# BSE Odisha 8th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(f)

Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(f) Textbook Exercise Questions and Answers.

## BSE Odisha Class 8 Maths Solutions Algebra Chapter 3 ବୀଜଗାଣିତିକ ପରିପ୍ରକାଶ ଓ ଅଭେଦ Ex 3(f)

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(i) (a + 2)² = a² + (………)a + 2² (2, 2a, 4, 4a)
(ii) (3 + y)² = 9 + 3 (………..) + y² (y, 2y, 3y, 4y)
(iii) (4 y)² = 16 + 2 (……….) + y² (-2, -2y, -4, -4y)
(iv) (2x-3y)² = 4x² – 3 (………..) + 9y² (2xy, 3xy, 4xy, 12xy)
(v) (x + a)(x – b) = x² + (………..)x – ab (a + b, a – b, b -a, -(a + b)
ସମାଧାନ :
(i) 4,
(ii) 2y,
(iii) -4y,
(iv) 4xy,
(v) a – b

Question 2.
ସୂତ୍ର ପ୍ରୟୋଗ କରି ନିମ୍ନଲିଖତ ରାଶିର ବର୍ଗ ନିର୍ଣ୍ଣୟ କର ।
(i) b+c
(ii) (4 + b)
(iii) r – 10
(iv) 3n+ 2
(v) 2m + n
(vi) 7p – q
(vii) 2x + 3y
(viii) 2m – 3n – p
(ix) xy + 4z
(x) a + 2b + 3c
ସମାଧାନ :
(i) (b + c)²
= (b)² + 2.b.c + (c)²
= b² + 2bc + c²

(ii) (4 + b)²
= (4)² + 2.4.b + (b)²
= 16 + 8b + b²

(iii) (r – 10)²
= (r)² – 2.r.10 + (10)²
= r² – 20r + 100

(iv) (3n+ 2)²
= (3n)²+ 2.3n.2 + (2)²
= 9n² + 12n + 4

(v) (2m + n)²
= (2m)² + 2.2m.n + (n)²
= 4m2 + 4mn + n2

(vi) (7p – q)²
= (7p)² – 2.7p.q + (q)²
= 49p² – 14pq + q²

(vii) (2x + 3y)²
= (2x)² + 2.2x.3y+ (3y)²
= 4x² + 12xy + 9y²

(viii) (2m 3n – p)²
= {2m + (-3n) + (-p)}²
= (2m)² + (-3n)² + (-p)² + 2.2m.(-3n) + 2.(-3n).(-p) + 2.(-p).2m
= 4m² + 9n² + p² – 12mn + 6np – 4pm

(ix) (xy+4z)²
= {x + (y) + 4z}²
= (x)² + (-y)² + (4z)² + 2.x.(y) + 2.(y).4z + 2.42.x
= x² + y² + 16z² – 2xy – 8yz + 8zx

(x) (a + 2b+3c)²
= (a)² + (2b)² + (3c)² + 2.a.2b + 2.2b.3c + 2.3c.a
= a² + 4b² + 9c² + 4ab + 12bc + 6ca

Question 3.
ଆବଶ୍ୟକ ସୂତ୍ର ପ୍ରୟୋଗ କରି ନିମ୍ନସ୍ଥ ସଂଖ୍ୟାମାନଙ୍କର ବର୍ଗ ନିର୍ଣ୍ଣୟ କର ।
(i) 102
(ii) 304
(iii) 1003
(iv) 4001
ସମାଧାନ :
(a+b)² = a² + 2ab +b²

(i) (100 + 2)² = (100)² + 2.(100).2 + (2)² = 10000 + 400 + 4 = 10404

(ii) (300 + 4)² = (300)² + 2.300.4 +(4)² = 90000 + 2400 + 16 = 92416

(iii) (1000 + 3)² = (1000)² + 2.1000.3 + (3)² = 1000000 + 6000 +9 = 1006009

(iv) (4001)² = (4000 + 1)² = 16000000 + 2.4000.1 + (1)²
= 16000000 + 8000 + 1= 16008001

