Mathematics Part i Solutions Solutions for Class 8 Math Chapter 5 Algebra are provided here with simple step-by-step explanations. These solutions for Algebra are extremely popular among class 8 students for Math Algebra Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part i Solutions Book of class 8 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part i Solutions Solutions. All Mathematics Part i Solutions Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Question 1:

Can’t you find these products on your own?

(i) (p + q) (2m + 3n)

(ii) (4x + 3y) (2a + 3b)

(iii) (4a + 2b) (5c + 3d)

(iv) (m + n) (5a + b)

(v) (2x + 3y) (x + 2y)

(vi) (3a + 2b) (x + 2y)

By the general principle, we know that:

(x + y) (u + v) = xu + xv + yu + yv

(i)

(p + q) (2m + 3n) = (p × 2m) + (p × 3n) + (q × 2m) + (q × 3n)

= 2pm + 3pn + 2qm + 3qn

(ii)

(4x + 3y) (2a + 3b) = (4x × 2a) + (4x × 3b) + (3y × 2a) + (3y × 3b)

= 8ax + 12bx + 6ay + 9by

(iii)

(4a + 2b) (5c + 3d) = (4a × 5c) + (4a × 3d) + (2b × 5c) + (2b × 3d)

= 20ac + 12ad + 10bc + 6bd

(iv)

(m + n) (5a + b) = (m × 5a) + (m × b) + (n × 5a) + (n × b)

= 5am + bm + 5an + bn

(v)

(2x + 3y) (x + 2y) = (2x × x) + (2x × 2y) + (3y × x) + (3y × 2y)

= 2x2 + 4xy + 3xy + 6y2

(vi)

(3a + 2b) (x + 2y) = (3a × x) + (3a × 2y) + (2b × x) + (2b × 2y)

= 3ax + 6ay + 2bx + 4by

#### Question 1:

Find these products on your own:

(i) (x + 3y) (2ab)

(ii) (3x + 5y) (3m − 2n)

(iii) (2r − 3s) (tu)

(iv) (ab) (4x − 3y)

(v) (3a − 5b) (2cd)

(vi) (2p + 5q) (3r − 4s)

By the general principle, we know that:

(x + y) (u v) = xu xv + yu yv

(x y) (u v) = xu xv yu + yv

(x y) (u + v) = xu + xv yu yv

(i)

(x + 3y) (2a b) = (x × 2a) (x × b) + (3y × 2a) (3y × b

= 2ax bx + 6ay 3by

(ii)

(3x + 5y) (3m 2n) = (3x × 3m) (3x × 2n) + (5y × 3m) (5y × 2n)

= 9mx 6nx + 15my 10ny

(iii)

(2r 3s) (t u) = (2r × t) (2r × u) (3s × t) + (3s × u)

= 2rt 2ru 3st + 3su

(iv)

(a b) (4x 3y) = (a × 4x) (a × 3y) (b × 4x) + (b × 3y)

= 4ax 3ay 4bx + 3by

(v)

(3a 5b) (2c d) = (3a × 2c) (5b × 2c) (3a × d) + (5b × d)

= 6ac 10bc 3ad + 5bd

(vi)

(2p + 5q) (3r 4s) = (2p × 3r) (2p × 4s) + (5q × 3r) (5q × 4s

= 6pr 8ps + 15qr 20qs

#### Question 1:

Find the squares of the following algebraic expressions

(i) 2x + 3y

(ii) x + 2

(iii) 2x + 1

We know that (x + y)2 = x2 + y2 + 2xy

(i)

Square of 2x + 3y = (2x + 3y)2

= (2x)2 + (3y)2 + 2 × 2x × 3y

= 4x2 + 9y2 + 12xy

(ii)

Square of x + 2 = (x + 2)2

= (x)2 + (ii)2 + 2 × 2 × x

= x2 + 4 + 4x

(iii)

Square of 2x + 1 = (2x + 1)2

= (2x)2 + (i)2 + 2 × 2x × 1

= 4x2 + 1+ 4x

#### Question 2:

Compute the squares of these numbers mentally.

