By using Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 4 Exploring Algebraic Identities Extra Questions, students can improve their problem-solving skills.
Class 9 Exploring Algebraic Identities Extra Questions
Extra Questions on Exploring Algebraic Identities Class 9
Class 9 Ganita Manjari Chapter 4 Extra Questions
Question 1.
Find the product of the foIIoing binomials:
(\(\frac{1}{2}\)x – \(\frac{1}{5}\)y)(\(\frac{1}{2}\)x – \(\frac{1}{5}\)y)
Solution:
Here, (\(\frac{1}{2}\)x – \(\frac{1}{5}\)y)(\(\frac{1}{2}\)x – \(\frac{1}{5}\)y) = (\(\frac{1}{2}\)x – \(\frac{1}{5}\)y)2
[∵ (a – b)(a – b) = (a – b)2]
= (\(\frac{1}{2}\)x)2 – 2(\(\frac{1}{2}\)x)(\(\frac{1}{5}\)y) + (\(\frac{1}{5}\)y)2
[∵ (a – b)2 = a2 – 2ab + b2]
= (\(\frac{1}{2}\))2x2 – 2 × \(\frac{1}{2}\) × \(\frac{1}{5}\)xy + (\(\frac{1}{5}\))2y2
= \(\frac{1}{4}\)x2 – \(\frac{1}{5}\)xy + \(\frac{1}{25}\)y2
Exploring Algebraic Identities Class 9 Very Short Question Answer
Question 2.
Using identities, evaluate:
(i) 78 × 82
Solution:
78 × 82 = (80 – 2) × (80 + 2)
= 802 – 22
= 80 × 80 – 2 × 2
[∵(a + b)(a – b) = a2 – b2]
= 6400 – 4 = 6396
(ii) 1.73 × 1.73 – 0.27 × 0.27
Solution:
1.73 × 1.73 – 0.27 × 0.27
= 1.732 – 0.272 [∵ xx = x2]
= (1.73 + 0.27) (1.73 – 0.27)
[∵ a2 – b2 = (a + b) (a – b)]
= 2.00 × 1.46
= 2 × 1.46
= 2.92
Question 3.
If x + \(\frac{1}{x}\) = 20. find the value of x2 + \(\frac{1}{x^2}\).
Solution:
Given x + \(\frac{1}{x}\) = 20
Squaring both sides,
(x + \(\frac{1}{x}\))2 = 202
⇒ x2 + 2x × \(\frac{1}{x}\) + (\(\frac{1}{x}\))2 = 20 × 20
⇒ x2 + 2 + \(\frac{1}{x^2}\) = 400
[∵ x × \(\frac{1}{x}\) = \(\frac{x}{x}\) = 1]
Shifting 2 to R.H.S
x2 + \(\frac{1}{x^2}\) = 400 – 2 = 398
Exploring Algebraic Identities Class 9 Short Question Answer
Question 1.
Using identities, find:
(i) (\(\frac{3}{2}\)m + \(\frac{2}{3}\)n)(\(\frac{3}{2}\)m – \(\frac{2}{3}\)n)
Solution:
(\(\frac{3}{2}\)m + \(\frac{2}{3}\)n)(\(\frac{3}{2}\)m – \(\frac{2}{3}\)n) = (\(\frac{3}{2}\)m)2 – (\(\frac{2}{3}\)n)2
[∵ (a + b)(a – b) = a2 – b2]
= \(\frac{3^2 m^2}{2^2}-\frac{2^2 n^2}{3^2}=\frac{9 m^2}{4}-\frac{4 n^2}{9}\)
(ii) (2x + 3y) (2x – 3y)
Solution:
(2x + 3y) (2x – 3y) = (2x)2 – (3y)2
[∵(a + b)(a – b) = a2 – b2]
= 22x2 – 32y2
= 4x2 – 9y2
(iii) (\(\frac{3}{4}\)x + \(\frac{5}{6}\)y)(\(\frac{3}{4}\)x – \(\frac{5}{6}\)y)
Solution:
(\(\frac{3}{4}\)x + \(\frac{5}{6}\)y)(\(\frac{3}{4}\)x – \(\frac{5}{6}\)y)
= (\(\frac{3}{4}\)x)2 – (\(\frac{5}{6}\)y)2
[∵(a + b)(a – b) = a2 – b2]
= \(\frac{3^2 x^2}{4^2}-\frac{5^2 y^2}{6^2}\) [∵ \(\left(\frac{a b}{c}\right)^2=\frac{a^2 b^2}{c^2}\)]
= \(\frac{9 x^2}{16}-\frac{25 y^2}{36}\)
(iv) (a2 + b2)(- a2 + b2)
Solution:
(a2 + b2)(- a2 + b2) = (b2 + a2)(b2 – a2)
(Interchanging the terms in both the factors)
= (b2)2 – (a2)2 [∵ (a + b) (a – b) = a2 – b2]
= b4 – a4 [∵ (am)n = amn]
Question 2.
Simplify:
(i) (7m – 8n)2 + (7m + 8n)2
Solution:
(7m – 8n)2 = (7m)2 – 2 x 7m x 8n + (8n)2
= 49m2 – 112mn + 64n2 …(1)
and (7m + 8n)2 = (7m)2 + 2 x 7m x 8n + (8n)2
= 49m2 + 112mn + 64n2 …(2)
Adding (1) and (2), we get
(7m – 8n)2 + (7m + 8n)2 = (49 + 49)m2 + (-112 + 112)mn + (64 + 64)n2
= 98m2 + 0mn + 128n2
= 98m2 + 128n2
(ii) (4m + 5n)2 + (5m + 4n)2
Solution:
(4m + 5n)2 = (4m)2 + 2(4m)(5n) + (5n)2
= 16m2 + 40mn + 25n2 …(1)
and (5m + 4n)2 = (5m)2 + 2(5m)(4n) + (4n)2
= 25m2 + 40mn + 16n2 …(2)
Adding (1) and (2) columnwise, we get
(4m + 5n)2 + (5m + 4n)2
= (16 + 25)m2 + (40 + 40)mn + (25 + 16)n2
= 41m2 + 80mn + 41n2
(iii) (2x + 5)2 – (2x – 5)2
Solution:
(2x + 5)2 = (2x)2 + 2(2x)5 + 52
= 4x2 + 20x + 25 ………..(1)
and (2x – 5)2 = (2x)2 = 2(2x)5 + 52
= 4x2 – 20x + 25 …….(2)
Subtracting columnwise,
(2x + 5)2 – (2x – 5)2
= (4 – 4)x2 + (20 + 20)x + 25 – 25
= 0x2 + 40x + 0
= 40x
Question 3.
Using a2 – b2 = (a + b) (a – b), find
(i) 512 – 492
Solution:
512 – 492 = (51 + 49)(51 – 49) = 100 × 2 = 200
(ii) (1.02)2 – (0.98)2
Solution:
(1.02)2 – (0.98)2 = (1.02 + 0.98)(1.02 – 0.98)
= 2 × 0.04
= 0.08
(iii) 1532 – 1472
Solution:
1532 – 1472 = (153 + 147) (153 – 147)
= 300 × 6
= 1800
(iv) 12.12 – 7.92
Solution:
12.12 – 7.92 = (12.1 + 7.9) × (12.1 – 7.9)
= 20 × (4.2)
= \(\frac{20}{1} \times \frac{42}{10}=\frac{10 \times 2 \times 42}{10}\)
= 2 × 42
= 84
Question 4.
If x – y = 7 and xy = 9. find the value of x2 + y2.
Solution:
Given: x – y = 7 …(1)
and xy = 9
Squaring both sides of equation (1), we have
(x – y)2 = 72
⇒ x2 – 2xy + y2 = 49
Putting xy = 9, we have
x2 – 18 + y2 = 49
Shifting-18 to R.H.S.;
x2 + y2 = 49 + 18 = 67
Question 5.
Use the identity (x + a) (x + b) = x2 + (a + b)x + ab, to find the following products:
(i) (x + 3) (x + 7)
Solution:
(x + 3) (x + 7) [here a = 3, b = 7]
= x2 + (3 + 7)x + 3 × 7
= x2 + 10x + 21
(ii) (4x + 5) (4x + 1)
Solution:
(4x + 5) (4x + 1)
Putting 4x = y
= (y + 5) (y + 1) [here a = 5, b = 1]
= y2 + (5 + 1)y + 5 x 1
= y2 + 6y + 5
Replacing y by4x,
= (4x)2 + 6(4x) + 5
= 42x2 + 24x + 5
= 16x2 + 24x + 5
(iii) (4x – 5) (4x – 1)
Solution:
(4x – 5) (4x – 1)
Putting 4x =y,
= (y – 5)(y – 1) = [y + (- 5)] [y + (-1)]
= y2 + (-5 – 1)y + (-5)(-1)
= y2 – 6.y + 5
Replacing y by 4x,
= (4x)2 – 6(4x) + 5
= 16x2 – 24x + 5
Question 6.
