \times 3 2x 3 2x+2 + 2 3 \cdot 3 2x+2

Combine numerator base 3 components. Factor out $3 2x+2$ from the denominator:

= 2 \times 3 2x+3 3 2x+2 (1 + 2 3)

= 2 \times 3 2x+3 3 2x+2 (1 + 8)

= 2 \times 3 2x+3 3 2x+2 \cdot 9

Express 9 as $32$ in denominator:

= 2 \times 3 2x+3 3 2x+2 + 2

= 2 \times 3 2x+3 3 2x+4

Subtract powers to divide: $3 2x+3 - (2x+4) = 3 -1$ :

= 2 \times 3 (2x+3) - (2x+4)

= 2 \times 3 -1

= 23

(iii)

\sqrt (216) 2/3 \times (25) 1/2 (0.04) -3/2

Solution

Convert negative denominator exponent to positive in numerator:

= \sqrt (216) 2/3 \times (25) 1/2 \times (0.04) 3/2

Express bases in power forms ( $216 = 6 3 = 2 3 \cdot 3 3$ , $25 = 5 2$ , $0.04 = 4100 = 125 = 5 -2$ ):

= \sqrt (6 3) 2/3 \times (5 2) 1/2 \times (125) 3/2

= \sqrt 62 \times 5 1 \times (152) 3/2

= \sqrt 36 \times 5 \times 153

= \sqrt 36 \times 5125

= \sqrt 3625

Evaluate square root of numerator and denominator:

= \sqrt 36 \sqrt 25 = 65

(iv)

( a 1/3 + b 2/3)( a 2/3 - a 1/3 b 2/3 + b 4/3)

Solution

Let $x = a 1/3$ and $y = b 2/3$ . The expression can be rewritten as:

= (x + y)( x 2 - xy + y 2)

Apply the sum of cubes algebraic identity: $(x + y)(x 2 - xy + y 2) = x 3 + y 3$ :

= x 3 + y 3

Substitute back $x = a 1/3$ and $y = b 2/3$ :

= (a 1/3) 3 + (b 2/3) 3

= a 1/3 \times 3 + b 2/3 \times 3

= a + b 2

Interactive Sandbox Rationalize Helper

Configure coefficients to rationalize denominators of the form:

Numerator (A) Constant (B) \pm Coefficient (C) \sqrt Root (D)

Numerator (A)

Constant (B)

Sign (±)

Coeff (C)

Root (D)

Mathematics 9

Real and Complex Numbers

Step-by-Step Rationalization: