## AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.5 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.5

Question 1.
Use suitable identities to find the following products.
i) (x + 5) (x + 2)
Solution:
(x + 5) (x + 2)
= x2 + (5 + 2)x + 5 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= x2 + 7x + 10

ii) (x – 5) (x – 5)
Solution:
(x – 5) (x – 5)
= (x – 5)2 = x2 – 2(x) (5) + 52
[ ∵(x – y)2 = x2 – 2xy + y2]
= x2 – 10x + 25

iii) (3x + 2) (3x – 2)
Solution:
(3x + 2) (3x – 2) = (3x)2 – (2)2
[∵ (x + y) (x – y) =x2 – y2]
= 9x2 – 4

iv) $$\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{2}-\frac{1}{x^{2}}\right)$$
Solution:

v) (1 + x) (1 + x)
Solution:
(1 + x) (1 + x)
= (1 + x)2 = 12 + 2 (1) (x) + x2
[∵(x + y)2 = x2 + 2xy + y2]
= 1 + 2x + x2

Question 2.
Evaluate the following products with¬out actual multiplication.
i) 101 x 99
Solution:
101 x 99
= (100 + 1) (100 – 1)
= 1002 – 12
= 10000 – 1
= 9999

ii) 999 x 999
Solution:
999 x 999
= 9992
= (1000 – 1)2
= 10002 – 2 x (1000) x 1 + 12
= 1000000-2000 + 1
= 998001

iii) $$50 \frac{1}{2} \times 49 \frac{1}{2}$$
Solution:

iv) 501 x 501
Solution:
501 x 501
= (500 + 1) (500 + 1)
= (500 + 1)2
= 5002 + 2 x (500) x 1 + 12
= 250000 + 1000 + 1 = 251001

v) 30.5 x 29.5 = (30 + 0.5) (30 – 0.5)
= 302 – (0.5)2
= 900 – 0.25
= 899.75

Question 3.
Factorise the following using appro-priate identities.
i) 16x2 + 24xy + 9y2
Solution:
16x2 + 24xy + 9y2
= (4x)2 + 2 (4x) (3y) + (3y)2
= (4x + 3y)2 = (4x + 3y) (4x + 3y)
[ ∵ (x + y)2 = x2 + 2xy + y2]

ii) 4y2 – 4y + 1
Solution:
4y2 – 4y + 1
= (2y)2 – 2 (2y) (1) + (1)2
[ ∵ (x -y)2 = x2 – 2xy + y2]
= (2y -1)2 = (2y – 1) (2y-1)

iii) $$4 x^{2}-\frac{y^{2}}{25}$$
Solution:

iv) 18a2 – 50
Solution:
18a2 – 50 = 2 (9a2 – 25)
= 2[(3a)2 – (5)2]
[ ∵ x2 – y2 = (x + y) (x – y)]
= 2 (3a + 5) (3a – 5)

v) x2 + 5x + 6
Solution:
x2 + 5x + 6 = x2 + (3 + 2) x + 3 x 2
[ ∵ (x + a) (x + b) = x2 + (a + b) x + a . b]
= (x + 3) (x + 2)

vi) 3p2 – 24p + 36
Solution:
3p2 – 24p + 36
= 3[p2 – 8p + 12]
= 3[p2 + (- 6 – 2)p + (- 6) (- 2)]
[ ∵ (x + a) (x + b) = x2 + (a + b) x + ab]
= 3 (p – 6) (p – 2)

Question 4.
Expand each of the following, using suitable identities.
i) (x + 2y + 4z)2
(x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2(x) (2y) + 2 (2y) (4z) + 2 (4z) (x)
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx]
= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

ii) (2a – 3b)3
Solution:
(2a – 3b)3 = (2a)3 – 3 (2a)2 (3b) + 3 (2a) (3b)2 – (3b)3
[ ∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 8a3 – 3(4a2) (3b) + 3 (2a) (9b2) – 27b3
= 8a3 – 36a2b + 54ab2-27b3
(or)
∵ (a – b)3 = a3 – b3– 3ab (a – b)]
= (2a)3 – (3b)3 – 3(2a) (3b) (2a – 3b)
= 8a3 – 27b3 – 18ab (2a – 3b)

iii) (- 2a + 5b – 3c)2
Solution:
(- 2a + 5b – 3c)2
= (- 2a)2 + (5b)2 + (- 3c)2 + 2 (- 2a) (5b) + 2 (5b) (- 3c) + 2 (- 3c) (- 2a)
= 4a2 + 25b2 + 9c2 – 20ab – 30bc + 12ca
[ ∵ (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz +2za]

iv) $$\left[\frac{a}{4}-\frac{b}{2}+1\right]^{2}$$
Solution:

v) (p + 1)3
Solution:
(p + 1)3
= (P)3 + 3 (p)2 (1) + 3 (p) (1)2 + (1)3
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= p3 + 3p2 + 3p + 1

vi) $$\left(x-\frac{2}{3} y\right)^{3}$$
Solution:

Question 5.
Factorise
i) 25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
Solution:
25x2 + 16y2 + 4z2 – 40xy + 16yz – 20xz
= (5x)2 + (- 4y)2 + (- 2z)2 + 2(5x) (- 4y) + 2 (- 4y) (- 2z) + 2 (- 2z) (5x)
= (5x – 4y – 2z)2 = (- 5x + 4y +, 2z)2

ii) 9a2 + 4b2 + 16c2 + 12ab – 16bc – 24ca
Solution:
9a2 + 4b2 + 16c2 + 12ab – 16bc -24ca
= (3a)2 + (2b)2 + (- 4c)2+ 2 (3a) (2b) + 2 (2b) (- 4c) + 2(- 4c) (3a)
= (3a + 2b – 4c)2