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab, evaluate the following:
(i) 103 × 104
Solution:
103 × 104 = (100 + 3) (100 + 4)
= (x + a)(x + b) where x = 100, a = 3, b = 4
= x2 + (a + b)x + ab
= (100)2 + (3 + 4) × 100 + 3 × 4
= 100 × 100 + 7 × 100 + 12
= 10000 + 700 + 12
= 10712
(ii) 5.1 × 5.2
Solution:
5.1 × 5.2 = (5 + 0.1) × (5 + 0.2)
= (x + a) × (x + b)
where x = 5, a = 0.1, b = 0.2.
= x2 + (a + b)x + ab
= 52 + (0.1 +0.2) × 5 + (0.1) × (0.2)
= 25 + (0.3) × 5 + 0.02
= 25 + 1.5 + 0.02 × o.2
= 26.50 + 0.02
= 26.52
Exploring Algebraic Identities Class 9 Long Question Answer
Question 1.
Find the continued product:
(i) (x + 2) (x – 2) (x2 + 4)
Solution:
(x + 2) (x – 2) (x2 + 4)
= [(x + 2) (x – 2)] (x2 + 4)
= (x2 – 22) (x2 + 4)
[∵ (a + b)(a – b) = a2 – b2]
= (x2 – 4) (x2 + 4) = (x2)2 – 42
[∵ (a – b) (a + b) = a2 – b2]
= x4 – 16 [∵ (am)n = anm]
(ii) (x – 1) (x + 1) (x2 + 1) (x4 + 1)
Solution:
(x – 1) (x + 1) (x2 + 1) (x4 + 1)
= [(x – 1) (x + 1)] (x2 + 1) (x4 + 1)
= (x2 – 12)(x2 + 1)(x4 + 1)
[∵ (a – b) (a + b) = a2 – b2]
= (x2 – 1) (x2 + 1) (x4 + 1)
= [(x2)2 – 12] (x4 + 1)
[∵ (a – b)(a + b) = a2 – b2]
= (x4 – 1) (x4 + 1) = (x4)2 – 12
[∵ (a – b) (a + b) = a2 – b2]
= x8 – 1
Question 2.
Simplify:
(i) (ab + bc)2 – 2ab2c
Solution:
(ab + bc)2 – (2ab2c)
= (ab)2 + 2ab be + (bc)2 – 2ab2c
[∵ (a + b)2 = a2 + 2 ab + b2]
= a2b2 + 2ab2c + b2c2 – 2ab2c
Grouping like terms,
= a2b2 + b2c2 + 2ab2c – 2ab2c
= a2b2 + b2c2 + (2-2) ab2c
= a2b2 + b2c2 + 0 ab2c
= a2b2 + b2c2
(ii) (m2 – n2m)2 + 2m3n2
Solution:
(m2 – n2m)2 + 2m2n2
= (m2)2 – 2m2n2m + (n2m)2 + 2m2n2
[∵ (a – b)2 = a2 – 2ab + b2]
= m4 – 2m2mn2 + (n2)2 m2 + 2m3n2
= m4 – 2m3n2– + n4m2 + 2m2n2
Grouping like terms,
= m4 + n4m2 – 2m3n2 + 2m3n2
= m4 + n4m2 + (- 2 + 2)m3n2
= m4 + n4m22 + 0m3n2
= m4 + n4m2
Question 3.
Show that
(\(\frac{4}{3}\)m – \(\frac{3}{4}\)n)2 + 2mn = \(\frac{16}{9}\)m2 + \(\frac{9}{16}\)n2
Solution:
Grouping like terms,
Question 4.
