Vedanta Excel in Mathematics Book - 9 Final

Commission and Taxation

b) Mrs. Anjali Subba is a medical doctor in a government hospital. Her monthly salary is
Rs 36,900 and she receives one month's salary as festival bonus. 10% of her salary is
deducted as citizen investment trust. How much income tax does she pay in a year?

c) After deducting 10% provident fund a married person draws Rs 40,500 salary per
month and one month's salary as festival bonus. He/she pays Rs 14,500 annually
as the premium of his/her insurance. Calculate the annual income tax paid by the
person.

d) Mr. Sayad Sharma an unmarried employee of a UN Project draws monthly salary
of Rs 51,000 after deducting 10% salary in his provident fund and 5% in citizen
investment trust. He also receive a Dashain bonus of one month's salary. He pays Rs
22,000 annually as the premium of his life insurance. How much income tax does he
pay in a year?

Project work

4. a) Let's ask the monthly salary of your Mathematics, Science and English teachers, then
complete the table given below.

Name of Marital Monthly Festival Provident Insurance
teachers status Salary allowance fund

Now, calculate the annual income tax paid by each subject teacher.

b) If your parents are involving in any government or private service, let's ask their
monthly salary, bonus, provident fund, insurance, etc. Then calculate the income tax
paid by them in a year.

3.5 Dividend

A dividend is a payment made by a corporation to it's shareholders, usually as a
distribution of profits. A dividend is allocated as a fixed amount per share with
shareholders receiving a dividend in proportion to their shareholding.

For example, a person has bought 1,500 shares out of total of 1,00,000 shares of
Rs 100 per share from a hydroelectricity corporation. If the corporation earned a net
profit of Rs 3,50,28,000 in a certain year and it decided to distribute 20% of the net
profit to its shareholders, the amount received by the person as per his/her number of
shares is called his/her dividend.

Worked-out examples

Example 1: Mrs. Bhatta bought 750 shares out of 20,000 shares from a Business
Company. If the company earned a net profit of Rs 90,00,000 and it
declared to distribute 10% dividend to its shareholders, how much money
did Mrs. Bhatta receive?

49 Vedanta Excel in Mathematics - Book 9

Commission and Taxation

Solution: = Rs 90,00,000
Here, the net profit of the company
= 10% × Rs 90,00,000
Now, 10% of Rs 90,00,000 100

= Rs 9,00,000

Again,

The profit of 20,000 shares = Rs 9,00,000

The profit of 1 share = Rs 9,00,000
20,000

= Rs 45

The profit of 750 shares = 750 × Rs 45 = Rs 33,750
Hence, she received Rs 33,750 as her dividend.

Example 2: A commercial bank sold 30,000 shares. The bank earned a net profit of
Solution: Rs 2,50,00,000 in a year and distributed a certain percent of profit as
dividend. If a shareholder, who has bought 400 shares, received Rs 50,000
dividend, find what percent of profit was distributed as divided by the
bank?

Here, the dividend of 400 shares = Rs 50,000

the dividend of 1 share = Rs 50,000 = Rs 125
400

the dividend of 30,000 shares = Rs 30,000 × Rs 125 = Rs 37,50,000

∴ The total dividend = Rs 37,50,000

The net profit = Rs 2,50,00,000

Now, the percent of net profit as dividend = Dividend × 100%
Net profit

= Rs 37,50,000 × 100%
Rs 2,50,00,000

= 15%

Hence 15% of the net profit was distributed as dividend by the bank.

Alternative process:

The dividend of 400 shares = Rs 50,000

The dividend of 1 share = Rs 50,000
400

= Rs 125

Also, the net profit for 30,000 shares = Rs 2,50,00,000

the net profit for 1 share = Rs 2,50,00,000 = Rs 2500
30,000 3

Vedanta Excel in Mathematics - Book 9 50

Commission and Taxation

Now, the percent of profit as dividend = Dividend for 1 share × 100%
Profit for 1 share

= Rs 125 × 100%
Rs 2500/3

= 15%

Example 3: Harka Magar bought 500 shares out of 50,000 shares sold by a hydropower
Solution: company. When the company distributed 25% of its net profit, he received
Rs 43,850 as his dividend in a year. Calculate the net profit of the company.

Here, the profit of 500 shares = Rs 43,850

the profit of 1 share = Rs 43,850
500

= Rs 87.70

the profit of 40,000 shares = 50,000 × Rs 87.70

= Rs 43,85,000

Let, the net profit of the bank be Rs x.

Now, 25% of Rs x = Rs 43,85,000

or, 25x = Rs 43,85,000
100

∴ x = Rs 1,75,40,000

Hence, the net profit of the company is Rs 1,75,40,000.

EXERCISE 3.3
General section

1. a) A business company makes a net profit of Rs 80,00,000 in a year. The Board of
Directors declares 12% dividend from the net profit. If the company has sold 1000
shares, calculate the dividend of each share.

b) A development bank made a net profit of Rs 2,40,00,000 in a year and the management
announced to distribute 24% dividend from the net profit. If the bank has sold 2500
shares, find the divided of each share.

2. a) A finance company earned a net profit of Rs 45,20,000 in a year. The company declared
to distribute a certain percent of profit as dividend. If the total dividend amount to
Rs 9,04,000, what percent of net profit was distributed as dividend?

b) A man has 200 shares out of 1000 total shares of a business company. He received
Rs 96,000 as dividend in a year which was a certain percent of net profit of
Rs 24,00,000. At what percent of the net profit was the dividend distributed?

51 Vedanta Excel in Mathematics - Book 9

Commission and Taxation

3. a) A publication house distributed 21% of dividend to it's shareholders from the net
profit of a year. If the amount of distributed dividend was Rs 23,52,000, calculate the
net profit made by the publication house.

b) A small hydropower company distributed the total amount of dividend of
Rs 14,58,000 to it's shareholders which was 30% of the net profit earned by the
company. Find the net profit earned by the company.

Creative section - A

4. a) Mrs. Rai bought 250 shares out of 10,000 shares from a finance company. The
company earned a net profit of Rs 85,20,000 and declared 17% dividend to it's
shareholders. Calculate the amount of dividend received by Mrs. Rai.

b) A Business Company sold 2,500 shares at Rs 1,200 per share. Bishwant bought 450
shares. If the company earned a net profit of Rs 39,00,000 in a year and it announced
to distribute 18% dividend from the net profit to its shareholders, find the amount
of dividend received by Bishwant.

5. a) A Cable Car Company sold 3,000 shares to the local people. The company earned
a net profit of Rs 1,20,00,000 in a year and distributed a certain percent of profit as
dividend. If a shareholder who has bought 125 shares received Rs 1,10,000 dividend,
what percent of profit was distributed as dividend?

b) Suntali bought 200 shares out of 5,000 shares sold by a hydropower company to
the local people. The company earned a net profit of Rs 75,00,000 in a year and it
declared to distribute a certain percent dividend to its shareholders. If she received
Rs 81,000, what percent of the net profit was distributed as dividend?

6. a) Mr. Dhurmus bought 500 shares out of 10,000 shares sold by a commercial bank.
The bank earned some profit and it distributed 14% of the net profit as dividend in
a year. If Dhurmus received Rs 1,03,600 in the year, find the net profit of the bank.

b) A Life Insurance Company earned some profit and announced to distribute 40%
dividend from it's net profit to its shareholders. If a shareholder who bought 300
shares out of 12,000 shares sold by the company received Rs 1,25,400, calculate the
net profit of the company.

Project work

7. a) Take the information from website or national level of daily news papers or from any
business papers and make a report about the share values of any five companies,
banks or any business organisations. Present your report in the class.

b) Is your any family member involving member involving in any business organisation
as one of the shareholders? If so discuss with her/him and learn more about bonus,
dividend, etc. Then prepare a report and present in the class.

Vedanta Excel in Mathematics - Book 9 52

Unit Algebraic Expressions

6

6.1 Factors and Factorisation - Review x2 x

In the given figure, x unit is the length of a square.

So, the area of the square = x2 sq. unit. x
Let, the length of the square is increased by 2 units. x+2
Then, the length of the rectangle = (x + 2) units.

Now, the area of the rectangle = x × (x + 2) sq. units x2 2x x
= (x2 + 2x) sq. units

Here, x2 + 2x is the product of x and (x + 2).
Therefore, x and (x + 2) are the factors of the expression x2 + 2x. x 2

Thus, factorisation is the process of expressing a polynomial as the product of two or

more polynomials.

When we factorise an expression, we write it as the product of its factors. The process

of getting the factors becomes easier if we apply the selected method of factorisation

for the particular type of expression. So, it is important and very much useful to know

about the types of expressions which are to be factorised.

(i) Expression having a common factor in each of its term

Let's take an expression ax + ay. Here, both terms contain a common term which a.
In such an expression, the factor which is present in all terms of the expression
is taken out as common and each term of the expression should be divided by the
common factor to get another factor.