Question 6.
If a + b + c = 9 and ab + be + ca = 26, find a2 + b2 + c2.
Solution:
Given that a + b + c = 9
Squaring on both sides,
(a + b + c)2 = 92
⇒ a2+ b2 + c2+ 2 (ab + be + ca) = 81 ⇒ a2 + b2 + c2 = 81 – 2 (ab + be + ca)
(by problem)
= 81 – 2 x 26
= 81 – 52 = 29

Question 7.
Evaluate the following by using suit¬able identities. m EachgM)
i) (99)3
Solution:
(99)2 = (100 – 1)3
= 1003 – 3 (100)2 (1) + 3 (100) (1)2 – 13
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 + y3]
= 1000000 – 30000 + 300 – 1
= 970299

ii) (102)3
Solution:
(102)3 = (100 + 2)3
= 1003 + 3 (100)2 (2) + 3 (100) (2)2 + 23
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3]
= 1000000 + 60000 + 1200 + 8
= 1061208

iii) (998)3
Solution:
(998)3 =(1000 – 2)3
[ ∵ (x – y)3 = x3 – 3x2y + 3xy2 – y3] = 10003– 3(1000)2(2) + 3(1000)(2)2– 23
= 1000000000 – 6000000 + 12000 – 8
= 994011992

iv) (1001)3
Solution:
(1001)3 = (1000 + 1)3 .
[ ∵ (x + y)3 = x3 + 3x2y + 3xy2 + y3] = 10003 + 3(1000)2(1) + 3(1000) (1)2 + 13
= 1000000000 + 3000000 + 3000 + 1
= 1003003001

Question 8.
Factorise each of the following.
i) 8a3 + b3 + 12a2 b + 6ab2
Solution:
8a3 + b3 + 12a2 b + 6ab2
= (2a)3 + (b)3 + 3 (2a)2 (b) + 3 (2a) (b)2
= (2a + b)3

ii) 8a3 – b3 – 12a2 b + 6ab2
Solution:
8a3 – b3 – 12a2 b + 6ab2
= (2a)3 – (b)3 – 3 (2a)2 (b) + 3 (2a) (b)2
= (2a – b)3

iii) 1 – 64a3 -12a + 48a2
Solution:
1 – 64a3 – 12a + 48a2
= (1)3 – (4a)3 – 3(1)2 (4a) + 3(1) (4a)2
= (1 – 4a)3

iv) $$8 p^{3}-\frac{12}{5} p^{2}+\frac{6}{25} p-\frac{1}{125}$$
Solution:

Question 9.
Verify i) x3 + y3 = (x + y) (x2 – xy + y2);
ii) x3 – y3 = (x – y) (x2 + xy + y2)
Using some non-zero positive integers and check by actual multiplication. Can you
call these as identities ?
i) x3 + y3 = (x + y) (x2 – xy + y2)
Solution:
Given x3 + y3 = (x + y) (x2 – xy + y2)
L.H.S = x3 + y3
R.H.S = (x + y) (x2 – xy + y2)
= x (x2 – xy + y2) + y (x2 – xy + y2)
= x3 -x2y + xy2 + x2y – xy2 + y3
= x3 + y3
= L.H.S
∴ L.H.S = R.H.S

Take x = 3, y = 2
L.H.S = 33 + 23 = 27 + 8 = 35
R.H.S = (3 + 2) (32 – 3 x 2 + 22)
= 5 x (9 – 6 + 4)
= 5 x 7 = 35
∴ L.H.S = R.H.S

ii) x3 – y3 = (x – y) (x2 + xy + y2)
Solution:
Given that x3 – y3 = (x – y) (x2 + xy + y2)
L.H.S = x3 – y3
R.H.S = (x – y) (x2 + xy + y2)
= x (x2 + xy + y2) – y (x2 + xy + y2)
= x3 + x2y + xy2 – x2y – xy2 – y3
= x3 – y3= L.H.S

L.H.S = 33 – 23 = 27 – 8 = 19
R.H.S = (3 – 2) (32 + 3 x 2 + 22)
= 1 x (9 + 6 + 4)
= 1 x 19 = 19
∴ L.H.S = R.H.S
We can call the above two expressions as identities

Question 10.
Factorise by using the above results (identities).
i) 27a3 + 64b3
Solution:
27a3+ 64b3 = (3a)3 + (4b)3
= (3a + 4b) {(3a)2 – (3a) (4b) + (4b)2}
= (3a + 4b) (9a2 – 12ab + 16b2)

ii) 343y3 – 1000
Solution:
343y3 – 1000 = (7y)3 – (10)3
= (7y – 10) [(7y)2 + (7y) (10) + (10)2]
= (7y – 10) (49y2 + 70y + 100)

Question 11.
Factorise 27x3 + y3 + z3 – 9xyz using identity.
Solution:
Given 27x3 + y3 + z3 – 9xyz
= (3x)3 + (y)3 + (z)3 – 3 (3x) (y) (z)
= (3x + y + z)
[(3x)2 + y2 + z2 – (3x) (y) – (y) (z) – (z) (3x)]
[ ∵ (x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2– xy – yz – zx)
= (3x + y + z) (9x2 + y2 + z2 – 3xy – yz – 3xz)

Question 12.
Verify that x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
(OR)
Verify that
p3 + q3 + r3 – 3pqr = 1/2 (p + q + r)
[(p – q)2 + (q – r)2 + (r – p)2]
Solution:
Given x3+ y3 + z3 – 3xyz = 1/2 (x + y + z) [(x – y)2 + (y – z)2 + (z – x)2 ]
R-H.S = 1/2 (x + y + z) [(x – y)2 + (y – z)2+ (z – x)2]
= 1/2 (x + y + z) [x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2xz]
= 1/2 (x + y + z) [2x2 + 2y2 + 2z2 – 2xy – 2yz – 2zx]
= 1/2 (x + y + z) (2) [x2 + y2 + z2 – xy – yz – zx]
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= L.H.S
Hence proved.

