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Chapter 9 Algebraic expressions and identities Class 8 Mathematics NCERT Solutions
Class 8 Mathematics students should refer to the following NCERT questions with answers for Chapter 9 Algebraic expressions and identities in Class 8. These NCERT Solutions with answers for Class 8 Mathematics will come in exams and help you to score good marks
Chapter 9 Algebraic expressions and identities NCERT Solutions Class 8 Mathematics
Exercise 9.1
Q.1) Identify the terms, their coefficients for each of the following expressions:
i) 5๐ฅ๐ฆ๐ง2 โ 3๐ง๐ฆ ii) 1 + ๐ฅ + ๐ฅ2 iii) 4๐ฅ2๐ฆ2 โ 4๐ฅ2๐ฆ2๐ง2 + ๐ง2
iv) 3 โ ๐๐ + ๐๐ โ ๐๐ v) ๐ฅ/2 + ๐ฆ/2 โ ๐ฅ๐ฆ vi) 0.3๐ โ 06๐๐ + 0.5๐
Sol.1) I) Terms : 5๐ฅ๐ฆ๐ง2 And โ3๐ง๐ฆ
Coefficient in 5๐ฅ๐ฆ๐ง2 Is 5 and in โ3๐ง๐ฆ is โ3.
ii) terms : 1, ๐ฅ and ๐ฅ2
Coefficient of ๐ฅ and coefficient of ๐ฅ2 in 1.
iii) terms : 4๐ฅ2๐ฆ2, โ4๐ฅ2๐ฆ2๐ง2 and ๐ง2
Coefficient in 4๐ฅ2๐ฆ2 is 4, coefficient of โ4๐ฅ2๐ฆ2๐ง2 is โ4 and coefficient of ๐ง2 is 1.
iv) terms : 3, โ๐๐, ๐๐, โ๐๐
Coefficient of โ ๐๐ is โ1, coeficient of ๐๐ is 1 and coefficient of โ ๐๐ is โ1.
v) terms: ๐ฅ/2, ๐ฆ/2 and โ ๐ฅ๐ฆ
coefficient of ๐ฅ/2 is 1/2, coefficient of ๐ฆ/2 is โ1 and coefficient of โ ๐ฅ๐ฆ is โ1
vi) Terms: 0.3๐, โ06๐๐ and 0.5๐
coefficient of 0.3๐ is 0.3, coeficient of โ0.6๐๐ is โ0.6 and coefficient of 0.5๐ is 0.5.
Q.2) Classify the following polynomials as monomials, binomials, trinomials. Which polynomials do not fit in any of these three categories:
i) ๐ฅ + ๐ฆ ii) 1000 iii) ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 iv) 7 + ๐ฆ + 5๐ฅ
v) 2๐ฆ โ 3๐ฆ2 vi) 2๐ฆ โ 3๐ฆ2 + 4๐ฆ3 vii) 5๐ฅ โ 4๐ฆ + 3๐ฅ๐ฆ viii) 4๐ง โ 15๐ง2
ix) ๐๐ + ๐๐ + ๐๐ + ๐๐ x) ๐๐๐ xi) ๐2๐ + ๐๐2 xii) 2๐ + 2๐
Sol.2) (i) Since ๐ฅ + ๐ฆ contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one term. Therefore it is monomial.
(iii) Since ๐ฅ + ๐ฅ2 + ๐ฅ3 + ๐ฅ4 contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(iv) Since 7 + ๐ฆ + 5๐ฅ contains three terms. Therefore it is trinomial.
(v) Since 2๐ฆ โ 3๐ฆ2 contains two terms. Therefore it is binomial.
(vi) Since 2๐ฆ โ 3๐ฆ2 + 4๐ฆ3 contains three terms. Therefore it is trinomial.
(vii) Since 5๐ฅ โ 4๐ฆ + 3๐ฅ๐ฆ contains three terms. Therefore it is trinomial.
(viii) Since 4๐ง โ 15๐ง2 contains two terms. Therefore it is binomial. -415 x z
(ix) Since ๐๐ + ๐๐ + ๐๐ + ๐๐ contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(x) Since ๐๐๐ contains one term. Therefore it is monomial.
(xi) Since ๐2๐ + ๐๐2 contains two terms. Therefore it is binomial.
(xii) Since 2๐ + 2๐ contains two terms. Therefore it is binomial.
Q.3) Add the following.
(i) ๐๐ โ ๐๐, ๐๐ โ ๐๐, ๐๐ โ ๐๐ (ii) ๐ โ ๐ + ๐๐, ๐ โ ๐ + ๐๐, ๐ โ ๐ + ๐๐
(iii) 2๐ ๐ โ 3๐๐ + 4, 5 + 7๐๐ โ 3๐ ๐ (iv) ๐ + ๐ , ๐ + ๐ , ๐ + ๐ , 2๐๐ + 2๐๐ + 2๐๐
Sol.3) (i) ๐๐ โ ๐๐, ๐๐ โ ๐๐, ๐๐ โ ๐๐
๐๐ โ ๐๐
+๐๐ โ ๐๐
โ๐๐ + ๐๐
0 + 0 + 0
Hence, the sum is 0.