Worked-out examples

Example 1: Factorise (i) 4ab + 6ac (ii) 2x2y – 4xy2 + 6xy
Solution:
(iii) 2a (x – y) + 7b (y – x) (iv) 3p2q (a – b) – 6pq2 (b – a)

(i) 4ab + 6ac = 2a (2b + 3c)

(ii) 2x2y – 4xy2 + 6xy = 2xy (x – 2y + 3)

(iii) 2a (x – y) + 7b (y – x) = 2a (x – y) – 7b (x – y)

= (x – y) (2a – 7b)

(iii) 3p2q (a – b) – 6pq2 (b – a) = 3p2q(a – b) + 6pq2(a – b)

= 3pq (a – b) (p + 2q)

91 Vedanta Excel in Mathematics - Book 9

Algebraic Expressions

(ii) Expression having common factors in the groups of terms

Let's take an express ax + by – ay – bx. It can be regroup as ax – ay – bx + by
(or ax – bx – ay + by). In ax – ay – bx + by, the group ax – ay has the common factor
a and the group – bx + by has the common factor b. Then,

ax – ay – bx + by = a(x – y) –b(x – y) = (x – y) (a – b)

Similarly, ax – bx – ay + by = x(a – b) –y (a – b) = (a– b) (x – y)

In this way, in such expressions, the terms are to be arranged in suitable groups such that
each group has a common factor.

Example 2: Factorise (i) x2 – ax + ab – bx (ii) ac (b2 + 1) + b (a2 + c2)

Solution:

(i) x2 – ax + ab – bx = x2 – ax – bx + ab = x (x – a) – b (x – a) = (x – a) (x – b)
(ii) ac (b2 + 1) + b (a2 + c2) = ab2c + ac + a2b + bc2

= ab2c + a2b + bc2 + ac

= ab (bc + a) + c (bc + a) = (bc + a) (ab + c)

Example 3: Simplify x2 2x2 – 6x – 6
Solution: + 2x – 3x

2x2 – 6x = 2x (x – 3) = (x 2x (x – 3) = x 2x
x2 + 2x – 3x – 6 x (x + 2) – 3 (x + 2) + 2) (x – 3) +2

EXERCISE 6.1

General section

1. In each of the following figures, write the polynomial as the product of it's factors.

a) b) 2 c) 1 d) 3

xx xa

x1 x x2 a3

Creative section

2. Factorise b) 4p2 – 6p c) 6a2b + 9ab2
a) 2ax + 4bx e) 6x3y2 + 9x2y3 – 3x2y2 f) 2x (a + b) + 3y (a + b)
d) 2px2 – 4px + 6p2x

g) 3a (x – y) – (x – y) h) x (x + 2) + 3x + 6 i) 2t (t – 1) – t + 1

3. Resolve into factors.

a) ax + by + ay + bx b) pm – qn + pn – qm c) a2 + ab + ca + bc
f) x2 + 4x + 3x + 12
d) mx2 + my2 – nx2 – ny2 e) xy – 2y + 3x – 6 i) a2 – a (b + c) + bc

g) p2 – 8p – p + 8 h) 16x2 – 4x – 4x + 1

j) x2 – (y – 3) x – 3y k) pq (r2 + 1) – r (p2 + q2) l) y (x + z) + z (x + y) + y2 + z2

4. Simplify.

a) a2 + a b) 3x2 – 6xy c) 3p2 6p2 – 2p – 1
2ab + 2b 2xy – 4y2 – p + 3p

d) 2x2 – xy + 2xy – y2 e) x2 + 2x + x + 2 f) x2 – 4x – 3x + 12
6xy – 3y2 x2 – 3x + 2x – 6 x2 – 4x + 2x – 8

Vedanta Excel in Mathematics - Book 9 92

Algebraic Expressions

(iii) Expression having the difference of two squared terms
(Expression of the form a2 – b2)

Let's take a square sheet of paper of length a units. From a corner of the sheet, another

square of length b units is cut out.

a aa b a b a
D C
a–b a–b
ab a bb
a a2 b b2 A B

a+b
Area = (a + b) (a – b)
Area = a2 a–b
Area of shaded region is a2 – b2

Here, area of the rectangle ABCD is a2 – b2, which is the product of length (a + b)
and breadth (a – b)

∴ length × breadth = area of rectangle

i.e. (a + b) (a – b) = a2 – b2

Here, the expression a2 – b2 is the difference of two squared terms a2 and b2 and it is
the product of (a + b) and (a – b). So, (a + b) and (a – b) are the factors of a2 – b2.

Thus, to factorise an expression of the form a2 – b2, we should use the formula,

a2 – b2 = (a + b) (a – b)

Worked-out examples

Example 1: Factorise (i) 8x3y – 18xy3 (ii) 81ax5 – 16ax
Solution:

(i) 8x3y – 18xy3 = 2xy (4x2 – 9y2) = 2xy[(2x)2 – (3y)2]
= 2xy (2x + 3y) (2x – 3y)

(ii) 81ax5 – 16ax = ax (81x4 – 16)
= ax [(9x2)2 – (4)2]
= ax (9x2 + 4) (9x2 – 4)
= ax (9x2 + 4) [(3x2) – 22]
= ax (9x2 + 4) (3x + 2) (3x – 2)

Example 2: Resolve into factorise. a) 1 – 9 (a – b)2 b) a2 – b2 + 2b – 1
Solution:

a) 1 – 9 (a – b)2 = 12 – [3 (a – b)]2
= [1 + 3 (a – b)] [1 – 3 (a – b)]
= (1 + 3a – 3b) (1 – 3a + 3b)

93 Vedanta Excel in Mathematics - Book 9

Algebraic Expressions

b) a2 – b2 + 2b – 1 = a2 – (b2 – 2b + 1)
= a2 – (b – 1)2 Using (a – b)2 = a2 – 2ab + b2
= (a + b – 1) [a – (b – 1)]
= (a + b –1) (a – b + 1)

Example 3: Factorise x2 – 6x – 40 + 14b – b2
Solution:

x2 – 6x – 40 + 14b – b2 = (x2 – 6x) – 40 + 14b – b2
= (x2 – 2.x.3 + 32 – 32) – 40 + 14b – b2
= (x – 3)2 – 9 – 40 + 14b – b2
= (x – 3)2 – (49 – 14b + b2)
= (x – 3)2 – (72 – 2.7.b + b2)
= (x – 3)2 – (7 – b)2 Using a2 – 2ab + b2 = (a – b)2
= (x – 3 + 7 – b) (x – 3 – 7 + b)
= (x – b + 4) (x + b – 10)

Example 4: Factorise (w2 – x2) (y2 – z2) – 4wxyz

Solution:

(w2 – x2) (y2 – z2) – 4wxyz = w2y2 – w2z2 – x2y2 + x2z2 – 4wxyz

= (wy)2 – 2.wxyz + (xz)2 – (wz)2 – 2wxyz – (xy)2

= (wy)2 – 2.wy.xz + (xz)2 – [(wz)2 + 2.wz.xy + (xy)2]

= (wy – xz)2 – (wz + xy)2

= (wy – xz + wz + xy) (wy – xz – wz – xy)

= (wy + wz + xy – xz) (wy – wz – xy – xz)

(iv) Expression of the form a4 + a2b2 + b4

The expressions of the form a4 + a2b2 + b4 are also factorised by using the similar

method of factorisation of the expression of a2 – b2 form. Following formulae are

useful while factorising these types of expressions.

a2 + 2ab + b2 = (a + b)2 a2 + b2 = (a + b)2 – 2ab

a2 – 2ab + b2 = (a – b)2 a2 + b2 = (a – b)2 + 2ab

a2 – b2 = (a + b) (a – b)

Example 5: Factorise a) a4 + a2b2 + b4 b) x4 – 3x2y2 + y4

Solution:

a) a4 + a2b2 + b4 = (a2)2 + (b2)2 + a2b2

= (a2 + b2)2 – 2a2b2 + a2b2 Using a2 + b2 = (a + b)2 – 2ab

= (a2 + b2)2 – (ab)2

= (a2 + b2 + ab) (a2 + b2 – ab) Using a2 – b2 = (a + b) (a – b)

= (a2 + ab +b2) (a2 – ab + b2)

Vedanta Excel in Mathematics - Book 9 94

Algebraic Expressions

b) x4 – 3x2y2 + y4 = (x2)2 + (y2)2 – 3x2y2
= (x2 – y2)2 + 2x2y2 – 3x2y2
= (x2 – y2)2 – (xy)2 Using a2 + b2 = (a – b)2 + 2ab
= (x2 – y2 + xy) (x2 – y2 – xy)
= (x2 + xy – y2) (x2 – xy – y2)