Question 13.
If x + y + z = 0, show that x3 + y3 + z3 = 3xyz
Solution:
Given x + y + z = 0
To prove x3 + y3 + z3 = 3xyz
We have an identity
(x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz
Substituting x + y + z = 0in the above equation, we get
0 x (x2 + y2 + z2 -xy-yz-zx)
= x3 + y3 + z3 – 3xyz
⇒ x3 + y3 + z3 – 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz

Question 14.
Without actual calculating the cubes, find the value of each of the following.
i) (- 10)3 + 73 + 33
Solution:
Given (-10)3 + 73 + 33
Sum of the bases = -10 + 7 + 3 = = 0
∴ (- 10)3 + 73 + 33
= 3 (- 10) x (7) x 3
= -630
[ ∵ x + y + z = 0 then x3 + y3 + z3 = 3xyz]

ii) (28)3 + (- 15)3 + (- 13)3
Solution:
Given (28)3 + (- 15)3+ (- 13)3
Sum of the bases = 28 + (- 15) + (- 13) = 0
∴ (28)3 + (- 15)3 + (- 13)3
= 3 x 28 x (- 15) x (- 13)
= 16380

iii) $$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$$ read it as $$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}+\left(\frac{-5}{6}\right)^{3}$$
Solution:

iv) (0.2)3 – (0.3)3 + (0.1)3
Solution:
Given that (0.2)3 – (0.3)3 + (0.1)3
= (0.2)3 + (- 0.3)3 + (0.1)3
Sum of the bases = 0.2 – 0.3 + 0.1 = 0
∴ (0.2)3 + (-0.3)3 + (0.1)3
= 3 x (0.2) (- 0.3) (0.1)
= -0.018

Question 15.
Give possible expressions for the length and breadth of the rectangle whose area is given by
i) 4a2 + 4a – 3
Given that area = 4a2 + 4a – 3
= 4a2 + 6a – 2a – 3
= 2a (2a + 3) – 1 (2a + 3)
= (2a – 1) (2a + 3)
∴ Length = (2a + 3); breadth = (2a – 1).

ii) 25a2 – 35a + 12
Solution:
Given that area = 25a2 – 35a +12
= 25a2 – 20a – 15a + 12
= 5a (5a – 4) – 3 (5a – 4)
= (5a – 4) (5a – 3)
∴ (5a – 4) (5a – 3) are the length and breadth.

Question 16.
What are the possible polynomial expressions for the dimensions of the cuboids whose volumes are given below ?
i) 3x3 – 12x
Solution:
Volume = 3x3 – 12x
= 3x (x2 – 4)
= 3x (x + 2) (x – 2) are the dimensions.

ii) 12y2 + 8y – 20
Solution:
Given that volume = 12y2 + 8y – 20
= 4 (3y2 + 2y – 5)
= 4 [3y2 + 5y – 3y – 5]
= 4 [y (3y + 5) – 1 (3y + 5)]
= 4 (3y + 5) (y – 1)
Hence 4, (3y + 5) and (y – 1) are the dimensions.

Question 17.
Show that if 2 (a2 + b2 ) = (a + b)2 then a = b
Solution:
Given that 2 (a2 + b2 ) = (a + b)2
To prove a = b
As 2 (a2 + b2 ) = (a + b)2
We have
2a2 + 2b2 = a2 + 2ab + b2
2a2 – a2 + 2b2 – b2 = 2ab
a2 + b2 = 2ab
This is possible only when a = b
∴ a = b

## AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.
Determine which of the following polynomials has (x + 1) as a factor.
i) x3 – x2 – x + 1
Solution:
f(- 1) = (- 1)3 – (- 1)2 – (- 1) + 1
= -1 – 1 + 1 + 1 = 0
∴ (x + 1) is a factor.

ii) x4 -x3 +x2 – x + 1
Solution:
f(- 1) = (- 1)4 – (- 1)3 + (- 1)2 – (- 1) + 1
= 1 + 1 + 1 + 1 + 1= 5
∴ (x + 1) is not a factor.

iii) x4 + 2x3 + 2x2 + x + 1
Solution:
f(- 1) = (-1)4 + 2 (- 1)3 + 2 (- 1)2 + (-1) + 1
= 1 – 2 + 2 – 1 + 1 = 1
∴ (x + 1) is not a factor.

iv) x3 – x2 – (3 – √3)x + √3
Solution:
f(- 1) = (- 1)3 – (- 1)2 – (3 – √3)(-1) + √3
= – 1 – 1 + 3 – √3 + √3 = 1
∴ (x + 1) is not a factor.

Question 2.
Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
i) f(x) = 5x3 + x2 – 5x – 1; g(x) = x + 1
[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]
Solution:
g(x) = x+ 1 = x- a say
∴ a = – 1
f(a) = f(- 1) = 5 (- 1)3 + (- 1)2 – 5 (- 1) – 1
= -5 + 1 + 5 – 1 = 0
∴ x + 1 is a factor of f(x).

ii) f(x) = x3 + 3x2 + 3x + 1; g(x) = x + 1
Solution:
g(x) = x + 1 = x – a
∴ a = – 1
f(a) = f(- 1) = (- 1)3 + 3 (- 1)2 + 3(-1) + 1
= -1 + 3 – 3 + 1 =0
∴ f(x) is a factor of g(x).

iii) f(x) = x3 – 4x2 + x + 6;
g(x) = x – 2
Solution:
g(x) = x- 2 = x- a
∴ a = 2
f(a) = f(2) = 23 – 4(2)2 + 2 + 6
= 8 – 16 + 2 + 6 = 0
∴ g(x) is a factor of f(x).