(i) 102

(ii) 202

(iii) 1001

(iv) 2002

(v) 205

(vi) 10.3

We know that (x + y)2 = x2 + y2 + 2xy

(i)

102 = 100 + 2

(102)2 = (100 + 2)2

= (100)2 + 2 × 100 × 2 + 22

= 10000 + 400 + 4

=10404

(ii)

202 = 200 + 2

(202)2 = (200 +2)2

= (200)2 + 2 × 200 × 2 + 22

= 40000 + 800 + 4

= 40804

(iii)

1001 = 1000 + 1

(1001)2 = (1000 + 1)2

= (1000)2 + 2 × 1000 × 1 + 12

= 1000000 + 2000 + 1

= 1002001

(iv)

2002 = 2000 + 2

(2002)2 = (2000 + 2)2

= (2000)2 + 2 × 2000 × 2 + 22

= 4000000 + 8000 + 4

= 4008004

(v)

205 = 200 + 5

(205)2 = (200 + 5)2

= (200)2 + 2 × 200 × 5 + 52

= 40000 + 2000 + 25

= 42025

(vi)

10.3 = 10 + 0.3

(10.3)2 = (10 + 0.3)2

= (10)2 + 2 × 10 × 0.3 + (0.3)2

= 100 + 6 + 0.09

=106.09

#### Question 3:

Prove that x2 + 6x + 9 is a perfect square for all natural numbers x. What can you say about its square root?

To show that the expression x2 + 6x + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.

x2 + 6x + 9 = x2 + 2 × 3 × x + 32

= (x + 3)2 {(x + y)2 = x2 + y2 + 2xy}

x2 + 6x + 9 = (x + 3)2

Hence, x2 + 6x + 9 is a perfect square.

Square root of x2 + 6x + 9 =

=

= x + 3

#### Question 4:

Prove that in the sequence 1, 2, 3, ….. of natural numbers, 1 added to the product of any two alternatives numbers perfect square.

Let us assume a number a in the sequence of natural numbers 1, 2, 3,…

Alternate number after a = a + 2

Product of the two alternate numbers = (a + 2) × a

= a2 + 2a

If one is added to this expression, it will become a2 + 2a + 1.

Now, a2 + 2a + 1 = a2 + 2 × a × 1 + 12

= (a + 1)2 {(x + y)2 = x2 + y2 + 2xy}

Thus, when 1 is added to the product of any two alternate natural numbers, the result obtained is a perfect square.

#### Question 1:

Find the squares of the algebraic expressions given below:

(i) 2x − 3y

(ii) x − 2

(iii) 2x − 1

We know that (x y)2 = x2 + y2 2xy

(i)

Square of 2x 3y = (2x 3y)2

= (2x)2 + (3y)2 2 × 2x × 3y

= 4x2 + 9y2 12xy

(ii)

Square of x 2 = (x 2)2

= x2 + 22 2 × x × 2

= x2 + 4 4x

(iii)

Square of 2x 1 = (2x 1)2

= (2x)2 + (i)2 2 × 2x × 1

= 4x2 + 1 4x

#### Question 2:

Mentally compute the squares of the numbers given below:

(i) 98

(ii) 198

(iii) 999

(iv) 1998

(v) 195

(vi) 9.7

We know that (x y)2 = x2 + y2 2xy

(i)

98 = 100 2

(98)2 = (100 2)2

= (100)2 + 22 2 × 100 × 2

= 10000 + 4 400

= 9604

(ii)

198 = 200 2

(198)2 = (200 2)2

= (200)2 + 22 2 × 200 × 2

= 40000 + 4 800

= 39204

(iii)

999 = 1000 1

(999)2 = (1000 1)2

= (1000)2 + 12 2 × 1000 × 1

= 1000000 + 1 2000

= 998001

(iv)

1998 = 2000 2

(1998)2 = (2000 2)2

= (2000)2 + 22 2 × 2000 × 2

= 4000000 + 4 8000

= 3992004

(v)

195 = 200 5

(195)2 = (200 5)2

= (200)2 + 52 2 × 200 × 5

= 40000 + 25 2000

= 38025

(vi)

9.7 = 10 0.3

(9.7)2 = (10 0.3)2

= (10)2 + (0.3)2 2 × 10 × 0.3

= 100 + 0.09

= 94.09

#### Question 3:

Prove that x2 − 6x + 9 is a perfect square for all natural numbers x. What can you say about its square root?