Example 6: Resolve into factors a) 9x4 + 14x2 + 25 b) x4 + x2 +1 c) a4 + 4b4
Solution: y4 y2

a) 9x4 + 14x2 + 25 = (3x2)2 + (5)2 + 14x2

= (3x2 + 5)2 – 2.3x2.5 + 14x2

= (3x2 + 5)2 – 16x2

= (3x2 + 5)2 – (4x)2

= (3x2 + 4x + 5) (3x2 – 4x + 5)

b) x4 + x2 +1 = x2 2 + (1)2 + x2 c) a4 + 4b4 = (a2)2 + (2b2)2
y4 y2 y2 y2

= x2 + 1 2 – 2. x2 .1 + x2 = (a2 + 2b2)2 – 2a2.2b2
y2 y2 y2 = (a2 + 2b2)2 – (2ab)2
=(a2 + 2ab + 2b2) (a2 – 2ab + 2b2)
= x2 + 1 2 x2
y2 y
–

= x2 + x +1 x2 – x +1
y2 y y2 y

EXERCISE 6.2
General section

1. Factorise:

a) 9x2 – 4 b) 25a2b2 – 1 c) 48ax2 – 75ay2 d) x4 – y4

e) 16x5y – 81xy5 f) 625a4 – 256b4 g) 4 – (m – n)2 h) 1 – (a – b)2

i) 16 – 25(p – q)2 j) (2a – b)2 – (a – 2b)2 k) a2 + 2ab + b2 – c2 l) p2 – q2 – r2 – 2qr

m) a2 – b2 + 4b – 4 n) 16a4 – 4a2 – 4a – 1 o) ax2 – ay2 – x – y p) a2 – (a – b)x – b2

2. Resolve into factors.

a) x2 + x2y2 + y4 b) a4 + a2 + 1 c) 9x4 + 2x2y2 + y4

d) m4 + 4m2n2 + 16n4 e) 4x4 – 8x2y2 + 49y4 f) 9p4 – 34p2q2 + 25q4

g) 36y4 – 25y2 + 4 h) a4 + 4 i) 64x4 + 1

j) 100a4 – 45a2 + 81 k) 169b4 – 35b2x2 + 961x4 l) a4 + a2 + 1
b4 b2

m) p4 + q4 + 1 n) x4 – 7x2 +1 o) b4 – 15db22 + 9
q4 p4 y4 y2 d4

95 Vedanta Excel in Mathematics - Book 9

Algebraic Expressions

3. Let’s find the area of the shaded region using a2 – b2 = (a + b) (a – b).

a) 10 cm b) 15 cm c) 18 m

10 cm 3 cm 15 cm 18 m 5m
5m
2 cm

2 cm 3 cm f)
d) 21 m e) 25 ft

21 m 9 ft 25 ft 12 m 30 m

7m7m 9 ft 12 m
30 m
Creative section
4. Factorise: b) a2 – 10a + 16 - 6b – b2
d) x4 + 8x2 – 65 + 18y – y2
a) x2 + 6x + 5 – 4y – y2 f) 625y2 + 400y – 36 + 20z – z2
c) p2 – 12p – 28 + 16q – q2 h) 25x2 – 20xy – 21y2 + 10yz – z2
e) 9a2 – 30a + 24 – 8x – 16x2
g) 16p2 – 72pq + 80q2 – 6qr – 9r2

5. Resolve into factors. b) (x2 – 1) (y2 – 1) – 4xy
a) (a2 – b2) (c2 – d2) + 4abcd d) (9 – x2) (100 – y2) – 120xy
c) (p2 – 4) (9 – q2) + 24pq

Project work
6. a) Write any three expressions of your own in the form of a2 – b2. Then factorise your

expressions.
b) Write any three expressions of your own in the form of a4 + a2b2 + b4. Then factorise

your expressions.
c) Write any three pairs of trinomial expressions of your own and find the product of

each pair. Then factorise each product. Check, are your answers correct?
[E.g. (a + b – c) (a – b + c) = a2 – b2 – c2 + 2bc, etc.]

(v) Expression of the form a3 + b3 or a3 – b3

Let's find the product of the expressions (a + b) and (a2 – ab + b2).
(a + b) (a2 – ab + b2) = a (a2 – ab + b2) + b (a2 – ab + b2)

= a3 – a2b + ab2 + a2b – ab2 + b3 = a3 + b3

Similarly, (a – b) (a2 + ab + b2) = 3 – b3

So, (a + b) and (a2 – ab + b2) are the factors of a3 + b3.

Also, (a – b) and (a2 + ab + b2) are the factors of a3 – b3.

Thus, to factorise the expressions of the forms a3 + b3 and a3 – b3, we should use the
following formulae.

a3 + b3 = (a + b) (a2 – ab + b2) a3 – b3 = (a – b) (a2 + ab + b2)

Vedanta Excel in Mathematics - Book 9 96

Algebraic Expressions

Worked-out examples

Example 1: Factorise a) 8a4 + 125a b) x6 – y6 c) a7 + 1
Solution: a5

a) 8a4 + 125a = a (8a3 + 125)

= a [(2a)3 + 53]

= a (2a + 5) [(2a)2 – 2a.5 + 52] Using a3 + b3 = (a + b) (a2 – ab + b2)

= a (2a + 5) (4a2 – 10a + 25)

b) x6 – y6 = (x3)2 – (y3)2
= (x3 + y3) (x3 – y3)
= (x + y) (x2 – xy + y2) (x – y) (x2 + xy + y2)
= (x + y) (x – y) (x2 – xy + y2) (x2 + xy + y2)

c) a7 – 1 = a7 – a × 1 = a(a6 – a16) = a[(a3)2 – 1 2 a3 + 1 a3 – 1
a5 a6 a3 a3 a3
]=a

1 11 1 11
= a a + a (a2 – a × a + a2) a – a (a2 + a × a + a2)

11 1 1
= a a + a a – a a2 – 1+ a2 a2 + 1+ a2

11 1 12 1
=a a + a a – a a2 – 1+ a2 a+ a –2×a×a+1

=a a + 1 a – 1 a2 – 1+ 1 a+ 1 2
a a a a
– (1)2

=a a + 1 a – 1 a2 – 1+ 1 a + 1+ 1 a– 1+ 1
a a a a a

Example 2: Resolve into factors 8x3 – 20x2y + 30xy2 – 27y3.
Solution:

8x3 – 20x2y + 30xy2 – 27y3 = (2x)3 – (3y)3 – 20x2y + 30xy2

= (2x – 3y) [(2x)2 + 2x.3y + (3y)2] – 10xy (2x – 3y)

= (2x – 3y) (4x2 + 6xy + 9y2 – 10xy)

= (2x – 3y) (4x2 – 4xy + 9y2)

(vi) Expression of the form ax2 + bx + c, where a ≠ 0.

x2 + 5x + 6, 2x2 + x – 28, etc. are the trinomial expressions of the form
ax2 + bx + c. To factorise such expressions, we need to find the numbers p and q such
that p + q = b and pq = ac. Then, the trinomial expression is expanded to four terms
and factorisation is performed by grouping.

97 Vedanta Excel in Mathematics - Book 9

Algebraic Expressions

Example 3: Factorise a) x2 + 7x + 12 b) 2x2 – 5x + 2 c) 2x2 – x – 6

Solution: 12 × 1 = 12
4 × 3 = 12
x2 x2 x x x x x + 3
a) x2 + 7x + 12 = x2 + (4 + 3) x + 12 x x
x
= x2 + 4x + 3x + 12 2x xx 12 34
x–2x2 x2 x 56 78
= x (x + 4) + 3 (x + 4) xx 9 101112

= (x + 4) (x + 3) x+4

2×2=4 2x – 1
4×1=4
x2 x2 x
b) 2x2 – 5x + 2 = 2x2 – (4 + 1) x + 2 2 xx xx 21
= 2x2 – 4x – x + 2
= 2x (x – 2) – 1 (x – 2) 2x – 1
= (x – 2) (2x – 1) x–2

2 × 6 = 12 2x + 3 2x + 3
4 × 3 = 12 x2 x2 x x x
x x(x–2) x(x–2)
c) 2x2 – x – 6 = 2x2 – (4 – 3) x – 6 x–2 2 xx xx
1 23
x–2 4 56
x–2
= 2x2 – 4x + 3x – 6 x–2
xx3

= 2x (x – 2) +3 (x – 2) 2x + 3

= (x – 2) (2x + 3) x(x–2) x(x–2) 3x–6

Example 4: Resolve into factors.

Solution: a) a2 – 3 + 2b2 b) 9 (x + y)2 + x + y – 8
b2 a2

a) a2 – 3 + 2b2 = a2 – 3. a . b + 2b2
b2 a2 b2 b a a2

= a2 – a . b – 2 ab . ab + 2b2
b2 b a a2

= a (ab – ab ) – 2b (ab – b )
b a a

=( – b ) ( a – 2b )
a b a

b) Let x + y = a

Then, 9 (x + y)2 + x + y – 8 = 9a2 + a – 8

= 9a2 + 9a – 8a – 8

= 9a (a + 1) – 8 (a + 1)

= (a + 1) (9a – 8)

= (x + y + 1) [9 (x + y) – 8]

= (x + y + 1) (9x + 9y – 8)

Vedanta Excel in Mathematics - Book 9 98

Algebraic Expressions

Example 5: Factorise 2a6 – 19a3 + 24
Solution:
2a6 – 19a3 + 24 = 2a6 – 16a3 – 3a3 + 24

= 2a3 (a3 – 8) – 3 (a3 – 8)
= (a3 – 8) (2a3 – 3)
= (a3 – 23) (2a3 – 3)
= (a – 2) (a2 + 2a + 4) (2a3 – 3)
= (a – 2)(2a3 – 3) (a2 + 2a + 4)

EXERCISE 6.3
General section

1. In each of the following figures write the polynomial as the product of it's factors.

a) x 3 b) x 3 c) x

x
xx

22 3

d) x x 1 e) x x 4 x
f) x

x x2 x2 x x
3 2
2 1
3 3

Creative section - A

2. Resolve into factors.

a) 8x3 + y3 b) 1 + 27a3 c) 128t3 – 2t d) x3y – 64y4

e) a6 + b3 f) 64x6y3 – 125 g) a6 – 64 h) 64x6 – y6
j) (x + 2)3 – 27 k) (x – y)3 – 8 (x + y)3
i) (a + b)3 + 1 l) p3 + 1
p3

m) a 3– b3 n) 27x3 – 30x2y + 40xy2 – 64y3 o) 8 – 6a – 9a2 + 27a3
b a

p) x4 + 1 q) p7 + 1
x2 p5

3. Factorise.

a) x2 + 4x + 3 b) x2 – 7x – 8 c) a2 – 27a + 180

d) 2x2 + 7x + 6 e) 3p2 – 7p – 6 f) 2x2 + 3xy – 5y2
h) 9a3bx + 12a2b2x – 5ab3x
g) 3a2 – 16ab + 13b2 k) 2 (x + y)2 + 9 (x + y) + 7 i) 12 a2 + a – 20
n) 2x6 + 17x3 + 8 b2 b
x2 3y2
j) y2 – 2 – x2 l) 3 (x – y)2 – 10 (x – y) + 8

m) 8a4 – 14a2 – 9 o) 3x6 – 79x3 – 54

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Creative section - B

4. Simplify.

a) a2 – 4 b) x2 x2 – 9 15 c) 4a2 – 1
a2 – a – 6 + 2x – 2a2 – 3a – 2

d) a2 – 9 e) x3 + 3x2 – 4x f) 2x2 + 3x – 2
a3 + 27 x4 – x 2x2 – 5x + 2

g) a3 – a2b + ab2 – b3 h) (x2 – x – 6) (x2 + 4x + 3) i) (x2 – 16) (x + 2)
a4 – b4 (x2 – 9) (x + 1) (x2 – 2x – 8) (x2 + 5x + 4)

5. a) The area of a rectangular field is (x2 + 8x + 15) sq. m.
(i) Find the length and breadth of the field.
(ii) Find the perimeter of the field.

b) The area of a rectangular plot of land is (x2 + 13x + 40) sq. m.

(i) Find the length and breadth of the land.

(ii) If the length and breadth of the land are reduced by 2/2 m respectively, find the
new area of the land.

c) A rectangular ground has area (2x2 + 11x + 12) sq. m. If the length of the ground
is decreased by 2 m and the breadth is increased by 2 m, find the new area of the
ground. (Take a longer side of the rectangles as length)

Project work

6. a) Write three different expressions of your own each in the form of a3 + b3 and a3 – b3.
Then factorise your expressions.

b) Write an expression in each of the following forms, then factorise your expressions

(i) x2 + ax + b (ii) x2 – ax – b (iii) x2 + ax – b (iv) x2 – ax + b

c) Write an expression in each of the following forms, then factorise your expressions.

(i) ax2 + bx + c (ii) ax2 – bx – c (iii) ax2 + bx – c (iv) ax2 – bx + c

6.2 Highest Common Factor (H.C.F.) of algebraic expressions

To find the H.C.F. of monomial expressions, at first should find the H.C.F. of the
numerical coefficients. Then, the common variable with the least power is taken as
the H.C.F. of the monomial expressions.

To find the H.C.F. of the polynomial expressions, they are to be factorised and a
common factor or the product of common factors is obtained as their H.C.F.

Worked-out examples

Example 1: Find the H.C.F. of a2 – 9 and a2 + 4a – 21.

Solution:

1st expression = a2 – 9 a2 – 9 a2 + 4a – 21
2nd expression = a2 – 32 = (a + 3) (a – 3)
= a2 + 4a – 21 (a + 3) (a – 3) (a + 7)
∴ H.C.F. = a2 + 7a – 3a – 21
= a (a + 7) – 3 (a + 7)
= (a + 7) (a – 3)
= (a – 3)

Vedanta Excel in Mathematics - Book 9 100

Algebraic Expressions

6.3 Lowest Common Multiple (L.C.M.) of algebraic expressions

The L.C.M. of monomial expressions is the common variable with the highest power.
To find the L.C.M. of the given polynomial expressions, they are to be factorised and
the product of common factors and the factors which are not common is taken as their
L.C.M.

Example 2: Find the L.C.M. of x2 – 25, x3 – 125 and x2 – 8x + 15

Solution:

1st expression = x2 – 25 = a2 – 52 = (a + 5) (a – 5)

2nd expression = x3 – 125 = x3 – 53 = (x – 5) (x2 + 5x + 25)

3rd expression = x2 – 8x + 15

= x2 – 5x – 3x + 15

= x (x – 5) – 3 (x – 5) = (x – 5) (x – 3)

∴ L.C.M. = (x – 5) (x + 5) (x – 3) (x2 + 5x + 25)

Example 3: Find the L.C.M. of a2 – 3a + 2, a2 – 5a + 6 and a2 – 4a + 3

Solution:

1st expression = a2 – 3a + 2 = a2 – 2a – a + 2 = a (a – 2) – 1 (a – 2) = (a – 2) (a – 1)

2nd expression = a2 – 5a + 6 = a2 – 3a – 2a + 6 = a (a – 3) – 2 (a – 3) = (a – 3) (a – 2)

3rd expression = a2 – 4 + 3 = a2 – 3a – a + 3 = a (a – 3) –1 (a – 3) = (a – 3) (a – 1)

∴ L.C.M. = (a – 1) (a – 2) (a – 3)

6.4 Simplification of rational expressions 3x , 25y , a + b,
7
Rational expressions can be expressed in the form of , where q ≠ 0.

x+y , etc. are a few examples of rational expressions.
x–y

When we multiply or divide rational expressions, the numerators and denominators
of each expression are factorised (if necessary). Then we simplify the expressions to
the simplest forms.

Worked-out examples

Example 1: Simplify a) a2 + 3a +2 × a2 – 9 6 b) 4x2 – 81y2 ÷ 2x – 9y
Solution: a2 + a –6 a2 – a – 1 – 4a2 a – 2a2

a) a2 + 3a + 2 × a2 – 9 6 = a2 + 2a + a + 2 × a2 – a2 – 32 – 6
a2 + a – 6 a2 – a – a2 + 3a – 2a – 6 3a + 2a

= a (a + 2) + 1 (a + 2) × (a + 3) (a – 3)
a (a + 3) – 2 (a + 3) a (a – 3) + 2 (a – 3)

= (a + 2) (a + 1) × (a + 3) (a – 3)
(a + 3) (a – 2) (a – 3) (a + 2)

= a+1
a–2

101 Vedanta Excel in Mathematics - Book 9

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b) 4x2 – 81y2 ÷ 2x – 9y = 4x2 – 81y2 × a – 2a2 = (2x)2 – (9y)2 × a (1 – 2a)
1 – 4a2 a – 2a2 1 – 4a2 2x – 9y 12 – (2a)2 2x – 9y

= (2x + 9y) (2x – 9y) × a (1 – 2a)
(1 + 2a) (1 – 2a) 2x – 9y

= a (2x + 9y)
1 + 2a

When we add or subtract rational expressions with unlike denominators, we should find the
L.C.M. of denominators. Then, the simplification is performed as like the simplification of
fractions.