Q.4) (a) Subtract 4๐ โ 7๐๐ + 3๐ + 12 from 12๐ โ 9๐๐ + 5๐ โ 3
(b) Subtract 3๐ฅ๐ฆ + 5๐ฆ๐ง โ 7๐ง๐ฅ from 5๐ฅ๐ฆ โ 2๐ฆ๐ง โ 2๐ง๐ฅ + 10๐ฅ๐ฆ๐ง
(c) Subtract 4๐2๐ โ 3๐๐ + 5๐๐2 โ 8๐ + 7๐ โ 10 from 18 โ 3๐ โ 11๐ + 5๐๐ โ 2๐๐2 + 5๐2๐
Sol.4)
Exercise 9.2
Q.1) Find the product of the following pairs of monomials:
(i) 4 , 7๐ (ii) โ 4๐, 7๐ (iii) โ4๐, 7๐๐ (iv) 4๐3, โ3๐ (v) 4๐, 0
Sol.1) (i) 4 , 7๐
4 ร 7 ๐ = 28๐
(ii) โ 4๐, 7๐
4๐ ร 7๐ = โ28๐2
(iii) โ4๐, 7๐๐
4๐ ร 7๐๐ = โ28๐2๐
(iv) 4๐3, โ3๐
4๐3๐ ร โ 3๐ = โ12๐4๐
(v) 4๐, 0
4๐ ร 0 = 0
Q.2) Find the areas of rectangles with the following pairs of monomials as their lengths and breadths respectively:
(๐, ๐); (10๐, 5๐); (20๐ฅ2, 5๐ฆ2); (4๐ฅ, 3๐ฅ2); (3๐๐, 4๐๐)
Sol.2) Area = Length ร breadth
(i) ๐ ร ๐ = ๐๐
(ii) 10๐ ร 5๐ = 50๐๐
(iii) 20๐ฅ2 ร 5๐ฆ2 = 100๐ฅ2๐ฆ2
(iv) 4๐ฅ ร 3๐ฅ2 = 12๐ฅ3
(v) 3๐๐ ร 4๐๐ = 12๐๐2๐
Q.3) Complete the following table of products:
Q.4) Obtain the volume of rectangular boxes with the following length, breadth and height respectively.
(i) 5๐, 3๐2, 7๐4 (ii) 2๐, 4๐, 8๐ (iii) ๐ฅ๐ฆ, 2๐ฅ2๐ฆ, 2๐ฅ๐ฆ2 (iv) ๐, 2๐, 3๐
Sol.4) We know that,
Volume = Length ร Breadth ร Height
(i) Volume = 5๐ ร 3๐2 ร 7๐4 = 5 ร 3 ร 7 ร ๐ ร ๐2 ร ๐4 = 105 ๐7
(ii) Volume = 2๐ ร 4๐ ร 8๐ = 2 ร 4 ร 8 ร ๐ ร ๐ ร ๐ = 64๐๐๐
(iii) Volume = ๐ฅ๐ฆ ร 2๐ฅ2๐ฆ ร 2๐ฅ๐ฆ2 = 2 ร 2 ร ๐ฅ๐ฆ ร ๐ฅ2๐ฆ ร ๐ฅ๐ฆ2 = 4๐ฅ4๐ฆ4
(iv) Volume = ๐ ร 2๐ ร 3๐ = 2 ร 3 ร ๐ ร ๐ ร ๐ = 6๐๐๐
Q.5) Obtain the product of
(i) ๐ฅ๐ฆ, ๐ฆ๐ง, ๐ง๐ฅ (ii) ๐, โ ๐2, ๐3 (iii) 2, 4๐ฆ, 8๐ฆ2, 16๐ฆ3
(iv) ๐, 2๐, 3๐, 6๐๐๐ (v) ๐, โ ๐๐, ๐๐๐
Sol.5) (i) ๐ฅ๐ฆ ร ๐ฆ๐ง ร ๐ง๐ฅ = ๐ฅ2๐ฆ2๐ง2
(ii) ๐ ร (โ ๐2) ร ๐3 = โ ๐6
(iii) 2 ร 4๐ฆ ร 8๐ฆ2 ร 16๐ฆ3 = 2 ร 4 ร 8 ร 16 ร ๐ฆ ร ๐ฆ2 ร ๐ฆ3 = 1024 ๐ฆ6
(iv) ๐ ร 2๐ ร 3๐ ร 6๐๐๐ = 2 ร 3 ร 6 ร ๐ ร ๐ ร ๐ ร ๐๐๐ = 36๐2 ๐2๐2
(v) ๐ ร (โ ๐๐) ร ๐๐๐ = โ ๐3๐2๐
Exercise 9.3
Q.1) Carry out the multiplication of the expressions in each of the following pairs.
(i) 4๐, ๐ + ๐ (ii) ๐๐, ๐ โ ๐ (iii) ๐ + ๐, 7๐2๐2
(iv) ๐2 โ 9, 4๐ (v) ๐๐ + ๐๐ + ๐๐, 0
Sol.1) (i) (4๐) ร (๐ + ๐) = (4๐ ร ๐) + (4๐ ร ๐) = 4๐๐ + 4๐๐
(ii) (๐๐) ร (๐ โ ๐) = (๐๐ ร ๐) + [๐๐ ร (โ ๐)] = ๐2๐ โ ๐๐2
(iii) (๐ + ๐) ร (7๐2 ๐2) = (๐ ร 7๐2 ๐2) + (๐ ร 7๐2 ๐2) = 7๐3๐2 + 7๐2๐3
(iv) (๐2 โ 9) ร (4๐) = (๐2 ร 4๐) + (โ 9) ร (4๐) = 4๐3 โ 36๐
(v) (๐๐ + ๐๐ + ๐๐) ร 0 = (๐๐ ร 0) + (๐๐ ร 0) + (๐๐ ร 0) = 0
Q.2) Complete the table
Q.4) (a) Simplify 3๐ฅ (4๐ฅ โ 5) + 3 and find its values for (i) ๐ฅ = 3, (ii) ๐ฅ = 1/2
(b) ๐ (๐2 + ๐ + 1) + 5 and find its values for (i) ๐ = 0, (ii) ๐ = 1, (iii) ๐ = โ 1.