iv) f(x) = 3x3+ x2 – 20x +12; g(x) = 3x – 2
Solution:
g(x) = 3x – 2 = $$x-\frac{2}{3}$$ = x – a
∴ a = 2/3

v) f(x) = 4x3+ 20x2+ 33x + 18; g(x) = 2x + 3
Solution:
g(x) = 2x + 3 = x + $$\frac{3}{2}=$$ = x – a
∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.
Show that (x – 2), (x + 3) and (x – 4) are factors of x3 – 3x2 – 10x + 24.
Solution:
Given f(x) = x3 – 3x2 – 10x + 24
To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)
f(2) = 23 – 3(2)2 – 10(2) + 24
= 8- 12-20 + 24 = 0
∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)3 – 3(- 3)2– 10(- 3) + 24
= – 27 – 27 + 30 + 24 = 0
∴ (x + 3) is a factor of f(x).

f(4) = (4)3 – 3 (4)2 – 10 (4) + 24
= 64 – 48 – 40 + 24
= 88 – 88
= 0
∴ (x – 4) is a factor of f(x).

Question 4.
Show that (x + 4), (x – 3) and (x – 7) are factors of x3 – 6x2 – 19x + 84.
Solution:
Let f(x) = x3 – 6x2 – 19x + 84
To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)
f(- 4) = (- 4)3 – 6 (- 4)2 – 19 (- 4) + 84
= -64 – 96 + 76 + 84
= 0 .
∴ (x + 4) is a factor of f(x).

f(3) = 33 – 6(3)2 – 19(3) + 84
= 27 – 54 – 57 + 84
= 0
∴ (x – 3) is a factor of f(x).

f(7) = 73 – 6(7)2 – 19(7) + 84
= 343 – 294 – 133 + 84
= 427 – 427
= 0
∴ (x – 7) is a factor of f(x).

Question 5.
If both (x – 2) and $$\left(x-\frac{1}{2}\right)$$ of px2 + 5x + r, show that p = r.
Solution:
Let f(x) = px2+ 5x + r
As (x – 2) and $$\left(x-\frac{1}{2}\right)$$ are factor of f(x), we have f(2) = 0 and f(1/2) = 0
∴ f(2) = p(2)2 + 5(2) + r
= 4p + 10 + r = 0
= 4p + r
= – 10 ………………(1)

⇒ p + 10 + 4r = 0
⇒ p + 4r = – 10 ………………. (2)
From (1) and (2);
4p + r = p + 4r
4p – p = 4r – r
3p = 3r
∴ P = r

Question 6.
If (x2 – 1) is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0.
Solution:
Let f(x) = ax4 + bx3 + cx2 + dx + e
As (x – 1) is a factor of f(x) we have
x2 – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0
f(1) = a + b + c + d + e = 0 ……………. (1)
and f(-1) = a- b + c- d + e = 0
⇒ a + c + e = b + d
Substitute this value in equation (1)
a + c + e + b + d=0
b + d + b + d=0
2 (b + d) = 0
⇒ b + d = 0
∴ a + c + e = b + d = 0

Question 7.
Factorise
i) x3 – 2x2 – x + 2
Solution:
Let f(x) = x3 – 2x2 – x + 2
By trial, we find f(l) = 13 – 2(1)2 – 1 + 2
= 1 – 2 – 1 + 2
= 0 .
∴ (x – 1) is a factor of f(x).
[by factor theorem]
Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x2 – x – 2)
= (x – 1) [x2 – 2x + x- 2]
= (x – 1) [x (x – 2) + 1 (x – 2)]
= (x – 1) (x – 2) (x + 1)

ii) x3 – 3x2 – 9x – 5
Solution:
Let f(x) = x3 – 3x2 – 9x – 5By trial,
f(- 1) = (- 1)3 – 3(- 1)2 – 9(- 1) – 5
=-1 – 3 + 9 – 5
=0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem]
Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x2 – 4x – 5)
But x2– 4x – 5 = x2 – 5x + x – 5
= x (x – 5) + 1 (x – 5)
=(x – 5)(x + 1)
∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x3 + 13x2 + 32x + 20
Solution:
Let f(x) = x3 + 13x2 + 32x + 20
Let f(- 1)
= (- 1)3 + 13 (- 1)2 + 32 (- 1) + 20
= – 1 + 13 – 32 + 20 = 33 – 33 = 0
∴ (x + 1) is a factor of f(x).
[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y3 + y2 – y – 1
Let f(y) = y3 + y2 – y – 1
f(1) = 13+ 12– 1 – 1 = 0
(y – 1) is a factor of f(y).
Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x2 + 12x + 20)
But (x2 + 12x + 20) = x2+ 10x + 2x + 20
=x(x + 10)+2(x + 10)
=(x + 10)(x + 2)
∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.
If ax2 + bx + c and bx2 + ax + c have a common factor x + 1 then show that c = 0 and a = b.
Solution:
Let f(x) = ax2 + bx + c and g(x) = bx2 + ax + c given that (x + 1) is a common factor for both f(x) and g(x).
∴ f(-1) = g(- 1)
⇒a(- 1)2 + b(- 1) + c
= b(- 1)2 + a (- 1) + c
⇒ a – b + c = b – a + c
⇒ a + a = b + b
⇒ 2a = 2b
⇒ a = b
Also f(- 1) = a – b + c = 0
⇒ b – b + c = 0
⇒ c = 0

Question 9.
If x2 – x – 6 and x2 + 3x – 18 have a common factor x – a then find the value of a.
Solution:
Let f(x) = x2 – x – 6 and
g(x) = x2 + 3x – 18
Given that (x – a) is a factor of both f(x) and g(x).
f(a) = g(a) = 0
⇒ a2 – a – 6 = a2 + 3a – 18
⇒ – 4a = – 18 + 6
⇒ – 4a = – 12
∴ a = 3

Question 10.
If (y – 3) is a factor of y3– 2y2– 9y + 18, then find the other two factors.
Solution:
Let f(y) = y3– 2y2 – 9y + 18
Given that (y – 3) is a factor of f(y).
Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)
But y2 + y – 6
= y2 + 3y – 2y – 6
= y (y + 3) – 2 (y + 3)
= (y + 3) (y – 2)
∴ f(y) = (y – 2)(y – 3)(y + 3)
The other two factors are (y – 2) and (y + 3).

## AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3

Question 1.
Find the remainder when
x3 + 3x2 + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each
Solution:
Let f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is f (- 1)
f (- 1) = (- 1)3 + 3(- 1)2 + 3(- 1) + 1
= – 1 + 3 – 3 + 1 = 0

ii) $$x-\frac{1}{2}$$
f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is $$\mathrm{f}\left(\frac{1}{2}\right)$$

iii) x
Solution:
f(x) = x3 + 3x2 + 3x + 1
The remainder is f(0)
∴ f(0) = 03 + 3(0)2 + 3(0) + 1 = 1

iv) x + π
Solution:
f(x) = x3 + 3x2 + 3x + 1
By remainder theorem, the remainder is f(- π)
f(- π) = (- π)3 + 3(-π)2 + 3 (-π) + 1 .
= – π3 + 3π2 – 3π + 1

v) 5 + 2x
f(x) = x3 + 3x2 + 3x + 1
The remainder is $$\mathrm{f}\left(\frac{-5}{2}\right)$$

Question 2.
Find the remainder when x3 – px2 + 6x – p is divided by x – p.
Solution:
Let f(x) = x3 – px2 + 6x – p
(x – a) = x – p)
By Remainder theorem, the remainder is f(p)
∴ f(P) = P3 – P(P)2 + 6p – p
= p3 – p3 + 5p = 5p

Question 3.
Find the remainder when 2x2 – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.
Solution:
Let f(x) = 2x2 – 3x + 5 and
x – a = 2x – 3 = x – $$\frac{3}{2}$$
By Remainder theorem f(x) when divided by (x – $$\frac{3}{2}$$ ) leaves a remainder $$\mathrm{f}\left(\frac{3}{2}\right)$$

As the remainder is 5 we say that (2x – 3) is not a factor of f(x).

Question 4.
Find the remainder when 9x3 – 3x2 + x – 5 is divided by x – $$\frac{2}{3}$$
Solution:
Let f(x) = 9x3 – 3x2 + x – 5
x-a = x – $$\frac{2}{3}$$
Remainder theorem the remainder is $$\mathrm{f}\left(\frac{2}{3}\right)$$

Question 5.
If the polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x + a leave the same remainder, when divided by x – 2, find the value of a.
Solution:
Let f(x) = 2x3 + ax2 + 3x – 5
g(x) = x3 + x2 – 4x + a
Given that f(x) and g(x) divided x – 2
give same remainder.
i e., f(2) = g(2)
By Remainder theorem.
But f(2) = 2(2)3 + a(2)2 + 3(2) – 5
= 2 x 8 + 4a + 6 – 5
= 17 +4a
g(2) = 23 + 22 – 4(2) + a .
= 8 + 4 – 8 + a = 4 + a
i.e., 4 + a = 17 + 4a
∴ a – 4a = 17 – 4
– 3a = 13
a = -13/3

Question 6.
If the polynomials x3 + ax2 + 5 and x3 – 2x2 + a are divided by (x + 2) leave the same remainder, find the value of a.
Solution:
Let f(x) = x3 + ax2 + 5
g(x) = x3 – 2x2 + a
Given that when f(x) and g(x) divided by (x + 2) leaves the same remainder.
i.e.,f(-2) = g(-2)
By Remainder theorem
f(- 2) = (- 2)3 + a(- 2)2 + 5
= -8 + 4a + 5 = 4a – 3
g(- 2) = (- 2)3 – 2(- 2)2 + a
= -8 – 8 + a = a – 16
By problem,
4a – 3 = a – 16
4a – a = – 16 + 3
⇒ 3a = – 13 ⇒ a = -13/3

Question 7.
Find the remainder when f(x) = x4 – 3x2 + 4 is divided by g(x) = x – 2 and verify the result by actual division.
Solution:
Given f(x) = x4 – 3x2 + 4
g(x) = x – 2
The remainder when f(x) is divided by g(x) is f(2).
f(2) = 24 – 3(2)2 + 4
= 16 – 12 + 4
= 8
Actual division

∴ The remainder either by Remainder theorem or by actual division is the same.

Question 8.
Find the remainder when p(x) = x3 – 6x2 + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.
Solution:
Given p(x) = x3 – 6x2 + 14x – 3
g(x) = 1 – 2x
By Remainder theorem when p(x) is divided by g(x) is p(1/2).

Question 9.
When a polynomial 2x3 + 3x2 + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.
Solution:Let f(x) = 2x3 + 3x2 + ax + b
The remainder when f(x) is divided by (x – 2) is 2.
i.e., f(2) = 2
⇒ 2(2)3 + 3(2)2+ a(2) + b = 2
⇒ 16 + 12 + 2a +b = 2
⇒ 2a + b = – 26 …………………..(1)

Also the remainder when f(x) is divided by (x + 2) is – 2.
i.e., f(- 2) = – 2
⇒ 2(- 2)3 + 3(- 2)2 + a (- 2) + b = – 2
⇒ -16 + 12 – 2a + b = – 2
– 2a + b = 2 ………………..(2)
Solving (1) and (2),

b = – 12
and 2a – 12 = – 26
2a = -26+ 12
a = -14/2 = -7,
a = -7, b = – 12

## AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2

Question 1.
Find the value of the polynomial 4x2 – 5x + 3, when
(i) x = 0
Solution:
The value at x = 0 is
4(0)2 – 5(0) + 3
= 3