Example 2: Simplify a) x x 2 – x x 2 b) x2 + ax + a2 + x2 – ax + a2
Solution: + – x+a x–a

a) x x 2 – x = x (x – 2) – x (x + 2) = x2 – 2x – x2 – 2x = –4x
+ x–2 (x + 2) (x – 2) x2 – 4 x2 – 4

b) x2 + ax + a2 + x2 – ax + a2 = (x – a) (x2 + ax + a2) + (x + a) (x2 – ax + a2)
x+a x–a (x + a) (x – a)

= x3 – a3 + x3 + a3 = 2x3
x2 – a2 x2 – a2

Example 3: Simplify x–1 + (x + 3 – 1) + (1 – 1 – 2x)
Solution: (2x – 1) (x + 2) 2) (x x) (1

x–1 + 3 1) – (1 – 1 – 2x)
(2x – 1) (x + 2) (x + 2) (x – x) (1

= x–1 + 3 – 1
(2x – 1) (x + 2) (x + 2) (x – 1) [– (x – 1)] [– (2x – 1)]

= (x – 1) (x – 1) + 3 (2x – 1) – (x + 2)
(2x – 1) (x + 2) (x – 1)

= x2 – 2x + 1 + 6x – 3 – x – 2
(2x – 1) (x + 2) (x – 1)

= x2 + 3x – 4
(2x – 1) (x + 2) (x – 1)

= (2x x2 + 4x – x – 4 1) = x (x + 4) – 1 (x + 4) = (2x (x + 4) (x – 1) 1) = (2x x+4 2)
– 1) (x + 2) (x – (2x – 1) (x + 2) (x – 1) – 1) (x + 2) (x – – 1) (x +

Example 4: Simplify p–1 + p2 p–2 6 + p2 p– 5 15
Solution: p2 – 3p + 2 – 5p + – 8p +

p–1 + p2 p–2 6 + p2 p–5 15
p2 – 3p + 2 – 5p + – 8p +

= p2 – p–1 + 2 + p2 – p – 2 + 6 + p2 – p–5 + 15
2p – p 3p – 2p 5p – 3p

=p (p – p – 1 (p – 2) + p (p – p – 2 (p – 3) + p (p – p – 5 (p – 5)
2) – p 3) – 2 5) – 3

= (p p–1 + p–2 + (p – p–5 – 3)
– 2) (p – 1) (p – 3) (p – 2) 5) (p

= p 1 2 + p 1 3 + p 1 3 = p – 3+ p–2 +p – 2 = (p 3p –7 3)
– – – (p – 2) (p – 3) – 2) (p –

Vedanta Excel in Mathematics - Book 9 102

Algebraic Expressions

Example 5: Simplify m+n + m2 m–n n2 – m4 + 2n3 + n4
Solution: m2 + mn + n2 – mn + m2n2

m2 m+n n2 + m2 m–n n2 – m4 + 2n3 + n4
+ mn + – mn + m2n2

= (m + n) (m2 – mn + n2) + (m – n) (m2 + mn + n2) – (m2 + n2)2 2n3 + m2n2
(m2 + mn + n2) (m2 – mn + n2) – 2m2n2

= m3 + n3 + m3 – n3 – 2n3
(m2 + mn + n2) (m2 – mn + n2) (m2 + n2)2 – (mn)2

= (m2 + mn + 2m3 – mn + n2) – (m2 + mn + 2n3 – mn + n2)
n2) (m2 n2) (m2

= (m2 + mn 2m3 – 2n3 mn + n2)
+ n2) (m2 –

= 2 (m – n) (m2 + mn + n2)
(m2 + mn + n2) (m2 – mn + n2)

= 2( m– n)
m2 – mn + n2

Example 6: Simplify (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
Solution: a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

(a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

= (a – b + c) (a – b – ) + (b – c + a) (b – c – a) + (c – a + b) (c – a – b)
(a + b + c) (a – b – c) (b + c + a) (b – c – a) (c + a + b) (c – a – b)

= a–b+c + b–c+a + c–a+b
a+b+c a+b+c a+b+c

= a – b + c +b–c+a+ c – a + b
a+b+c

= a + b + c = 1
a + b + c

EXERCISE 6.4
General section

1. Find the H.C.F. and the L.C.M. of the following factors.
a) (x + 2) (x – 1) and (x + 1) (x + 2) b) (a – 5) (a – 4) and (a + 4) (a – 5)
c) (x + y) (2x – y) and (2x – y) (x – y) d) (a + b) (a2 – ab + b2) and (a – b) (a + b)

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Creative section - A
2. Find the H.C.F.

a) ax2 + 2ax, ax3 – 4ax b) x2 – 1, x3 – 1

c) a3 – 8, a2 + 2a + 4 d) (x – 1)2, x2 – 3x + 2

e) x2 + 5x + 6, x2 – 4, x3 + 8 f) 2x2 + 5x + 2, 3x2 + 8x + 4, 2x2 + 3x – 2

g) m3 – 1, m4 + m2 + 1, m2 + m + 1 h) 2x3 – x2 – x, 4x3 – x, 8x4 + x

i) x3 – y3, x6 – y6, x4 + x2y2 + y4 j) a3 + b3, a4 + a2b2 + b4, a3 – a2b + ab2

3. Find the L.C.M.

a) (x + 1) (x + 2), (x + 2) (x – 2) b) (x – 1) (x – 2), (x – 2) (x – 3), (x – 3) (x – 1)

c) x2 – 1, x3 – 1 d) x2 – 9, 3x + 9

e) 4x2 – 25, 2x2 – x – 15 f) x4 + x2y2 + y4, x3 – y3

g) 2x2 – 3x – 9, 4x2 – 5x – 21, x3 – 9x h) 8x3 + y3, 8x3 – y3, 16x4 + 4x2y2 + y4

i) x2 – 5x + 6, x2 – 4x + 3, x2 – 3x + 2 j) x3 + 27, 2x3 – 6x2 + 18x, x2 – 3x + 9

4. Simplify.

a) x2 – a2 × 7a b) 4y2 – 9z2 × y– 2 c) 4a2 – 9b2 × x2y + xy2
ax + a2 x–a y2 – 4 2y – 3z x2 – y2 4a – 6b

d) a2 – b2 ÷ a2b – ab2 e) 2x + 6 ÷ 3x2 + 9x f) a2 – 4b2 ÷ a + 2b
a2b + ab2 a2b2 x2 – 9 2x2 – 6x a2 – 9x2 a – 3x

5. Simplify.

a) a2 b + b2 b) p 4 + 4 2 c) x2 x y2 + y2 y x2
a– b–a p2 – – p2 – –

d) 1 1 + 1 1 e) 1 – 1 f) y 1 3 – 2 1
2a + 2a – x–2 x–1 – 2y –

g) a 1 b – a+b h) p+q + q 1 p i) x3 + y3
– a2 – b2 p2 – q2 – x–y y–x

j) a–b + b–c + c–a k) x+2 – x–2 l) x2y – xy2
ab bc ca x–2 x+2 x–y x+y

m) y2 1 y) – x2 1 y) n) a2 + 2a – 1 o) 1 + 3
(x – (x – a–1 1–a y2 – 4 y2 + 5y + 6

p) a2 + ab + b2 + a2 – ab + b2 q) x2 + 2x + 4 + x2 – 2x + 4 r) (a 1 – a2 1 b2
a+b a–b x+2 x–2 – b)2 –

s) x+2 2 + 3 1 t) x–2 – x+ 1 u) x–2 – x2 x+1
x2 + x – x2 – x2 + 4x + 4 x2 – 4 x2 – 1 – 2x + 1

Vedanta Excel in Mathematics - Book 9 104

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Creative section - B
6. Simplify.

a) 1 – x 2 1 + x 1 2 b) x 1 1 – x 1 1 + 3 1
x + + + – x2 –

c) x x 2 + x – 4x d) 2xy – x–y + x+y
+ x–2 x2 – 4 x2 – y2 x+y x–y

e) x+1 + x–1 – 1 4 f) 2a 1 2x – 2a 1 2x – a2 x x2
2x3 – 4x2 2x3 + 4x2 x2 – – + +

7. Simplify.

a) 1 + 1 + 1
(x – 3) (x – 4) (x – 4) (x – 5) (x – 5) (x – 3)

b) ( 2 (a – 3) 5) + (3 – a–1 – 4) + (5 – a–2 – 3)
– 4) (a – a) (a a) (a

c) x2 – 2 + 6 – x2 – 2 + 3 + x2 – 1 + 2
5x 4x 3x

d) x2 x–1 2 + x2 x–2 6 + x2 x– 5 15
– 3x + – 5x + – 8x +

e) 2y + 5 + 11 – 16y
y2 + 6y + 9 y2 – 9 8y2 – 24y

8. Simplify.

a) x–y + x2 x + y y2 + 2y3
x2 – xy + y2 + xy + x4 + x2y2 + y4

b) a2 a–2 4 + a2 a +2 4 – a4 + 16 + 16
– 2a + + 2a + 4a2

c) x+3 9 + x–3 9 – 54 81
x2 + 3x + x2 – 3x + x4 + 9x2 +

d) 1 a +2 a2 – a–2 – 1 2a2 a4
+ a+ 1 – a + a2 + a2 +

9. Simplify.

a) (x – y)2 – z2 + (y – z)2 – x2 + (z – x)2 – y2
x2 – (y + z)2 y2 – (z + x)2 z2 – (x + y)2

b) a2 – (b – c)2 + b2 – (c – a)2 + c2 – (a – b)2
(c + a)2 – b2 (a + b)2 – c2 (b + c)2 – a2

105 Vedanta Excel in Mathematics - Book 9

Unit Indices

7

7.1 Indices – review

An index is a number that tells how many times a base is multiplied by itself. An
index is also called an exponent.

Let's take a few examples of repetitive multiplication of the same base.

a = a1, a × a = a2, a × a × a = a3, a × a × a × a = a4, etc.

Here, 1 is the index of a1, 2 is the index of a2, 3 is the index of a3, 4 is rthe index of a4,
and so on.

Thus, the index refers to the power to which a number is raised. For example, in
23 the base 2 is raised to the power 3. Indices is the plural of index.