Sol.4) (a) 3๐ฅ (4๐ฅ โ 5) + 3 = 12๐ฅ2 โ 15๐ฅ + 3
(i) For ๐ฅ = 3, 12๐ฅ2 โ 15๐ฅ + 3 = 12 (3)2 โ 15(3) + 3
= 108 โ 45 + 3
= 66
(ii) For ๐ฅ = 1/2, 12๐ฅ2 โ 15๐ฅ + 3 = 12 (1/2)2 โ 15 (1/2) + 3
= 6 โ 15/2
= 12 โ 15/2 = โ3/2
(b) ๐ (๐2 + ๐ + 1) + 5 = ๐3 + ๐2 + ๐ + 5
(i) For ๐ = 0, ๐3 + ๐2 + ๐ + 5 = 0 + 0 + 0 + 5 = 5
(ii) For ๐ = 1, ๐3 + ๐2 + ๐ + 5 = (1)3 + (1)2 + 1 + 5
= 1 + 1 + 1 + 5 = 8
(iii) For ๐ = โ1, ๐3 + ๐2 + ๐ + 5 = (โ1)3 + (โ1)2 + (โ1) + 5
= โ 1 + 1 โ 1 + 5 = 4
Q.5) (a) Add: ๐ (๐ โ ๐), ๐ (๐ โ ๐) and ๐ (๐ โ ๐)
(b) Add: 2๐ฅ (๐ง โ ๐ฅ โ ๐ฆ) and 2๐ฆ (๐ง โ ๐ฆ โ ๐ฅ)
(c) Subtract: 3๐ (๐ โ 4๐ + 5๐) from 4๐ (10๐ โ 3๐ + 2๐)
(d) Subtract: 3๐ (๐ + ๐ + ๐) โ 2๐ (๐ โ ๐ + ๐) from 4๐ (โ ๐ + ๐ + ๐)
Sol.5) a) ๐(๐ โ ๐) + ๐(๐ โ ๐) + ๐(๐ โ ๐)
= ๐2 โ ๐๐ + ๐2 โ ๐๐ + ๐2 โ ๐๐
= ๐2 + ๐2 + ๐2 โ ๐๐ โ ๐๐ โ ๐๐
b) 2๐ฅ(๐ง โ ๐ฅ โ ๐ฆ) + 2๐ฆ(๐ง โ ๐ฆ โ ๐ฅ)
= 2๐ฅ๐ง โ 2๐ฅ2 โ 2๐ฅ๐ฆ + 2๐ฆ๐ง โ 2๐ฆ2 โ 2๐ฅ๐ฆ
= 2๐ฅ๐ง โ 2๐ฅ๐ฆ โ 2๐ฅ๐ฆ + 2๐ฆ๐ง โ 2๐ฅ2 โ 2๐ฆ2
= โ2๐ฅ2 โ 2๐ฆ2 โ 4๐ฅ๐ฆ + 2๐ฆ๐ง + 2๐ง๐ฅ
c) 4๐ (10๐ โ 3๐ + 2๐) โ 3๐ (๐ โ 4๐ + 5๐)
= 40๐๐ โ 12๐๐ + 8๐2 โ 3๐2 + 12๐๐ โ 15๐๐
= 8๐2 โ 3๐2 โ 12๐๐ + 12๐๐ + 40๐๐ โ 15๐๐
= 5๐2 + 25๐๐
d) 4๐(โ๐ + ๐ + ๐) โ [3๐(๐ + ๐ + ๐) โ 2๐(๐ โ ๐ + ๐)]
= โ4๐๐ + 4๐๐ + 4๐2 โ [3๐2 + 3๐๐ + 3๐๐ โ 2๐๐ + 2๐2 โ 2๐๐
= โ4๐๐ + 4๐๐ + 4๐2 โ [3๐2 + 2๐2 + 3๐๐ โ 2๐๐ + 3๐๐ โ 2๐๐]
= โ4๐๐ + 4๐๐ + 4๐2 โ [3๐2 + 2๐2 + ๐๐ + 3๐๐ โ 2๐๐]
= โ4๐๐ + 4๐๐ + 4๐2 โ 3๐2 โ 2๐2 โ ๐๐ โ 3๐๐ + 2๐๐
= โ3๐2 โ 2๐2 + 4๐2 โ ๐๐ + 4๐๐ + 2๐๐ โ 4๐๐ โ 3๐๐
= โ3๐2 โ 2๐2 + 4๐2 โ ๐๐ + 6๐๐ โ 7๐๐
Exercise 9.4
Q.1) Multiply the binomials.
(i) (2๐ฅ + 5) and (4๐ฅ โ 3) (ii) (๐ฆ โ 8) and (3๐ฆ โ 4)
(iii) (2.5๐ โ 0.5๐) and (2.5๐ + 0.5๐) (iv) (๐ + 3๐) and (๐ฅ + 5)
(v) (2๐๐ + 3๐2)and (3๐๐ โ 2๐2) (vi) (3/4 ๐2 + 3๐2) and 4 (๐2 โ2/3 ๐2)
Sol.1) (i) (2๐ฅ + 5) ร (4๐ฅ โ 3)
= 2๐ฅ ร (4๐ฅ โ 3) + 5 ร (4๐ฅ โ 3)
= 8๐ฅ2 โ 6๐ฅ + 20๐ฅ โ 15
= 8๐ฅ2 + 14๐ฅ โ 15 (By adding like terms)
(ii) (๐ฆ โ 8) ร (3๐ฆ โ 4)
= ๐ฆ ร (3๐ฆ โ 4) โ 8 ร (3๐ฆ โ 4)
= 3๐ฆ2 โ 4๐ฆ โ 24๐ฆ + 32
= 3๐ฆ2 โ 28๐ฆ + 32 (By adding like terms)
(iii) (2.5๐ โ 0.5๐) ร (2.5๐ + 0.5๐)
= 2.5๐ ร (2.5๐ + 0.5๐) โ 0.5๐ (2.5๐ + 0.5๐)
= 6.25๐2 + 1.25๐๐ โ 1.25๐๐ โ 0.25๐2
= 6.25๐2 โ 0.25๐2
(iv) (๐ + 3๐) ร (๐ฅ + 5)
= ๐ ร (๐ฅ + 5) + 3๐ ร (๐ฅ + 5)
= ๐๐ฅ + 5๐ + 3๐๐ฅ + 15๐
(v) (2๐๐ + 3๐2) ร (3๐๐ โ 2๐2)
= 2๐๐ ร (3๐๐ โ 2๐2) + 3๐2 ร (3๐๐ โ 2๐2)
= 6๐2๐2 โ 4๐๐3 + 9๐๐3 โ 6๐4
= 6๐2๐2 + 5๐๐2 โ 6๐4
Q.2) Find the product.