(ii) x = – 1
Solution:
The value at x = – 1 is
4 (- 1)2 – 5 (- 1) + 3
= 4 + 5 + 3
= 12

iii) x = 2
Solution:
The value at x = 2 is
4(2)2 – 5(2) + 3
= 16 -10 + 3
= 9

iv) x = $$\frac{1}{2}$$
Solution:

Question 2.
Find p(0), p(1) and p(2) for each of the following polynomials.
i) p(x) = x2 – x + 1
Solution:
p(0) = 02 – 0 + 1 = 1
p(1) = 12 – 1 + 1 = 1
p(2) = 22 – 2 + 1 = 3

ii) P(y) = 2 + y + 2y2 – y3
Solution:
p(0) = 2 + 0 + 2(0)2 – 03 = 2
p(1) = 2+ 1 + 2(1)2 – 13 = 4
p(2) = 2 + 2 + 2(2)2 -23 = 4 + 8- 8 = 4

iii) P(z) = z3
Solution:
p(0) = 03 = 0
p(1) = 13 = 1
p(2) = 23 = 8

iv) p(t) = (t – 1)(t + 1) = t2 – 1
Solution:
p(0) = (0 – 1) (0 + 1) = – 1
p(1) = t2 – 1 = 12 – 1 = 0
p(2) = 22 – 1 = 4 – 1 = 3

v) p(x) = x2 – 3x + 2
Solution:
p(0) = 02 – 3(0) + 2 = 2
p(1) = 12 – 3(1) + 2 = 1 – 3 + 2 = 0
p(2) = 22 – 3(2) + 2 = 4- 6 + 2 = 0

Question 3.
Verify whether the values of x given in each case are the zeroes of the polynomial or not ?
i) p(x) = 2x + 1; x = $$\frac{-1}{2}$$
Solution:
The value of p(x) at x = $$\frac{-1}{2}$$ is
$$\mathrm{p}\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)+1$$
= -1 + 1 = 0
∴ x = $$\frac{-1}{2}$$ is a zero of p(x).

(ii) p(x) = 5x – π ; x = $$\frac{-3}{2}$$
Solution:
The value of p(x) at x = $$\frac{-3}{2}$$ is
$$\mathrm{p}\left(\frac{-3}{2}\right)=5\left(\frac{-3}{2}\right)-\pi=\frac{-15}{2}-\pi \neq 0$$
∴ x = $$\frac{-3}{2}$$ is not a zero of p(x).

iii) p(x) = x2 – 1; x = ±1
Solution:
The value of p(x) at x = 1 and – 1 is
p(1) = 12 – 1 = 0
p(-1) = (-1)2 -1 = 0
∴ x = ±1 is a zero of p(x).

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2
Solution:
The value of p(x) at x = – 1 is
p(-1) = (-1 – 1) (-1 + 2)
=-2 x 1 =-2 ≠ 0
Hence x = – 1 is not a zero of p(x).
And the value of p(x) at x = – 2 is
p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0
Hence, x = – 2 is a zero of p(x).

v) p(y) = y2; y = o
Solution:
The value of p(y) at y = 0 is p(0) = 02 = 0
Hence y = 0 is a zero of p(y).

vi) p(x) = ax + b ; x = $$\frac{-\mathbf{b}}{\mathbf{a}}$$
Solution:
The value of p(x) at x = $$\frac{-\mathbf{b}}{\mathbf{a}}$$ is
$$\mathrm{p}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)=\mathrm{a}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)+\mathrm{b}$$
= -b + b = 0
∴ x = $$\frac{-\mathbf{b}}{\mathbf{a}}$$ is a zero of p(x).

vii) f(x) = 3x2 – 1; x = $$\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$$
Solution:

viii) f(x) = 2x – 1; x = $$\frac{1}{2} ;-\frac{1}{2}$$
Solution:

Question 4.
Find the zero of the polynomial in each of the following cases.
i) f(x) = x + 2
Solution:
x + 2 = 0
x = – 2

ii) f(x) = x – 2
Solution:
x – 2 = 0
x = 2

iii) f(x) = 2x + 3
Solution:
2x + 3 = 0
2x = – 3
x = $$\frac{-3}{2}$$

iv) f(x) = 2x – 3
Solution:
2x – 3 = 0
2x = 3
x = $$\frac{3}{2}$$

v) f(x) = x2
Solution:
x2 = 0
x = 0

vi) f(x) = px, p ≠ 0
Solutin:
px = 0
x = 0

vii) f(x) = px + q; p ≠ 0; p, q are real numbers.
Solution:
px + q = 0
px = -q
x = $$\frac{-\mathrm{q}}{\mathrm{p}}$$

Question 5.
If 2 is a zero of the polynomial p(x) = 2x2 – 3x + 7a, find the value of
a.
Solution:
Given that 2 is a zero of p(x) = 2x2 – 3x + 7a
(i.e.) p(2) = 0
⇒ 2(2)2 – 3(2) + 7a = 0
⇒ 8 – 6 + 7a = 0
⇒ 2 + 7a = 0
⇒ 7a = – 2
⇒ a = $$\frac{-2}{7}$$

Question 6.
If 0 and 1 are the zeroes of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the values of a and b.
Solution:
Given that f(0) = 0; f(1) = 0 and
f(x) = 2x3 – 3x2 + ax + b
∴ f(0) = 2(0)3 – 3(0)2 + a(0) + b
⇒ 0 = b
Also f(1) = 0
⇒ 2(1)3 – 3(1)2 + a(1) + 0 = 0
⇒ 2 – 3 + a = 0 .
⇒ a = 1
Hence a = 1; b = 0

## AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.1 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.1

Question 1.
Find the degree of each of the polynomials given below,
i) x5 – x4 + 3
Solution:
Degree is 5.

ii) x2 + x – 5
Solution:
Degree is 2.

iii) 5
Solution:
Degree is 0.

iv) 3x6 + 6y3 – 7
Solution:
Degree is 6.

v) 4 – y2
Solution:
Degree is 2.

vi) 5t – √3
Solution:
Degree is 1.