Use the identity
(x + a) (x + b) = x2 + (a + b)x + ab to find the following products:
(i) (2x + 5y) (2x + 3y)
Solution:
(2x + 5y) (2x + 3y)
Putting 2x = t,
= (t + 5y) (t + 3y) [∵ a = 5y, b = 3y]
= t2 + (5y + 3y)t + (5y) x 3y
= t2 + 8yt + 15y2
Replacing t by 2x,
= (2x)2 + 8y (2x) + 15y2
= 22x2 + 16xy + 15y:
= 4x2 + 16xy + 15y2
(ii) (2a2 + 9) (2a2 + 5)
Solution:
(2a2 + 9) (2a2 + 5)
Putting, 2a2 = y,
= (y + 9)(y + 5)
= y2 + (9 + 5)y + 9 × 5
= y2 + 14y + 45
Replacing y by 2a2,
= (2a2)2 + 14(2a2) + 45
= 22 (a2)2 + 28a2 + 45
= 4a4 + 28a2 + 45
Exploring Algebraic Identities Class 9 Case Based Questions
A playground is in shape of a square. The area of the square PQRS is 256 m2 with each side (x + 2) m. One day Suraj along with his two friends Ajay and Aman went to play there with bicycle. Someone stole Suraj bicycle, but Ajay and Aman helped him by contributing? ₹ (4a + 60) and ₹ (6a + 10) respectively, to buy a new bicycle. The cost of bicycle was ₹ 4200.
On basis of this information given in passage answer following questions:
Question 1.
Find the value of x.
(a) 16
(b) 18
(c) 14
(d) 12
Solution:
(c) 14
Question 2.
Find the side of square shaped ground.
(a) 19 m
(b) 12 m
(c) 18 m
(d) 16 m
Solution:
(d) 16 m
Question 3.
What is the value of a ?
(a) 410
(b) 403
(c) 413
(d) 423
Solution:
(c) 413
Question 4.
What was the amount given by Ajay and Aman to Suraj?
Solution:
₹ 1712, ₹ 2488
Question 5.
What is the perimeter of the playground?
Solution:
64 m
Exploring Algebraic Identities Extra Questions for Practice
Very Short Answer Type Questions
Question 1.
If x – \(\frac{1}{x}\) = 9, find the value o x2 + \(\frac{1}{x^2}\)
Solution:
83
Question 2.
If x2 + \(\frac{1}{x^2}\) = 47 and x is positive, find x + \(\frac{1}{x}\)
Solution:
7
Question 3.
If x + y = 12 and xy = 14, find the value of x2 + y2.
Solution:
116
Question 4.
If 3x + 2y = 12 and xy = 6, find the value of 9x2 + 4y2.
Solution:
72
Question 5.
If x + y = 4 and xy = 2, find the value of x2 + y2.
Solution:
12
Short Answer Type Questions
Question 1.
Using identities, find the following:
(i) \(\left(x+\frac{a}{2}\right)^2\)
Solution:
x2 + xa + \(\frac{a^2}{4}\)
(ii) \(\left(y+\frac{y^2}{2}\right)^2\)
Solution:
y2 + y3 + \(\frac{a^4}{4}\)
(iii) (8a + 3b)2
Solution:
64a2 + 48ab + 9b2
Question 2.
Using identities, find the following products:
(i) (\(\frac{4}{3}\)x2 + 3)(\(\frac{4}{3}\)x2 + 3)
Solution:
\(\frac{16}{9}\)x4 + 8x2 + 9
(ii) (\(\frac{2}{3}\)x2 + 5y2)(\(\frac{2}{3}\)x2 + 5y2)
Solution:
\(\frac{4}{9}\)x4 + \(\frac{20}{3}\)x2y2 + 25y4
Question 3.
Using identities, evaluate:
(i) 1032
Solution:
10609
(ii) (4.9)2
Solution:
24.01
(iii) 9832 – 172
Solution:
966000
(iv) 194 × 206
Solution:
39964
Question 4.
Evaluate the following using identities:
(i) \(\frac{58^2-42^2}{16}\)
Solution:
100
(ii) \(\frac{198 \times 198-102 \times 102}{96}\)
Solution:
300
(iii) \(\frac{8.63 \times 8.63-1.37 \times 1.37}{0.726}\)
Solution:
100
Question 5.