(i) (5 โ 2๐ฅ) (3 + ๐ฅ) (ii) (๐ฅ + 7๐ฆ) (7๐ฅ โ ๐ฆ)
(iii) (๐2 + ๐) (๐ + ๐2) (iv) (๐2 โ ๐2) (2๐ + ๐)
Sol.2) (i) (5 โ 2๐ฅ) (3 + ๐ฅ)
= 5 (3 + ๐ฅ) โ 2๐ฅ (3 + ๐ฅ)
= 15 + 5๐ฅ โ 6๐ฅ โ 2๐ฅ2
= 15 โ ๐ฅ โ 2๐ฅ2
(ii) (๐ฅ + 7๐ฆ) (7๐ฅ โ ๐ฆ)
= ๐ฅ (7๐ฅ โ ๐ฆ) + 7๐ฆ (7๐ฅ โ ๐ฆ)
= 7๐ฅ2 โ ๐ฅ๐ฆ + 49๐ฅ๐ฆ โ 7๐ฆ2
= 7๐ฅ2 + 48๐ฅ๐ฆ โ 7๐ฆ2
(iii) (๐2 + ๐) (๐ + ๐2)
= ๐2(๐ + ๐2) + ๐ (๐ + ๐2)
= ๐3 + ๐2๐2 + ๐๐ + ๐3
(iv) (๐2 โ ๐2) (2๐ + ๐)
= ๐2(2๐ + ๐) โ ๐2(2๐ + ๐)
= 2๐3 + ๐2๐ โ 2๐๐2 โ ๐3
Q.3) Simplify.
(i) (๐ฅ2 โ 5) (๐ฅ + 5) + 25
(ii) (๐2 + 5) (๐3 + 3) + 5
(iii) (๐ก + ๐ 2) (๐ก2 โ ๐ )
(iv) (๐ + ๐) (๐ โ ๐) + (๐ โ ๐) (๐ + ๐) + 2 (๐๐ + ๐๐)
(v) (๐ฅ + ๐ฆ) (2๐ฅ + ๐ฆ) + (๐ฅ + 2๐ฆ) (๐ฅ โ ๐ฆ)
(vi) (๐ฅ + ๐ฆ) (๐ฅ2 โ ๐ฅ๐ฆ + ๐ฆ2)
(vii) (1.5๐ฅ โ 4๐ฆ) (1.5๐ฅ + 4๐ฆ + 3) โ 4.5๐ฅ + 12๐ฆ
(viii) (๐ + ๐ + ๐) (๐ + ๐ โ ๐)
Sol.3) (i) (๐ฅ2 โ 5) (๐ฅ + 5) + 25
= ๐ฅ2 (๐ฅ + 5) โ 5 (๐ฅ + 5) + 25
= ๐ฅ3 + 5๐ฅ2 โ 5๐ฅ โ 25 + 25
= ๐ฅ3 + 5๐ฅ2 โ 5๐ฅ
(ii) (๐2 + 5) (๐3 + 3) + 5
= ๐2(๐3 + 3) + 5 (๐3 + 3) + 5
= ๐2๐3 + 3๐2 + 5๐3 + 15 + 5
= ๐2๐3 + 3๐2 + 5๐3 + 20
(iii) (๐ก + ๐ 2) (๐ก2 โ ๐ )
= ๐ก (๐ก2 โ ๐ ) + ๐ 2 (๐ก2 โ ๐ )
= ๐ก3 โ ๐ ๐ก + ๐ 2๐ก2 โ ๐ 3
(iv) (๐ + ๐) (๐ โ ๐) + (๐ โ ๐) (๐ + ๐) + 2 (๐๐ + ๐๐)
= ๐ (๐ โ ๐) + ๐ (๐ โ ๐) + ๐ (๐ + ๐) โ ๐ (๐ + ๐) + 2 (๐๐ + ๐๐)
= ๐๐ โ ๐๐ + ๐๐ โ ๐๐ + ๐๐ + ๐๐ โ ๐๐ โ ๐๐ + 2๐๐ + 2๐๐
= (๐๐ + ๐๐ + 2๐๐) + (๐๐ โ ๐๐) + (๐๐ โ ๐๐) + (2๐๐ โ ๐๐ โ ๐๐)
= 4๐๐
(v) (๐ฅ + ๐ฆ) (2๐ฅ + ๐ฆ) + (๐ฅ + 2๐ฆ) (๐ฅ โ ๐ฆ)
= ๐ฅ (2๐ฅ + ๐ฆ) + ๐ฆ (2๐ฅ + ๐ฆ) + ๐ฅ (๐ฅ โ ๐ฆ) + 2๐ฆ (๐ฅ โ ๐ฆ)
= 2๐ฅ2 + ๐ฅ๐ฆ + 2๐ฅ๐ฆ + ๐ฆ2 + ๐ฅ2 โ ๐ฅ๐ฆ + 2๐ฅ๐ฆ โ 2๐ฆ2
= (2๐ฅ2 + ๐ฅ2) + (๐ฆ2 โ 2๐ฆ2) + (๐ฅ๐ฆ + 2๐ฅ๐ฆ โ ๐ฅ๐ฆ + 2๐ฅ๐ฆ)
= 3๐ฅ2 โ ๐ฆ2 + 4๐ฅ๐ฆ
(vi) (๐ฅ + ๐ฆ) (๐ฅ2 โ ๐ฅ๐ฆ + ๐ฆ2)
= ๐ฅ (๐ฅ2 โ ๐ฅ๐ฆ + ๐ฆ2) + ๐ฆ (๐ฅ2 โ ๐ฅ๐ฆ + ๐ฆ2)
= ๐ฅ3 โ ๐ฅ2๐ฆ + ๐ฅ๐ฆ2 + ๐ฅ2๐ฆ โ ๐ฅ๐ฆ2 + ๐ฆ3
= ๐ฅ3 + ๐ฆ3 + (๐ฅ๐ฆ2 โ ๐ฅ๐ฆ2) + (๐ฅ2๐ฆ โ ๐ฅ2๐ฆ)
= ๐ฅ3 + ๐ฆ3
(vii) (1.5๐ฅ โ 4๐ฆ) (1.5๐ฅ + 4๐ฆ + 3) โ 4.5๐ฅ + 12๐ฆ