7.2 Laws of Indices

While performing the various operations of indices, we apply different proven rules
like product rule, quotient rule, power rule, etc. These rules are well known as laws
of indices.

The table given below shows the laws of indices at a glance.

Name of laws Rules Examples
Product law am × an = am + n 23 × 25 = 23+5 = 28

Quotient law am ÷ an = am – n when m > n 37 ÷ 33 = 37 – 3 = 34 and
am ÷ an = 1 when m < n 11

an – m 33 ÷ 37 = 37 – 3 = 34

Power law (am)n = am × n, (ab)m = ambm, (24)2 = 24 × 2 = 28,
a am (2x)5 = 25 × x5, …
Negative index law b m = bm
Zero index law
Root laws of indices 1 1 5–3 = 1 or 53 = 1
a– m = am or am = a–m 53 5–3

a0 = 1, (ab)0 = 1, (a + b)0 = 1 20 = 1, 30 = 1, 990 = 1

1 = 2 31 or 3,

m n am or n am m 32

an = = an 2 = 3 32

33

(i) Product law of indices

If am and an are the two algebraic terms, where m and n are the positive integers,
then am × an = am + n

Vedanta Excel in Mathematics - Book 9 106

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Proof

We know that,

a2 = a × a (two factors)

a3 = a × a × a (three factors)

am = a × a × a × ... ('m' factors)

an = a × a × a × ... ('n' factors)

∴ am × an = (a × a × a × ... 'm' factors) × (a ×a × a × ... 'n' factors)
= a × a × a × ... ('m + n' factors)
= am + n

Thus, am × an = am + n
Similarly, if m, n, p, q, r, ... are the positive integers, then
am × an × ap × aq × ar × ... = am + n + p + q + r + ...

(ii) Quotient law of indices

If am and an are the two algebraic terms, where m and n are the positive integers,
then am ÷ an = am – n when m > n

am ÷ an = 1 m when n > m
an –
Proof
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors

= a × a × a × ... ('m – n' factors), where m > n

= am – n

Thus, am ÷ an = am – n, when m > n

But, in the case of n > m,
am a × a × a × ... 'm' factors
am ÷ an = an = a × a × a × ... 'n' factors

= a ×a × a × 1 – m' factors
... 'n

= 1 m
an –

Thus, am ÷ an = 1 m, when n > m
an –

(iii) Power law of indices

If am be an algebraic term, then (am)n = am × n = amn, where m and n are the

positive integers.

Proof

(am)n = am × am × am × ... (n) factors

= am + m + m + ... 'n' times 'm'

= amn

Thus, (am)n = amn

cor 1. amn = (am)n cor 2. amn = (an)m

cor 3. am = am cor 4. a m = am
bm b b bm

cor 5. (ab)m = ambm cor 6. ambm = (ab)m

107 Vedanta Excel in Mathematics - Book 9

Indices

(iv) Law of negative index

If a–m is an algebraic term, where m is a negative integer, then

a–m = 1 or, 1 = a–m or, am = 1
am am a–m

Proof

Here, a–m = am – 2m = am ÷ a2m = am = am = 1
a2m am × am am

Thus, a–m = 1
am

Similarly, 1 = a–m and am = 1
am a–m
(v) Law of zero index

If a0 is an algebraic term, where a ≠ 0, then a0 = 1.

Proof am
am
Here, a0 = am – m = am × a–m = =1

Thus, a0 = 1, where a ≠ 0

(vi) Root law of indices

If amn is an algebraic term, where m and n are the positive integers, then

amn = n am

Proof n

nth order of root in represented as 2

In this way 2nd order of root is represented as or only
3
3rd order of root is represented as and so on.

The square root of 32 = 3 = 322 = 2 32 or 32

The cube root of 53 = 5 = 533 = 3 53

Thus, 322 = 2 32

3 = 3 53

53

In general, amn = n am

Worked-out examples

Example 1: Find the value of: 243 – 52
32
Solution: a) (16)34 b) (9–3)16 c) d) (160.5)32 e) 3 729 f) 3 64–1

a) (16)43 = (24) 34 = (2)4 × 43 = 23 = 8

b) (9–3)16 = (32)–3 × 16 = 3–6 × 61 = 3–1= 1
3

c) 243 – 25 = 35 – 52 = 25 52 = 2 =5 × 52 2 2= 4
32 25 35 3 3 9

d) (160.5)23 = (24)0.5 × 32 = 24 × 0.5 × 23 = 23 = 8

Vedanta Excel in Mathematics - Book 9 108

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e) 3 729 = 3 93 = 933 = 9 = 32 = 3

1 1 2= 3 1 3 1 3 1
64 8 8 2 2
f) 3 64–1 = 3 = 3 = =

Example 2: Evaluate: a) 25 – 21 125 13 ÷ 16 – 41 b) 127 × 286
16 64 81 217 × 166
1 1 1
Solution: c) (1 – 6–9)–1 + (1 – 69)–1 d) (a – b)–1 + (b – c)–1 + (c – a)–1

a) 25 – 12 125 31 ÷ 16 – 14
16 64 81

= 16 21 125 13 ÷ 81 41
25 64 16

= 4 2 × 12 5 3× 13 ÷ 3 4× 14
5 4 2

= 4 5 ÷ 3 = 4 5 × 2 = 4 × 5 = 2
5 4 2 5 4 3 5 6 3

b) 127 × 286 = (22 × 3)7 × (22 × 7)6 = 214 × 37 × 212 × 76 = 214 + 12 × 76 = 226 – 24 = 4
217 × 166 (3 × 7)7 × (24)6 37 × 77 × 224 224 × 77 27 – 6 7

c) (1 – 6–9)–1 + (1 – 69)–1

= 1 – 1 –1 + (1 – 69)–1
69

= 69 – 1 –1 + (1 – 69)–1 = 69 + 1 = 69 – 1 = 69 – 1 = 1
69 69 – – 69 69 – 69 – 69 – 1
1 1 1 1

d) (a 1 + (b 1 + (c 1 = (a – b) + (b – c) + (c – a) = 0
– b)–1 – c)–1 – a)–1

Example 3: Prove that: a) 7m + 2 + 4 × 7m =1 b) 5x – 5x – 1 =1
7m + 1 × 8 – 3 × 7m 4 × 5x – 1

c) 273n + 1 × (243)–45n = 1 d) 3–p × 92p – 2 = 2
9n + 5 × 33n – 7 33p – 2 × (3 × 2)–1 3

Solution: 7m + 2 + 4 × 7m 7m × 72 + 4 × 7m 7m (49 + 4) 53
7m + 1 × 8 – 3 × 7m 7m × 71 × 8 – 3 × 7m 7m (7 × 8 – 3) 53
a) LHS = = = = = 1 = RHS

5x – 5x – 1 5x – 5x 5x 1 – 1 4
4 × 5x – 1 4× 5 5 5
b) LHS = = 5x = 5x = 4 = 1= RHS
c) LHS 5 4 × 5
5

= 273n + 1 × (243)–45n = (33)3n + 1 × (35)–45n = 39n + 3 × 3–4n
9n + 5 × 33n – 7 (32)n + 5 × 33n – 7 32n + 10 × 33n – 7

= 39n + 3 – 4n = 35n + 3 = 35n + 3 – (5n + 3) = 30 = 1 = RHS
32n + 10 + 3n – 7 35n + 3

3–p × 92p – 2
d) LHS = 33p – 2 × (3 × 2)–1

3–p × (32)2p – 2 3–p + 4p – 4 33p – 4 33p – 4 – 3p + 3 3–1 2
= 33p – 2 × 3–1 × 2–1 = 33p – 2 –1 × 2–1 = 33p – 3 × 2–1 = 2–1 = 2–1 = 3 = RHS

109 Vedanta Excel in Mathematics - Book 9

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Example 4: Simplify:

Solution: a) 3 50ba5c–2 × 3 20ab5c8 b) 3 27x3y6 ÷ 4 81x4y8 c) 3 (a + b)–8 × (a + b)32

a) 3 50ba5c–2 × 3 20ab5c8 = 3 1000a5+1 b1+ 5 c–2 + 8 = 3 1000a6 b6 c6 = (103a6b6c6)13 = 10a2b2c2

b) 3 27x3y6 ÷ 4 81x4y8 = (33x3y6)31 ÷ (34x4y8)41 = 3xy2 ÷ 3xy2 = 1.

c) 3 (a + b)–8 × (a + b)23 = (a + b)– 83 × (a + b)23 = (a + b)– 83 + 23 = (a + b) – 83+ 2
1
= (a + b)– 63 = (a + b)–2 = (a + b)2

Example 5: Simplify: xa a2 + ab + b2 × b b2 + bc + c2 c c2 + ca + a2
Solution: xb a
xx × xxc

xa a2 + ab + b2 × xb b2 + bc + c2 × xc c2 + ca + a2 = (xa – b)a2 + ab + b2 × (xb – c)b2 + bc + c2 × (xc )– a c2 + ca + a2
xb xc xa