Question 2.
Which of the following expressions are polynomials in one variable and which are not ? Give reasons for your answer.
i) 3x2 – 2x + 5
Solution:
Given expression is a polynomial in one variable.

ii) x2 + √2
Solution:
Given expression is a polynomial in one variable.

iii) p2 – 3p + q
Solution:
Given expression is not a polynomial in one variable. It involves two variables p and q.

iv) y + $$\frac{2}{\mathbf{y}}$$
Solution:
Given expression is not a polynomial. Since the second term contains the variable in its denominator.

v) $$5 \sqrt{x}+x \sqrt{5}$$
Solution:
Given expression is not a polynomial. Since the first term’s exponent is not an integer.

vi) x100 + y100
Solution:
Given expression has two variables. So it is not a polynomial in one variable.

Question 3.
Write the coefficient of x3 in each of the following.
i) x3 + x + 1
ii) 2 – x3+ x2
iii) $$\sqrt{2} x^{3}+5$$
iv) 2x3 + 5
v) $$\frac{\pi}{2} x^{3}+x$$
vi) $$-\frac{2}{3} x^{3}$$
vii) 2x2 + 5
viii) 4
Solution:
i) x3 + x + 1 : co-efficient of x3 is 1.
ii) 2 – x3+ x2 : co-efficient of x3 is – 1.
iii) $$\sqrt{2} x^{3}+5$$ co-efficient of x3 is √2
iv) 2x3 + 5 : co-efficient of x3 is 2.
v) $$\frac{\pi}{2} x^{3}+x$$ co-efficient of x3 is $$\frac{\pi}{2}$$
vi) $$-\frac{2}{3} x^{3}$$ co-efficient of x3 is $$-\frac{2}{3}$$
vii) 2x2 + 5 : co-efficient of x3 is ‘0’.
viii) 4 : co-efficient of x3 is ‘0’.

Question 4.
Classify the following as linear, quadratic and cubic polynomials.
i) 5x2+ x – 7 : degree 2 hence quadratic polynomial.
ii) x – x3 , : degree 3 hence cubic polynomial.
iii) x2 + x + 4 : degree 2 hence quadratic polynomial.
iv) x – 1 : degree 1 hence linear polynomial.
v) 3p : degree 1 hence linear polynomial.
vi) πr2 : degree 2 hence quadratic polynomial.

Question 5.
Write whether the following statements are True or False. Justify your answer.
i) A binomial can have at the most two terms
ii) Every polynomial is a binomial
iii) A binomial may have degree 3
iv) Degree of zero polynomial is zero
v) The degree of x2 + 2xy + y2 is 2
vi) πr2 is monomial
Solution :
i) A binomial can have at the most two terms -True
ii) Every polynomial is a binomial – False
[∵ A polynomial can have more than two terms]
iii) A binomial may have degree 3 – True
iv) Degree of zero polynomial is zero – False
v) The degree of x2 + 2xy + y2 is 2 – True
vi) πr2 is monomial – True

Question 6.
Give one example each of a monomial and trinomial of degree 10.
Solution :
– 7x10 is a monomial of degree 10.
3x2y8 + 7xy – 8 is a trinomial of degree 10.

## AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.4

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

Question 1.
Simple the following expressions.
i) (5 + √7) (2 + √5)
Solution:
(5 + √7) (2 + √5)
= 10 + 5√5 + 2√7 + √35

ii) (5 + √5) (5 – √5)
Solution:
(5 + √5) (5 – √5)
= 52 + (√5)2
= 25 – 5 = 20

(iii) (√3 + √7)2
Solution:
(√3 + √7)2
= (√3)2 + (√7)2 + 2(√3)(√7)
= 3 + 7 + 2√21
= 10 + 2√21

iv) (√11 – √7) (√11 + √7)
= (√11)2 – (√7)2
= 11 – 7 = 4

Question 2.
Classify the following numbers as rational or irrational.
i) 5 – √3
ii) √3 + √2
iii) (√2 – 2)2
iv) $$\frac{2 \sqrt{7}}{7 \sqrt{7}}$$
v) 2π
vii) (2 +√2) (2 – √2)
Solution:
i) 5 – √3 – irrational
ii) √3 + √2 – irrational
iii) (√2 – 2)2 – irrational
iv) $$\frac{2 \sqrt{7}}{7 \sqrt{7}}$$ – rational
v) 2π – Transcendental number. (not irrational)
vi) $$\frac{1}{\sqrt{3}}$$– irrational
vii) (2 +√2) (2 – √2) – rational

Question 3.
In the following equations, find whether variables x, y, z etc., represents rational or irrational numbers.
i) x2 = 7
ii) y2 = 16
iii) z2 = 0.02
iv) u2 = $$\frac{17}{4}$$
v) w2 = 27
vi) t4 = 256
Solution:
i) x2 = 7
⇒ x = √7 is an irrational number.
ii) y2 = 16 ⇒ y = 4 is a rational number.
iii) z2 = 0.02 ⇒ z = $$\sqrt{0.02}$$ is an irrational number.
iv) u2 = $$\frac{17}{4}$$ ⇒ x = $$\frac{\sqrt{17}}{2}$$ is an irrational number.
v) w2 = 27 ⇒ w = $$3 \sqrt{3}$$ an irrational number.
vi) t4 = 256 ⇒ t2 = $$\sqrt{256}$$ = 16
⇒ t = $$\sqrt{16}$$ = 4 is a rational number

Qeustion 4.
The ratio of circumference to the diameter of a circle c/d is represented by π. But we say that π is an irrational number. Why?