= 1.5๐ฅ (1.5๐ฅ + 4๐ฆ + 3) โ 4๐ฆ (1.5๐ฅ + 4๐ฆ + 3) โ 4.5๐ฅ + 12๐ฆ
= 2.25 ๐ฅ2 + 6๐ฅ๐ฆ + 4.5๐ฅ โ 6๐ฅ๐ฆ โ 16๐ฆ2 โ 12๐ฆ โ 4.5๐ฅ + 12๐ฆ
= 2.25 ๐ฅ2 + (6๐ฅ๐ฆ โ 6๐ฅ๐ฆ) + (4.5๐ฅ โ 4.5๐ฅ) โ 16๐ฆ2 + (12๐ฆ โ 12๐ฆ)
= 2.25๐ฅ2 โ 16๐ฆ2
(viii) (๐ + ๐ + ๐) (๐ + ๐ โ ๐)
= ๐ (๐ + ๐ โ ๐) + ๐ (๐ + ๐ โ ๐) + ๐ (๐ + ๐ โ ๐)
= ๐2 + ๐๐ โ ๐๐ + ๐๐ + ๐2 โ ๐๐ + ๐๐ + ๐๐ โ ๐2
= ๐2 + ๐2 โ ๐2 + (๐๐ + ๐๐) + (๐๐ โ ๐๐) + (๐๐ โ ๐๐)
= ๐22 + ๐2 โ ๐2 + 2๐๐
Exercise 9.5
Q.1) Use a suitable identity to get each of the following products.
(i) (๐ฅ + 3) (๐ฅ + 3) (ii) (2๐ฆ + 5) (2๐ฆ + 5) (iii) (2๐ โ 7) (2๐ โ 7)
(iv) (3๐ โ1/2) (3๐ โ 1/2) (v) (1.1๐ โ 0.4) (1.1 ๐ + 0.4) (vi) (๐2 + ๐2) (โ ๐2 + ๐2)
(vii) (6๐ฅ โ 7) (6๐ฅ + 7) (viii) (โ ๐ + ๐) (โ ๐ + ๐) (ix) (๐ฅ/2 + 3๐ฆ/4) (๐ฅ/2 + 3๐ฆ/4)
(x) (7๐ โ 9๐) (7๐ โ 9๐)
Sol.1) The products will be as follows.
(i) (๐ฅ + 3) (๐ฅ + 3)
= (๐ฅ + 3)2
= (๐ฅ)2 + 2(๐ฅ) (3) + (3)2 [(๐ + ๐)2
= ๐2 + 2๐๐ + ๐2]
= ๐ฅ2 + 6๐ฅ + 9
(ii) (2๐ฆ + 5) (2๐ฆ + 5) = (2๐ฆ + 5)2
= (2๐ฆ)2 + 2(2๐ฆ) (5) + (5)2 [(๐ + ๐)2
= ๐2 + 2๐๐ + ๐2]
= 4๐ฆ2 + 20๐ฆ + 25
(iii) (2๐ โ 7) (2๐ โ 7) = (2๐ โ 7)2
= (2๐)2 โ 2(2๐) (7) + (7)2 [(๐ โ ๐)2
= ๐2 โ 2๐๐ + ๐2]
= 4๐2 โ 28๐ + 49
(iv) (3๐ โ1/2) (3๐ โ 1/2)
= (3๐ โ 12)2
= (3๐)2 โ 2 ร 3๐ ร 1/2 + (1/2)2
[(๐ โ ๐)2
= ๐2 โ 2๐๐ + ๐2]
= 9๐2 โ 3๐ + 14
(v) (1.1๐ โ 0.4) (1.1 ๐ + 0.4)
= (1.1๐)2 โ (0.4)2 [(๐ + ๐) (๐ โ ๐) = ๐2 โ ๐22]
= 1.21๐2 โ 0.16
(vi) (๐2 + ๐2) (โ ๐2 + ๐2)
= (๐2 + ๐2) (๐2 โ ๐2)
= (๐2)2 โ (๐2)2 [(๐ + ๐) (๐ โ ๐) = ๐2 โ ๐2]
= ๐4 โ ๐4
(vii) (6๐ฅ โ 7) (6๐ฅ + 7)
= (6๐ฅ)2 โ (7)2 [(๐ + ๐) (๐ โ ๐) = ๐2 โ ๐2]
= 36๐ฅ2 โ 49
(viii) (โ ๐ + ๐) (โ ๐ + ๐)
= (โ ๐ + ๐)2
= (โ ๐)2 + 2(โ ๐) (๐) + (๐)2 [(๐ + ๐)2= ๐2 + 2๐๐ + ๐2]
= ๐2 โ 2๐๐ + ๐2
(x) (7๐ โ 9๐) (7๐ โ 9๐)
= (7๐ โ 9๐)2
= (7๐)2 โ 2(7๐)(9๐) + (9๐)2 [(๐ โ ๐)2
= ๐2 โ 2๐๐ + ๐2]
= 49๐2 โ 126๐๐ + 81๐2
Q.2) Use the identity (๐ฅ + ๐) (๐ฅ + ๐) = ๐ฅ2 + (๐ + ๐)๐ฅ + ๐๐ to find the following products.