Question 4.
ଆବଶ୍ୟକ ଅଭେଦ ପ୍ରୟୋଗ କରି ମୂଲ୍ୟ ନିରୂପଣ କର ।
(i) 99²
(ii) 998²
(iii) 297 × 303
(iv) 78 × 82
(v) 8.9²
(vi) 1.05 × 9.5
(vii) 51² – 49²
(viii) (1.02)² – (0.98)²
(ix) 153² – 147²
ସମାଧାନ :
(a – b)² = a² – 2ab +b²

(i) (99)² = (100 – 1)
= (100) – 2.100.1 + (1)²
=10000 – 200 + 1
= 980

(ii) (998)²
= (1000 – 2)²
= (1000)² – 2.1000.2 + (2)²
= 1000000 – 4000 + 4
= 96004

(iii) 297 × 303
= (300 – 3) (300 + 3)
= (300)² – (3)²
= 90000 – 9 = 89991

(iv) 78 × 82 = (80 – 2) (80 + 2)
= (80)² – (2)²
= 6400 – 4
= 6396

(v) 8.92 = (9 – 0.1)²
= (9)² – 2 × 9 × (0.1)+(0.1)²
= 81 – 1.8 + 0.01
= 81.01 – 1.8
= 79.21

(vi) 1.05 × 9.5
= (1 + 0.05) (1 – 0.05)
= (1)² – (0.05)²
= 1 – 0.0025
= 0.9975

(vii) 51² – 49²
= (51 + 49) (51 – 49)
= 100 × 2
= 200

(viii) (1.02)² – (0.98)²
= (1.02 + 0.98) (1.02 – 0.98)
= (2.00) × (0.04)
= 0.08

(ix) 153² – 147²
= (153 + 147) (153 – 147)
= (300) × (6)
= 1800

Question 5.
(x + a) (x + b) = x² + (a + b)x + ab ଅଭେଦ ପ୍ରୟୋଗ କରି ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର ।
(i) 103 × 104
(ii) 5.1 × 5.2
(iii) 103 × 98
(iv) 9.7 × 9.8
ସମାଧାନ :
(i) 103 × 104
=(100 + 3) (100 + 4)
= (100)² + (3 + 4) 100 + 3 × 4
= 10000 + 700 + 12
= 10712

(ii) 5.1 × 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

(iii) 103 × 98
=(100 + 3)(100 – 2)
=(100)² + (3 – 2)100 – 3 × 2
= 10000 + 100 – 6
= 10094

(iv) 9.7 × 9.8
= (9 + 0.7) (9 + 0.8)
= (9)²+ (0.7 + 0.8) 9 + 0.7 0215 0.8
= 81 + 1.5 × 9 + 0.56
= 81 + 13.5 + 0.56
= 95.06

Question 6.
ଆବଶ୍ୟକ ଅଭେଦ ପ୍ରୟୋଗକରି ଗୁଣଫଳ ନିର୍ଣ୍ଣୟ କର ।
(i) (x + 3) (x + 3)
(ii) (2y + 5) (2y + 5)
(iii) (2a — 7) (2a — 7)
(iv) (1.1m – 0.4) (1.1m + 0.4)
(v) (a² + b²) (-a² + b²)
(vi) (6x – 7) (6x + 7)
(vii) (p – 5) (p + 5)
(viii) (2x + 3y) (3y – 2x)
(ix) (x + 1)(x – 1)(x2 + 1)
(x) (2y + 3)(2y — 3)(4y2 + 9)
ସମାଧାନ :
(i) (x + 3) (x + 3) = (x + 3)²
= (x)² + 2.x.3 + (3)²
= x² + 6x + 9

(a + b)² = a² + 2ab + b²

(ii) (2y + 5)(2y + 5)
= (2y + 5)²
= (2y)² + 2 . 2y . 5 + (5)²
= 4y² + 20y + 25

(iii) (2a – 7) (2a – 7) = (2a – 7)²
= (2a)² – 2.2a. 7 + (7)²
= 4a² – 28 a + 49

(iv) (1.1m – 0.4)(1.1m + 0.4)
(a+b)(a-b) = a² – b²
= (1.1m)² – (0.4)²
= 1.21m² – 0.16