Find the continued product:
(i) (x – y)(x + y)(x2 + y2)(x4 + y4)
Solution:
x8 – y8
(ii) (2x – 1)(2x + 1)(4x2 + 1)(16x4 + 1)
Solution:
256x8 – 1
Question 6.
Use a suitable identity to find each of the followings:
(i) (- a + c) (- a + c)
Solution:
c2 – 2ac + a2
(ii) \(\left(\frac{x}{2}+\frac{3 y}{4}\right)\left(\frac{x}{2}+\frac{3 y}{4}\right)\)
Solution:
\(\frac{x^2}{4}+\frac{3 x y}{4}+\frac{9 y^2}{16}\)
(iii) (7a – 9b)(7a – 9b)
Solution:
49a2 – 126ab + 81b2
(iv) (1.1m – 0.4)(1.1m + 0.4)
Solution:
1.21m2 – 0.16
Using the identity (x + a) (x + b) = x2 + (a + b)x + ab, find the following products:
Question 7.
(x + 4) (x + 7)
Solution:
x2 + 11x + 28
Question 8.
(3x + 4) (3x – 5)
Solution:
9x2 – 3x – 20
Question 9.
(y2 – 4) (y2 – 3)
Solution:
y4 – 7y2 + 12
Question 10.
(3x2 – 4xy) (3x2 – 3xy)
Solution:
9x4 – 21x3y + 12x2y2
Question 11.
(3x – 4y) (2x – 4y)
Solution:
6x2 – 20xy + 16y2
Question 12.
102 × 106
Solution:
10812
Question 13.
994 × 1006
Solution:
999964
Question 14.
501 × 502
Solution:
251502
Question 15.
95 × 103
Solution:
9785
Long Answer Type Questions
Question 1.
Write the squares of each of the following binomials:
(i) (5b – \(\frac{1}{2}\))
Solution:
25b2 – 5b + \(\frac{1}{4}\)
(ii) 2x – 3y
Solution:
4x2 – 12xy + 9y2
(iii) (3x – \(\frac{1}{3x}\))
Solution:
9x2 – 2 + \(\frac{1}{9 x^2}\)
Question 2.
Using identities, find the products of the following binomials:
(i) (2x + y) (2x – y)
Solution:
\(\frac{16}{9}\)x4 + 8x2 + 9
(ii) (a2 + bc) (a2 – bc)
Solution:
\(\frac{4}{9}\)x4 + \(\frac{20}{3}\)x2y2 + 25y4
(iii) \(\left(\frac{4 x}{5}-\frac{3 y}{4}\right)\left(\frac{4 x}{5}+\frac{3 y}{4}\right)\)
Solution:
49a2 – 126ab + 81b2
(iv) \(\left(x^4+\frac{2}{x^2}\right)\left(x^4-\frac{2}{x^2}\right)\)
Solution:
1.21m2 – 0.16
Question 3.
Evaluate the following using the identities:
(i) (82)2 – (18)2
Solution:
6400
(ii) (467)2 – (33)2
Solution:
217000
(iii) 113 × 87
Solution:
9831
(iv) 178 × 178 – 22 × 22
Solution:
31200
Question 4.
If x + \(\frac{1}{x}\) = 4, find the value of
(i) x2 + \(\frac{1}{x^2}\)
Solution:
14
(ii) x4 + \(\frac{1}{x^4}\)
Solution:
194
Question 5.
Simplify the following:
(i) (x2 + x + 1) (x2 – x + 1)
Solution:
x4 + x2 + 1
(ii) (x2 + 2x + 2) (x2 – 2x + 2)
Solution:
x4 – 2x2 + 4
Question 6.
Find the value of x, if
(i) 6x = 232 – 172
Solution:
40
(ii) 4x = 982 – 882
Solution:
465
(iii) 14x = 472 – 332
Solution:
80
Question 7.
Use a suitable identity to find each of the following products:
(i) (x + 3) (x + 3)
Solution:
x2 + 6x + 9
(ii) (2y + 5) (2y + 5)
Solution:
4y2 + 20y + 25
(iii) (2a – 7) (2a – 7)
Solution:
4a2 – 28a + 49
(iv) (3a – \(\frac{1}{2}\))(3o + \(\frac{1}{2}\))
Solution:
9a2 – \(\frac{1}{4}\)
(v) (6x – 7)(6x + 7)
Solution:
36x2 – 49