(i) (๐ฅ + 3) (๐ฅ + 7) (ii) (4๐ฅ + 5) (4๐ฅ + 1) (iii) (4๐ฅ โ 5) (4๐ฅ โ 1)
(iv) (4๐ฅ + 5) (4๐ฅ โ 1) (v) (2๐ฅ + 5๐ฆ) (2๐ฅ + 3๐ฆ) (vi) (2๐2 + 9) (2๐2 + 5)
(vii) (๐ฅ๐ฆ๐ง โ 4) (๐ฅ๐ฆ๐ง โ 2)
Sol.2) The products will be as follows.
(i) (๐ฅ + 3) (๐ฅ + 7)
= ๐ฅ2 + (3 + 7) ๐ฅ + (3) (7)
= ๐ฅ2 + 10๐ฅ + 21
(ii) (4๐ฅ + 5) (4๐ฅ + 1)
= (4๐ฅ)2 + (5 + 1) (4๐ฅ) + (5) (1)
= 16๐ฅ2 + 24๐ฅ + 5
(iii) (4๐ฅ โ 5)(4๐ฅ โ 1)
= (4๐ฅ)2 + (โ5 โ 1)4๐ฅ + (โ5) ร (โ1)
= 16๐ฅ2 + (โ6) ร 4๐ฅ + 5 = 16๐ฅ2 โ 24๐ฅ + 5
(iv) (4๐ฅ + 5)(4๐ฅ โ 1)
= (4๐ฅ)2 + {5 ร (โ1)} ร 4๐ฅ + 5 ร (โ1)
= 16๐ฅ2 + (5 โ 1) ร 4๐ฅ โ 5
= 16๐ฅ2 + 4 ร 4๐ฅ โ 5
= 16๐ฅ2 + 16๐ฅ โ 5
(v) (2๐ฅ + 5๐ฆ) (2๐ฅ + 3๐ฆ)
= (2๐ฅ)2 + (5๐ฆ + 3๐ฆ) (2๐ฅ) + (5๐ฆ) (3๐ฆ)
= 4๐ฅ2 + 16๐ฅ๐ฆ + 15๐ฆ2
(vi) (2๐2 + 9) (2๐2 + 5)
= (2๐2)2 + (9 + 5) (2๐2) + (9) (5)
= 4๐4 + 28๐2 + 45
(vii) (๐ฅ๐ฆ๐ง โ 4) (๐ฅ๐ฆ๐ง โ 2)
= (๐ฅ๐ฆ๐ง)2 + (โ4 โ 2) ร ๐ฅ๐ฆ๐ง + (โ4) ร (โ2)
= ๐ฅ2๐ฆ2๐ง2 โ 6๐ฅ๐ฆ๐ง + 8
Q.3) Find the following squares by suing the identities.
(i) (๐ โ 7)2
(ii) (๐ฅ๐ฆ + 3๐ง)2
(iii) (6๐ฅ2 โ 5๐ฆ)2 (iv) (2/3 ๐ + 3/2 ๐)2
(v) (0.4๐ โ 0.5๐)2
(vi) (2๐ฅ๐ฆ + 5๐ฆ)2
Sol.3) (i) (๐ โ 7)22
= (๐)2 โ 2(๐) (7) + (7)2 [(๐ โ ๐)2
= ๐2 โ 2๐๐ + ๐2
= ๐2 โ 14๐ + 49
(ii) (๐ฅ๐ฆ + 3๐ง)22
= (๐ฅ๐ฆ)2 + 2(๐ฅ๐ฆ) (3๐ง) + (3๐ง)2 [(๐ + ๐)2
= ๐2 + 2๐๐ + ๐2]
= ๐ฅ2๐ฆ2 + 6๐ฅ๐ฆ๐ง + 9๐ง2
(iii) (6๐ฅ2 โ 5๐ฆ)2
= (6๐ฅ2)2 โ 2(6๐ฅ2) (5๐ฆ) + (5๐ฆ)2 [(๐ โ ๐)2
= ๐2 โ 2๐๐ + ๐2]
= 36๐ฅ2 โ 60๐ฅ2๐ฆ + 25๐ฆ2
(v) (0.4๐ โ 0.5๐)2
= (0.4๐)2 โ 2 (0.4๐) (0.5๐) + (0.5๐)2[(๐ โ ๐)2
= ๐2 โ 2๐๐ + b2]
= 0.16๐2 โ 0.4๐๐ + 0.25q2
(vi) (2๐ฅ๐ฆ + 5๐ฆ)2
= (2๐ฅ๐ฆ)2 + 2(2๐ฅ๐ฆ) (5๐ฆ) + (5๐ฆ)2 [(๐ + ๐)2
= ๐2 + 2๐๐ + b2]
= 4x2๐ฆ2 + 20๐ฅ๐ฆ2 + 25๐ฆ2
Q.4) Simplify.