(v) (a² + b²)(-a² + b²)
= (b² + a²)(b² – a²)
= (b²)² – (a²)² = b4 – a4

(vi) (6x – 7)(6x + 7)
= (6x)² – (7)²
= 36x² – 49

(vii) (p – 5) (p + 5)
= (p)² – (5)² = p² – 25

(viii) (2x + 3y) (3y – 2x)
= (3y + 2x) (3y – 2x)
= (3y)² – (7x)²
= 9y² – 4x².

(ix) (x + 1)(x – 1)(x² + 1)
= ((x + 1)(x – 1))(x² + 1)
= {(x)² – (1)²}(x2 + 1)
= (x² – 1)(x² + 1)
= (x²)² – (1²)² = x4 – 1

(x) (2y + 3) (2y – 3) (y² + 9)
= {(2y + 3) (2y – 3)} (4y² + 9)
= {(2y)² – (3)²) (4y² + 9)
= (4y² – 9) (4y² + 9)
= (4y²)² – (9)²
= l6y4 – 81

Question 7.
(x + a) (x + b) = x² + (a + b)x + ab ଅଭେଦ ପ୍ରୟୋଗ କରି ଗୁଣଫଳ ନିଶ୍ଚୟ କର ।
(i) (x + 3) (x + 7)
(ii) (4x + 5)(4x + 1)
(iii) (4x – 5)(4x – 1)
(iv) (4x + 5)(4x – 1)
(v) (2a² + 9)(2a² + 5)
(vi) (xyz – 4)(xyz – 2)
ସମାଧାନ :
(i) (x + 3) (x + 7)
= x² + (3 + 7)x + 3 × 7
= x² + lOx + 21

(ii) (4x + 5)(4x + 1)
=(4x)² + (5 + 1)4x + 5 × 1
= 16x² + 24x + 5

(iii) (4x – 5)(4x – 1)
= (4x)² + (-5 – 1)4x + (-5).(-1)
= 16x² – 24x + 5

(iv) (4x + 5)(4x — 1)
= (4x)² + (5 – 1)4x + 5(-1)
= 16x² + 16x – 5

(y) (2a² + 9)(2a² + 5)
= (2a²)² + (9 + 5)2a² + 9 × 5
=4a4 + 28a² + 45

(vi) (xyz – 4)(xyz – 2)
= (xyz)² + (-4 – 2)xyz + (-4).(-2)
= x²y²z² – 6xyz + 8

Question 8.
ସରଳ କର ।
(i) (a² – b²)² + (a² + b²)²
(ii) (2x + 5)² – (2x 5)²
(iii) (7m – 8n)² + (7m + 8n)²
(iv) (4m + 5n)² + (5m + 4n)²
(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
(vi) (ab + bc)² – 2ab²c
(vii) (m² – n²m)² + 2m²n²
(viii) (a + b – c)² + (a – b – c)²
(ix) (2a – 3b – c)² + (2a – b + 5c)²
(x) (3x – 4y + z)² – (x – y – z)²
ସମାଧାନ :
(i) (a² – b²)² + (a² + b²)²
= ((a²)² – 2.a².b² + (b²)²) + ((a²)² + 2.a².b²+ (b²)²)
= a4 – 2a²b² + b4 + a4 + 2a²b² + b4
= 2a4 + 2b4 = 2(a4 + b4)

(ii) (2x + 5)²- (2x – 5)² [a² – b² = (a+b)(a-b)]
= (2x + 5 + 2x – 5) (2x + 5 – 2x + 5) = 4x × 10 = 40x

(iii) (7m – 8n)² + (7m + 8n)²
= {(7m)² – 2.7m.8n + (8n)²) + ((7m)² + 2.7m.8n + (8n)²)
= (49m² – 11.2mn + mn²) + (49m² + 11.2mn + 64n²)
= 49m² – 11.2mn + 64n²+ 49m² + 11.2mn + mn²
= 98m² + 128n² = 2(49 m² + Mn²)