= xa3 – b3 × xb3 – c3 × xc3 – a3

= xa3 – b3 + b3 – c3 + c3 – a3 = x0 = 1

Example 6: Simplify: a+b xa2 × b+c xb2 × c+a xc2
xb2 xc2 xa2
Solution:
a2 – b2 b2 – c2 c2 – a2
xa2 xb2 xc2 = x a+b × x b+c × x c+a
a+b xb2 × b+c xc2 × c+a xa2

= xa – b × xb – c × xc – a = xa – b + b – c + c – a = x0 = 1

111
Example 7 : Simplify 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a
Solution:

111
1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a

= 1 + 1 + 1

1+ xa + xc 1+ xb + xa 1+ xc + xb
xb xb xc xc xa xa
1 1 1
= xb + xa + xc + xc + xb + xa + xa + xc + xb

xb xc xa

= xb + xc + xa = xb + xc + xa =1
xa + xb + xc xa + xb + xc xa + xb + xc xa + xb + xc

Example 8 : Simplify m + (mn2)31 + (m2n)31 × 1 – n13
m–n m31
Solution:

m + (mn2)13 + (m2n)31 × 1 – n31 = m +m31 n32 + m32 n13 × m13 – n13
m– n m31 m–n m31

= m31 (m32 + n32 + m13 n13 ) × m13 – n13
m– n m13

Vedanta Excel in Mathematics - Book 9 110

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= (m31 – n13) [(m13)2 + m13.n31+ (n13)2] = (m13)3 – (n13)3 = m–n =1
m–n m – n m–n

x2 – 1 x× x – 1 y–x
y2 y
Example 9 : Show that
Solution: y2 – 1 y× y + 1 x–y
x2 x

x2 – 1 x× x – 1 y–x x + 1 x x – 1 x x – 1 y–x
y2 y y y y
Here, LHS = =
1 y× 1 x–y 1 y 1 y 1 x–y
y2 – x2 y + x y + x y – x y+ x

x + 1 x x – 1 x+y–x x + 1 x x – 1 y
y y y y
= =
1 y+x–y 1 y 1 x 1 y
y + x y – x y + x y – x

x + 1 x x – 1 y xy + 1 x xy – 1 y
y y yy
= ×
y + 1 y – 1 = xy + 1 × xy – 1
x x xx

= xy + 1 × xy x 1 x× xy – 1 × x y= x x× x y= x x+y
y + y xy – 1 y y y

111
Example 10 : If pqr = 1, prove that 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1 = 1.
Solution:
1
Here, pqr = 1, then qr = p = p–1

111
Now, LHS = 1 + p + q–1 + 1 + q + r–1 + 1 + r + p–1
qr r 1
= qr(1 + p + q–1) + r(1 + q + r–1) + 1 + r + p–1

qr r 1
= qr + pqr + r + r + qr + 1 + 1 + r + qr

qr r 1 qr + 1 + r
= qr + 1 + r + qr + r + 1 + qr + r + 1 = qr + 1 + r = 1 = RHS

Example 11: If x – 1 = 223 + 231 , show that x3 – 3x2 – 3x = 1.

Solution:
Here, x – 1 = 232 + 213
21

or, (x – 1)3 = (23 + 23 )3 [Cubing on both sides]
3 213 3 + 3. 223 . 231
or, x3 – 3x2 + 3x – 1 = 232 223 + 213
+

or, x3 – 3x2 + 3x – 1 = 4 + 2 + 6 (x – 1)

or, x3 – 3x2 + 3x – 1 = 6 + 6x – 6

or, x3 – 3x2 + 3x – 6x = 1

or, x3 – 3x2 – 3x = 1

Hence, x3 – 3x2 – 3x = 1 proved.

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EXERCISE 7.1

General section

1. a) Express am × an as a single base.

b) What is the value of (5a)0, a ≠ 0?

c) Find the value of am – n × an – m.

d) What is the value of (p + q)0 + 1p + q?

e) Find the value of: (i) 1 – 1 – n + 1 – 1 – m (ii) (1 – 3–5)–1 + (1 – 35)–1
xm xn

f) Simplify: (i) (a + b)–1.(a–1 + b–1) (ii) y–1 + x–1 –1
x–1 y–1

2. Evaluate: b) 5–7 × 58 c) 11–5 ÷ 11–3 d) (25) 3
a) 29 × 2-6 2

e) (64)– 2 f) 1 1 g) 8 – 4 h) 169 – 1
3 64 6 3 2
27 196

i) (32–1)5–1 j) a0 – 2 k) (70.5)2 l) 9 0.5 × 32 0.2
125 3 25 243

m) 3 64–1 729 –1 3 1
64 4 100
n) 3 o) 100 × 4 p) 3 9 3 9 9

Creative section - A

3. Find the value of:

a) 8 – 1 ÷ 4 – 1 b) 125 – 2 ÷ 625 – 1
27 3 9 2 64 3 256 2

c) 27 – 13 81 41 ÷ 4 – 21 d) 25 – 12 125 13 ÷ 8 – 31
8 16 25 16 64 27

4. Simplify: 1 2
3 3
a) (8a3 ÷ 27x–3)– b) (125p3 ÷ 64q–3)– c) 146 × 155 d) 409 × 498
356 × 65 569 × 358

5. Simplify.

a) (xa)b – c × (xb)c – a × (xc)a – b b) (ax + y)x – y × (ay+z)y – z × (az – x)z + x

c) x2a + 3b × x3a – 4b d) 1 1 + 1 1
xa + 2b × x4a – 3b + ax – y + ay – x

6. Show that:

a) 3x + 1 + 3x = 1 b) 5n + 2 – 5n =1 c) 72p +1 –3× 49p = 1
4 × 3x 24 × 5n × 49p
4

d) 6m + 2 – 6m = 7 e) 7n + 2 + 4 × 7n = 1 f) 5a + 3 – 55 × 5a – 1 = 1
6m+1 – 6m 7n + 1 × 8 – 3 × 7n 5a + 2 + 89 × 5a

7. Simplify:

a) 3p – 3p –1 b) 5x – 5x – 1 c) 3 5 × 2m – 4 × 2m – 2 1
3p + 1 + 4× 5x – 1 × 2m + 2 – 5 × 2m +
3p

d) 2n +2 × (2n – 1)n + 1 ÷ 4n e) 5–n × 625n –1 f) 9x × 3x – 1 – 3x
2n(n – 1) 53n – 2 × (5 × 32x + 1 × 3x – 2 – 3x
2)–1

Vedanta Excel in Mathematics - Book 9 112

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8. Simplify: b) a6b–2c4 ÷ 4 a4b–4c8
a) 25a2b2 × 3 27a3
d) 3 56p7q4
c) 4 16x8y4 ÷ 3 8x6y3 3 7p4q7

e) 4 216m7n5 ÷ 4 6–1m–1n f) 3 (a + b)–7 1

× (a + b) 3

g) 3 (2x – y)–8 ÷ (2x – y)– 2 h) (a + b)–1 × (a – b) (a2 – b2)
3

9. Simplify:

a) xa a+b × xb b+c × xc c + a b) ax x–y × ay y–z × az z – x
xb xc xa a–y a–z a–x

c) xa ×a2 + ab + b2 xb ×b2 + bc + c2 xc d)c2 + ca + a2 xl2+m2 l – m× xm2+n2 m–n× xn2+l2 n–l
xb xc xa x–lm x–mn x–nl

e) xa + b c–a × xb + c a–b × xc + a b – c f) xm + n m–n× xn + p n–p × xp + m p – m
xa – b xb – c xc – a xp xm xn

+ 11 1 1 1 1
p (pq2) 3 + (p2q) 3 q 3 a–c
g) p– q × 1 – h) a b–a × b c–b × c
p 1
3 xb–c xc–a xa–b

x+ 1 a× 1 – x a a2 – 1 a× a – 1 b–a
y y b2 b
i) j)
1 a× 1 a 1 b× 1 a–b
y+ x x – y b2 – a2 b + a

10. Simplify:

a) yz ay × zx az × xy ax b) pq xr – p × pr xq – r × qr xp – q
az ax ay xr – q xq – p xp – r
1 1 1
c) x+y ax2 × y+z ay2 × z+x az2 d) xb xc xa ab
ay2 az2 ax2 bc × ca ×
xc xa xb
111
e) 1 + xa – b + xc – b + 1 + xb – c + xa – c + 1 + xc – a + xb – a

111
f) 1 + ax – y + ax – z + 1 + ay – z + ay – x + 1 + az – x + az – y

11. a) If a3 + b3 + c3 = 0, prove that (xa + b)a2 – ab + b2 × (xb + c)b2 – bc + c2 × (xc + a)c2 – ca + a2 = 1

b) If a = xq + r.yp, b = xr + p.yq and c = xp + q.yr, prove that aq – r × br – p × cp – q = 1

c) If xyz = 1, prove that: 1 + 1 + 1 = 1.
1 + x + y–1 1 + y + z–1 1 + z + x–1
1 1 1
d) If a + b + c = 0, prove that: 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1.