(i) (๐2 โ b2)2
(ii) (2๐ฅ + 5)2 โ (2๐ฅ โ 5)2
(iii) (7๐ โ 8๐)2 + (7๐ + 8๐)2
(iv) (4๐ + 5๐)2 + (5๐ + 4๐)2
(v) (2.5๐ โ 1.5๐)2 โ (1.5๐ โ 2.5๐)2 (vi) (๐๐ + ๐๐)2 โ 2๐b2๐
(vii) (m2 โ ๐2๐)2 + 2๐3๐2
Sol.4) (i) (๐2 โ b2)2
= (๐2)2 โ 2(๐2) (b2) + (b2)2 [(a โ b)2 = ๐2 โ 2ab + b2 ]
= a4 โ ๐2b2 + b4
(ii) (2x +5)2 โ (2x โ 5)2
= (2x)2 + 2(2x) (5) + (5)2 โ [(2x)2 โ 2(2x) (5) + (5)2] [(a โ b)2 = ๐2 โ 2ab + b2]
[(a + b)2 = ๐2 + 2ab + b2]
= 4x2 + 20x + 25 โ [4x2 โ 20x + 25]
= 4x2 + 20x + 25 โ 4x2 + 20x โ 25 = 40x
(iii) (7m โ 8n)2 + (7m + 8n)2
= (7m)2 โ 2(7m) (8n) + (8n)2 + (7m)2 + 2(7m) (8n) + (8n)2 [(a โ b)2 = ๐2 โ 2ab + b2 and
(a + b)2 = ๐2 + 2ab + b2]
= 49m2 โ 112mn + 64๐2 + 49m2 + 112mn + 64๐2
= 98m2 + 128๐2
(iv) (4m + 5n)2 + (5m + 4n)2
= (4m)2 + 2(4m) (5n) + (5n)2 + (5m)2 + 2(5m) (4n) + (4n)2 [(a + b)2 = ๐2 + 2ab + b2]
= 16m2 + 40mn + 25๐2 + 25m2 + 40mn + 16๐2
= 41m2 + 80mn + 41๐2
(v) (2.5p โ 1.5q)2 โ (1.5p โ 2.5q)2
= (2.5p)2 โ 2(2.5p) (1.5q) + (1.5q)2 โ [(1.5p)2 โ 2(1.5p)(2.5q) + (2.5q)2] [(a โ b)2 = ๐2 โ 2ab + b2 ]
= 6.25๐2 โ 7.5pq + 2.25q2 โ [2.25๐2 โ 7.5pq + 6.25q2]
= 6.25๐2 โ 7.5pq + 2.25q2 โ 2.25๐2 + 7.5pq โ 6.25q2]
= 4๐2 โ 4q2
(vi) (ab + bc)2 โ 2ab2c
= (ab)2 + 2(ab)(bc) + (bc)2 โ 2ab2c [(a + b)2 = ๐2 + 2ab + b2 ]
= ๐2b2 + 2ab2c + b2c2 โ 2ab2c
= ๐2b2 + b2c2
(vii) (m2 โ ๐2m)2 + 2๐3๐2
= (m2)2 โ 2(m2) (๐2m) + (๐2m)2 + 2๐3๐2 [(a โ b)2 = ๐2 โ 2ab + b2 ]
= m4 โ 2๐3๐2 + n4m2 + 2๐3๐2
= m4 + n4m2
Q.5) `Show that
(i) (3๐ฅ + 7)2 โ 84๐ฅ = (3๐ฅ โ 7)2
(ii) (9๐ โ 5๐)2 + 180๐๐ = (9๐ + 5๐)2
(iv) (4๐๐ + 3๐)2 โ (4๐๐ โ 3๐)2 = 48๐q2
(v) (๐ โ ๐) (๐ + ๐) + (๐ โ ๐) (๐ + ๐) + (๐ โ ๐) (๐ + ๐) = 0
Sol.5) (i) L.H.S = (3x + 7)2 โ 84x
= (3x)2 + 2(3x)(7) + (7)2 โ 84x
= 9x2 + 42x + 49 โ 84x
= 9x2 โ 42x + 49
R.H.S = (3x โ 7)2 = (3x)2 โ 2(3x)(7) +(7)2
= 9x2 โ 42x + 49
L.H.S = R.H.S
(ii) L.H.S = (9p โ 5q)2 + 180pq
= (9p)2 โ 2(9p)(5q) + (5q)2 โ 180pq
= 81p2 โ 90pq + 25q2 + 180pq
= 81p2 + 90pq + 25q2
R.H.S = (9p + 5q)2
= (9p)2 + 2(9p)(5q) + (5q)2
= 81p2 + 90pq + 25q2q2
L.H.S = R.H.S
(iv) L.H.S = (4pq + 3q)2 โ (4pq โ 3q)2
= (4pq)2 + 2(4pq)(3q) + (3q)2 โ [(4pq)2 โ 2(4pq) (3q) + (3q)2]
= 16p2q2 + 24pq2 + 9q2 โ [16p2q2 โ 24pq2 + 9q2]
= 16p2q2 + 24pq2 + 9q2 โ16p2q2 + 24pq2 โ 9q2
= 48pq2 = R.H.S
(v) L.H.S = (a โ b) (a + b) + (b โ c) (b + c) + (c โ a) (c + a)
= (a2 โ b2) + (b2 โ c2) + (c2 โ a2) = 0 = R.H.S
Q.6) Using identities, evaluate.