To show that the expression x2 6x + 9 is a perfect square, we need to show that it can be written as the square of a linear expression.

x2 6x + 9 = x2 2 × 3 × x + 9

= (x 3)2 {(x y)2 = x2 + y2 2xy}

x2 6x + 9 = (x 3)2

Hence, x2 6x + 9 is a perfect square.

Square root of x2 6x + 9 =

=

= x 3

#### Question 1:

Can’t you easily find the products below like this?

(i) 51 × 49

(ii) 98 × 102

(iii) 10.2 × 9.8

(iv) 7.3 × 6.7

We know that (x y)(x + y) = x2 y2

(i)

The number 51 can be written as (50 + 1) and 49 can be written as (50 1).

51 × 49 = (50 + 1) (50 1)

= 502 12

= 2500 1

= 2499

(ii)

The number 98 can be written as (100 2) and 102 can be written as (100 + 2).

98 × 102 = (100 2) (100 + 2)

= (100)2 22

= 10000 4

= 9996

(iii)

The number 10.2 can be written as (10 + 0.2) and 9.8 can be written as (10 0.2).

10.2 × 9.8 = (10 + 0.2) (10 0.2)

= (10)2 (0.2)2

= 100 0.04

= 99.96

(iv)

The number 7.3 can be written as (7 + 0.3) and 6.7 can be written as (7 0.3).

7.3 × 6.7 = (7 + 0.3) × (7 0.3)

= 72 (0.3)2

= 49 0.09

= 48.91

#### Question 1:

(i) 672 − 332

(ii) 1232 − 1222

(iii)

(iv) 0.272 − 0.232

We know that x2 y2 = (x + y) (x y)

(i)

672 332 = (67 + 33) (67 33)

= 100 × 34

= 3400

(ii)

1232 1222 = (123 + 122) (123 122)

= 245 × 1

= 245

(iii)

(iv)

(0.27)2 (0.23)2 = (0.27 + 0.23) (0.27 0.23)

= 0.5 × 0.04

= 0.02

#### Question 2:

The diagonal of a rectangle is 65 centimetres long and one of its sides is 63 centimeres long. What is the length of the other side?

Given: A rectangle ABCD with length of one side, DC = 63 cm and length of the diagonal, BD = 65 cm.

We know that each angle of a rectangle measures 90°.

In ΔBCD, BCD = 90°

Using Pythagoras theorem, we get:

BD2 = BC2 + CD2

(65)2 = BC2 + (63)2

BC2 = (65)2 (63)2

= (65 63) (65 + 63) {x2 y2 = (x y) (x + y)}

= 2 × 128

= 256

BC = 16

Thus, the length of the other side of the rectangle is 16 cm.

#### Question 1:

Prove that the difference of the squares of two consecutive natural is equal to their sum.

Let n be a natural number.

Its consecutive natural number = n + 1

Sum of these consecutive natural numbers = n + (n +1) = 2n +1

Difference of the squares of two consecutive natural numbers

= (n + 1)2 n2

= (n + 1 n) (n + 1 + n) {x2 y2 = (x y) (x + y)}

= 2n + 1

Thus, the difference of the squares of two consecutive natural numbers is equal to their sum.

#### Question 2:

In how many different ways you can write 15 as the difference of the squares of two natural numbers?

The factors of 15 are 1, 3, 5 and 15.

We can write 15 as:

15 = 1 × 15

= (8 7) (8 + 7) {x2 y2 = (x y) (x + y)}

= 82 72

So, 15 can be written as the difference of the squares of 8 and 7.

15 = 3 × 5

= (4 1) (4 + 1) {x2 y2 = (x y) (x + y)}

= 42 12

So, 15 can be written as the difference of the squares of 4 and 1.

Thus, there are two ways to write 15 as the difference of the squares of two natural numbers.

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