12. a) If x = 1 + 2– 1 , prove that: 2x3 – 6x = 5.
3
23

b) If a = 1 – p– 1 , prove that: a3 + 3a = p – 1 .
3 p
p3

12
c) If x – 2 = 3 3 + 3 3 , show that: x(x2 – 6x + 3) = 2.

113 Vedanta Excel in Mathematics - Book 9

Indices

7.3 Exponential equation

Let's take any two equations: 2x = 4 and 2x = 4.

In the equation 2x = 4, variable x is a base. It is a linear equation. However, in 2x = 4,

variable x is an exponent of the base 2. Such an equation in which variable appears

as an exponent of a base is known as exponential equation.

To solve an exponential equation, we need to have equations with the same base on

either side of the 'equal' sign. Then we compare the powers of equal base and solve

the equation.

Let's study, the following axioms which are used in solving the exponential equations.

(i) If ax = ab, then x = b (ii) If ax = 1, then ax = a0 and x = 0

In this way, while solving an exponential equation, we should simplify the equation
till the equation is obtained in the form ax = ab or ax = 1.

Worked-out examples

Example 1: Solve a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
Solution: 25

a) 2x = 32 b) 52x = 1 c) 7x – 2 = 1
or, 2x = 25 25 or, 7x – 2 = 70
∴ x=5 1 or, x – 2 = 0
or, 52x = 52 ∴ x=2

or, 52x = 5–2

or, 2x = – 2

or, x = – 1

Example 2: Solve a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 =10

Solution:

a) ( 2)x – 3 = ( 4 2)x + 1 b) 3x + 3x + 2 = 10

x–3 x+1 or, 3x + 3x × 32 = 10
or, 2 2 = 2 4

or, x – 3 = x + 1 or, 3x (1 + 9) = 10
2 4
or, 4x – 12 = 2x + 2 or, 3x = 10 =1
10
or, 2x = 14 or, 3x = 30

or, x = 7 or, x = 0

Example 3: Solve a) 2x + 1 × 3x – 2 = 48 b) 9y + 1 = 32y + 1 + 54

Solution: b) 9y + 1 = 32y + 1 + 54
a) 2x + 1 × 3x – 2 = 48
or, 32y + 2 – 32y + 1 = 54
or, 2x × 21 × 3x × 3–2 = 48 or, 32y × 32 – 32y × 3 = 54
or, 32y (32 – 3) = 54
or, 2x × 3x × 2 = 48 or, 32y =9
32 or, 32y = 32
or, (2 × 3)x = 216 or, 2y =2

or, 6x = 63 ∴y =1

∴ x =3

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Example 4: If ax = b, by = c and cz = a, prove that xyz = 1.

Solution: Another process:
by = c
Here, ax = b, by = c and cz = a
Now, ax = b or, axy = c [putting b = ax in by = c]
or, cxz = b [By putting a = cz in ax = b] or, cxyz = c [putting a = cz in axy = c]
or, bxyz = b [Putting c = by in cxz = b] or, xyz = 1
or, xyz = 1 proved.

Example 5: If 4p = 5q = 20–r, show that 1 = 1 + 1 = 0.
Solution: p q r

Another process:

Let, 4p = 5q = 20–r = k 4p = 5q, i.e. 4 = 5pq

Then, 4p = k ∴ 4 = kp1 20–r = 5q, i.e. 20 = 5– q
r

5q = k ∴ 5 = kq1 Now, 4 × 5 = 20
and 20–r = k 1 or, q q
or, = 5– r
∴ 20 = k r 5p × 5

Now, 4 × 5 = 20 5q +1 = 5– q
p r
11 1
or, = k– 2 q q
kp × kq p r
or, +1 = –
k1 + 1 k– 1
or, p q = r q q
p r
1 1 –1 or, +1 + =0
p q r
+ = or, q( 1 + 1 + 1 ) =0
p q r
∴ 1 + 1 + 1 = 0 Proved 1 1 1
p q r or, p + q + r =0

Example 6: Solve: 2x + 1 = 818 .
2x
Solution:
1 = 881
Here, 2x + 2x

or, 2x + 1 = 65
2x 8
1 65
Let 2x = a, then equation becomes a + a = 8

Now, a + 1 = 65
a 8
a2 + 1 65
or, a = 8

or, 8a2 + 8 = 65a

or, 8a2 – 65a + 8 = 0

or, 8a2 – 64a – a + 8 = 0

or, 8a(a – 8) –1(a – 8) = 0

or, (a – 8) (8a – 1) = 0

Either a – 8 = 0 i.e. a = 8 or, 2x = 23 ∴x=3
1 1
or, 8a – 1 = 0 i.e. a = 8 or, 2x = 23 = 2–3 ∴ x = –3
Hence, x = ±3.

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Indices

EXERCISE 7.2

General section
1. a) If (ap × aq) ÷ ar = ax then express x in terms of p, q and r.

b) If (bm ÷ bn) × bp = by then express y in terms of m, n and p.
c) If (xm)n = xm × xn, express m in terms of n.
d) If ax = a , find the value of x.
e) If am = 1, is the value of m?
f) If xx = 4, what is the value of x?

2. Solve. b) 23x = 8 c) 32x = 1 d) 53x = 1
a) 2x = 4 9 125

e) mx – 2 = m2 f) 52x + 3 = 1 g) 1 = 64 h) 1 = 1
43x 7– 2x 49

i) 2 x 8 j) 4 2x= 16 –2 k) 22x + 1 = 23x – 1 l) 32x + 1 = 92x – 1
3 27 5 25
2=

m) 3 = 3x n) 2x – 2 = 2 × 82 o) 33x – 2 = 27 × 92 p) 53x–1 = 25 × 5x+1
3x 3

q) 42x – 1 – 2x + 1 = 0 r) 9x +1 = 1 s) 252x – 3 = 1 t) 49x +2 = 1
27x 625 343–2x
3. a) If px ÷ p4 = 1, find the value of x. b) If 3m ÷ 34 = 27, find the value of m.

c) If 5t = 0.04, find the value of t. d) If 103r = 1 , find the value of r.
0.001

Creative section - A

4. Solve.

a) 2x+5 = 16 b) (35)x – 1 = 25 c) ( 2)3x – 1 = ( 4)x – 2 d) ( 9)x – 3 = ( 3)x +2
f) (0.3)35x = 0.027
e) (0.5) x = 0.25 g) 2 1 2x = 2x h) 9 1 = 27–x
2 × × 32x

5. Solve.

a) 2x + 1 – 2x = 8 b) 3x + 1 – 3x = 54 c) 2x + 2x + 2 = 5

d) 7x + 7x + 1 = 56 e) 11x + 1 + 11x = 12 f) 3x + 2 + 3x + 1 = 1 1
3

g) 2x + 2x – 1 = 3 h) 3x – 3x – 2 = 8 i) 3x + 5 = 3x + 3 + 8
3

j) 2x + 2 + 2x + 3 = 1 k) 3x + 3 + 3x + 4 = 162 l) 5x + 5x + 1 + 5x + 2 = 155
2 3

m) 32x + 3 – 2.9x + 1 = 1 n) 9x – 2 + 2 × 32x – 3 = 63 o) 23x – 1 + 3 × 8x – 1 = 56
9

6. Solve.

a) 2x + 3 × 3x + 4 = 18 b) 2x + 3 × 3x + 2 = 432
c) 2x – 5 × 5x – 4 = 5
e) 23x – 5 × ax – 2 = 2x – 2 × a1 – x d) 72x + 1 × 52x – 1 = 7
5

f) 75x – 4 × a4x – 3 = 72x – 3 × ax – 2

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Indices

7. a) If a = bc, b = ca and c = ab, prove that abc = 1.

b) If xa = y, yb = z and zc = x, prove that abc = 1.
c) If a1x = b13 and ab = 1, prove that x + 3 = 0.

d) If a = 7x, b = 7y and aybx = 49, show that xy = 1.

e) If ax = by and ay = bx, show that x = y.

f) ax = by = cz and b2 = ac, show that y = 2xz
x+z

Creative section - B

8. a) If (a–1 + b–1) (a + b)–1 = ambn, prove that am – n = 1.

b) If m–1n2 7÷ m3n–5 –5 = mxny, prove that mx – 2y = 1.
m2n4 m–2n3

9. Solve:

a) 3x + 1 = 9 1 b) 4x + 1 = 16 1
3x 9 4x 16

c) 4x – 6 × 2x + 1 + 32 = 0 d) 25x – 6 × 5x + 1 + 125 = 0

e) 2x + 21–x = 3 f) 3x – 1 + 32–x = 4

Project work

10. a) Take any base number and index number greater than 1 and less than 6. Then
verify the following laws of indices.

(i) Product law (ii) Quotient law

(iii) Power law (iv) Law of negative index

(v) Root law of indices (vi) Law of zero index

b) Write any five exponential equations of your own in the form of ax = b, where a is
positive integer greater than 1 and b is the integer of power of a. Then solve your
equations.

117 Vedanta Excel in Mathematics - Book 9