(i) 712 (ii) 992 (iii) 1022 (iv) 9982 (v) (5.2)2 (vi) 297 ร 303
(vii) 78 ร 82 (viii) 8.92 (ix) 1.05 ร 9.5
Sol.6) (i) 712 = (70 + 1)2
= (70)2 + 2(70) (1) + (1)2 [(๐ + ๐)2= a2 + 2๐๐ + b2 ]
= 4900 + 140 + 1 = 5041
(ii) 992 = (100 โ 1)2
= (100)2 โ 2(100) (1) + (1)2 [(๐ โ ๐)2 = ๐2 โ 2๐๐ + b2 ]
= 10000 โ 200 + 1 = 9801
(๐๐๐)1022 = (100 + 2)2
= (100)2 + 2(100)(2) + (2)2 [(๐ + ๐)2 = a2 + 2๐๐ + b2 ]
= 10000 + 400 + 4 = 10404
(๐๐ฃ)9982 = (1000 โ 2)2
= (1000)2 โ 2(1000)(2) + (2)2 [(๐ โ ๐)2 = a2 โ 2๐๐ + b2 ]
= 1000000 โ 4000 + 4 = 996004
(๐ฃ) (5.2)2 = (5.0 + 0.2)2
= (5.0)2 + 2(5.0) (0.2) + (0.2)2 [(๐ + ๐)2 = a2 + 2๐๐ + b2 ]
= 25 + 2 + 0.04 = 27.04
(๐ฃ๐) 297 ร 303 = (300 โ 3) ร (300 + 3)
= (300)2 โ (3)2 [(๐ + ๐) (๐ โ ๐) = a2 โ b2]
= 90000 โ 9 = 89991
(๐ฃ๐๐) 78 ร 82 = (80 โ 2) (80 + 2)
= (80)2 โ (2)2 [(๐ + ๐) (๐ โ ๐) = a2 โ b2]
= 6400 โ 4 = 6396
(๐ฃ๐๐๐) 8.92 = (9.0 โ 0.1)2
= (9.0)2 โ 2(9.0) (0.1) + (0.1)2 [(๐ โ ๐)2 = a2 โ 2๐๐ + b2]
= 81 โ 1.8 + 0.01 = 79.21
(๐๐ฅ) 1.05 ร 9.5 = 1.05 ร 0.95 ร 10
= (1 + 0.05) (1 โ 0.05) ร 10
= [(1)2 โ (0.05)2] ร 10
= [1 โ 0.0025] ร 10 [(๐ + ๐) (๐ โ ๐) = a2 โ b2]
= 0.9975 ร 10 = 9.975
Q.7) Using a2 โ b2 = (๐ + ๐) (๐ โ ๐), find
(i) 512 โ 492 (ii) (1.02)2 โ (0.98)2 (iii) 1532 โ 1472 (iv) 12.12 โ 7.92
Sol.7) (๐) 512 โ 492
= (51 + 49) (51 โ 49)
= (100) (2) = 200
(๐๐)(1.02)2 โ (0.98)2
= (1.02 + 0.98) (1.02 โ 0.98)
= (2) (0.04) = 0.08
(๐๐๐)1532 โ 1472
= (153 + 147) (153 โ 147)
= (300) (6) = 1800
(๐๐ฃ)12.12 โ 7.92
= (12.1 + 7.9) (12.1 โ 7.9)
= (20.0) (4.2) = 84
Q.8) Using (๐ฅ + ๐) (๐ฅ + ๐) = ๐ฅ2 + (๐ + ๐) ๐ฅ + ๐๐, find
(i) 103 ร 104 (ii) 5.1 ร 5.2 (iii) 103 ร 98 (iv) 9.7 ร 9.8
Sol.8) (๐) 103 ร 104
= (100 + 3) (100 + 4)
= (100)2 + (3 + 4) (100) + (3) (4)
= 10000 + 700 + 12 = 10712
(๐๐) 5.1 ร 5.2
= (5 + 0.1) (5 + 0.2)
= (5)2 + (0.1 + 0.2) (5) + (0.1) (0.2)
= 25 + 1.5 + 0.02 = 26.52
(๐๐๐) 103 ร 98
= (100 + 3) (100 โ 2)
= (100)2 + [3 + (โ 2)] (100) + (3) (โ 2)
= 10000 + 100 โ 6
= 10094
(๐๐ฃ) 9.7 ร 9.8
= (10 โ 0.3) (10 โ 0.2)
= (10)2 + [(โ 0.3) + (โ 0.2)] (10) + (โ 0.3) (โ 0.2)
= 100 + (โ 0.5)10 + 0.06 = 100.06 โ 5 = 95.06
NCERT Solutions Class 8 Mathematics Chapter 1 Rational Numbers |
NCERT Solutions Class 8 Mathematics Chapter 2 Linear Equations In One Variable |
NCERT Solutions Class 8 Mathematics Chapter 3 Understanding Quadrilaterals |
NCERT Solutions Class 8 Mathematics Chapter 4 Practical Geometry |
NCERT Solutions Class 8 Mathematics Chapter 5 Data Handling |
NCERT Solutions Class 8 Mathematics Chapter 6 Squares And Square Roots |
NCERT Solutions Class 8 Mathematics Chapter 7 Cube and Cube Roots |
NCERT Solutions Class 8 Mathematics Chapter 8 Comparing Quantities |
NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities |
NCERT Solutions Class 8 Mathematics Chapter 10 Visualising Solid Shapes |
NCERT Solutions Class 8 Mathematics Chapter 11 Mensuration |
NCERT Solutions Class 8 Mathematics Chapter 12 Exponents And Powers |
NCERT Solutions Class 8 Mathematics Chapter 13 Direct and Inverse Proportions |
NCERT Solutions Class 8 Mathematics Chapter 14 Factorisation |
NCERT Solutions Class 8 Mathematics Chapter 15 Introduction To Graphs |
NCERT Solutions Class 8 Mathematics Chapter 16 Playing with Numbers |
NCERT Solutions Class 8 Mathematics Chapter 9 Algebraic expressions and identities
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