Question 5.
Rationalise the denominators of the following.
i) $$\frac{1}{3+\sqrt{2}}$$
Solution:

ii) $$\frac{1}{\sqrt{7}-\sqrt{6}}$$
Solution:

iii) $$\frac{1}{\sqrt{7}}$$
Solution:

iv) $$\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}$$
Solution:

Question 6.
Simplify each of the following by rationalising the denominator.
i) $$\frac{6-4 \sqrt{2}}{6+4 \sqrt{2}}$$
Solution:

ii) $$\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$
Solution:

iii) $$\frac{1}{3 \sqrt{2}-2 \sqrt{3}}$$
Solution:

iv) $$\frac{3 \sqrt{5}-\sqrt{7}}{3 \sqrt{3}+\sqrt{2}}$$
Solution:

Question 7.
Find the value of $$\frac{\sqrt{10}-\sqrt{5}}{2 \sqrt{2}}$$ upto three decimal places. (take $$\sqrt{2}$$ = 1.414, $$\sqrt{3}$$ = 1.732 and $$\sqrt{5}$$ = 2.236).
Solution:

Question 8.
Find
i) 641/6
Solution:
= (26)1/6
= 6

ii) 321/5
Solution:
321/5
= (25)1/5
= 2

iii) 6251/4
6251/5
= (54)1/4
= 5

iv) 163/2
Solution:
163/2
= (42)3/2

v) 2432/5
Solution:
2432/5
= (35)2/5

vi) (46656)-1/6
Solution:

Question 9.
Simplify $$\sqrt[4]{81}-8 \sqrt[3]{343}+15 \sqrt[5]{32}+\sqrt{225}$$
Solution:

Question 10.
If ‘a’ and ‘b’ are rational numbers, find the values of a and b in each of the following equations.
i) $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}$$
Solution:
Given that $$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}$$
Rationalising the denominator we get

Comparing 5 + 2√6 with a + b√6
We have a = 5 and b = 2

ii) $$\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}$$
Solution:
Given that $$\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}$$
Rationalising the denominator we get

AP Board 9th Class Maths Solutions

## AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.3

Question 1.
Visualise $$2.8 \overline{74}$$ on the number line, using successive magnification.
Solution:

Question 2.
Visualise $$5 . \overline{28}$$ on the number line, upto 3 decimal places.
Solution:

## AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2

AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.2

Question 1.
Classify the following numbers as rational or irrational.
i) $$\sqrt{27}$$
ii) $$\sqrt{441}$$
iii) 30.232342345
iv) 7.484848
v) 11.2132435465
vi) 0.3030030003
Solution:
i) $$\sqrt{27}$$ – irrational number
ii) $$\sqrt{441}$$ = 21 – rational
iii) 30.232342345 – irrational number
iv) 7.484848 – rational number
v) 11.2132435465 – irrational number
vi) 0.3030030003 – irrational number

Question 2.
Explain with an example how irrational numbers differ from rational numbers ?
Solution:
Irrational numbers can’t be expressed in $$\frac { p }{ q }$$ form where p and q are integers and q ≠ 0.
E.g.$$\sqrt{2}, \sqrt{3} ; \sqrt{5}, \sqrt{7}$$ etc.
Where as a rational can be expressed in $$\frac { p }{ q }$$ form
E.g. :- -3 = $$\frac { -3 }{ 1 }$$ and $$\frac { 5 }{ 4 }$$ etc.

Question 3.
Find an irrational number between $$\frac { 5 }{ 7 }$$ and $$\frac { 7 }{ 9 }$$. How many more there may be ?
Solution :
The decimal forms of $$\frac { 5 }{ 7 }$$ and $$\frac { 7 }{ 9 }$$ are
$$\frac{5}{7}=0 . \overline{714285} \ldots ., \frac{7}{9}=0.7777 \ldots \ldots=0 . \overline{7}$$
∴ An irrational between $$\frac { 5 }{ 7 }$$ and $$\frac { 7 }{ 9 }$$ is 0.727543…………
There are infinitely many irrational numbers between $$\frac { 5 }{ 7 }$$ and $$\frac { 7 }{ 9 }$$.

Question 4.
Find two irrational numbers between 0.7 and 0.77.
Solution:
Two irrational numbers between 0.7 and 0.77 can take the form
0.70101100111000111…………. and 0.70200200022……………

Question 5.
Find the value of √5 uPto 3 decimal places.
Solution:

[√5 is not exactly equal to 2.2350679………….. as shown ¡n calculators]

Question 6.
Find the value of √7 upto six decimal places by long division method.
Solution:

Question 7.
Locate $$\sqrt{\mathrm{10}}$$ on number line.

Step – 1 : Draw a number line.
Step – 2 : Draw a rectangle OABC at zero with measures 3 x 1. i.e., length 3 units and breadth 1 unit.
Step – 3 : Draw the diagonal OB.
Step – 4 : Draw an arc with centre ‘O’ and radius OB which cuts the number line at D.
Step – 5 : ‘D’ represents $$\sqrt{\mathrm{10}}[latex] on the number line. Question 8. Find atleast two irrational numbers between 2 and 3. Solution: An irrational number between a and b is Tab [latex]\sqrt{\mathrm{ab}}$$ unless ab is a perfect square.
∴ Irrational number between 2 and 3 is √6

∴ Required irrational numbers are 61/2, 241/4

Method – II:
Irrational numbers between 2 and 3 are of the form 2.12111231234………….. and 3.13113111311113…….

Question 9.
State whether the following statements are true or false. Justify your answers.
Solution:

1. Every irrational number is a real number – True (since real numbers consist of rational numbers and irrational numbers)
2. Every rational number is a real number – True (same as above)
3. Every rational number need not be a rational number – False (since all rational numbers are real numbers).
4. $$\sqrt{n}$$ is not irrational if n is a perfect square – True. (since by definition of an irrational number).
5. $$\sqrt{n}$$ is irrational if n is not a perfect square – True. (same as above)
6. All real numbers are irrational – False (since real numbers consist of rational