(iv) (4m+ 5n)2 + (5m + 4n)²
= ((4m)² + 2.4m.5n + (5n)²) + ((5m)² + 2.5m.4n + (4n)²)
= 16m² + 40mn + 25n² + 25m² + 40mn + 16n² = 41m² + 80mn + 4m²

(v) (2.5p – 1.5q)² – (1.5p – 2.5q)²
= {(2.5 p – 1.5q) + (1.5 p – 2.5 q))} {(2.5 p – 1.5 q) – (1.5 p – 2.5 q)}
= (2.5p – 1.5q + 1.5p – 2.5q) (2.5p – 1.5q – 1.5p + 2.5q)
= (4p – 4.q)(p + q) = 4(p – q)(p + q) = 4(p² – q²)

(vi) (ab + bc)² – 2ab²c = (ab)² + 2ab.bc + (bc)² – 2ab²c
= a²b³ + 2ab²c + b²c² – 2ab²c = a²b² + b²c²

(vii) (m – n²m)² + 2m²n²
= (m²n – 2.m².n²m + (n²m)² + 2m²n² = m4 – 2m³n² + n4m² + 2m²n²

(ix) (2a – 3b – c)² + (2a – b + 5c)².
= {(2a)² + (-3b)² + (-c)² + 2.2a.(-3b) + 2.(-3b).(-c) + 2.(-c).2a}
+ {(2a)² + (-b)² + (5c)² + 2.2a.(-b) + 2(-b).(5c) + 2.5c.2a}
= 4a² + 9b² + c² – 12ab + 6bc – 4ca + 4a² + b² + 25c² – 4ab – 10bc + 20ca
= 8a² + 10b² + 26c² – 16ab – 4bc + 16ca

(x) (3x – 4y + z)² – (x – 2y – z)²
= {(3x – 4y + z) + (x – 2y – z)} {(3x – 4y + z) – (x – 2y – z)}
= (3x – 4y + z + x – 2y – z) (3x – 4y + z – x + 2y + z)
= (4x – 6y)(2x – 2y + 2z) = 2(2x – 3y).2(x – y + z)
= 2 × 2 (2x – 3y)(x – y + z) = 4(2x – 3y)(x – y + z).

Question 9.
ନିମ୍ନଲିଖୂତ ପଲିନୋମିଆଲ୍‌ଗୁଡ଼ିକୁ ପୂର୍ବବର୍ଗରେ ପରିଣତ କର ।
(i) 4x² + 12xy + 9y²
(ii) 64m² – 48mn + 9n²
(iii) 4x² – 4x + 1
(iv) x² + 4y² + z² + 4xy + 4yz + 2zx
(v) 4x² + y² + z² – 4xy + 2yz – 4xz
(vi) 9x² + 4y² + z² – 12xy – 4yz + 6zx
ସମାଧାନ :
(i) 4x² + 12xy + 9y² = (2x)² + 2.2x.3y + (3y)² = (2x + 3y)²

(ii) 64m² – 48mn + 9n² = (8m)² – 2.8m.3n + (3n)² = (8m – 3n)²

(iii) 4x² – 4x + 1 = (2x)² – 2.2x. 1 + (1)² = (2x – 1)²

(iv) x² + 4y² + z² + 4xy + 4yz + 2zx
= (x)² + (2y)² + (z)² + 2.x.2y + 2.2y.z + 2.z.x = (x + 2y + z)²

(v) 4x² + y² + z² – 4xy + 2yz – 4xz
= (2x)² + (-y)² + (-z)² + 2.2x.(-y) + 2.(-y).(-z) + 2.(-z).2x
= {(2x + (-y) + (-z)}² = (2x – y – z)²

(vi) 9x² + 4.y² + z² – 12xy – 4yz + 6zx
= (3x)² + (-2y)² + (z)² + 2.3x.(-2y) + 2.(-2y).(z) + 2.(z).3x
= (3x – 2y + z)²

Question 10.
(i) ଦର୍ଶାଅ ଯେ (a + b)² = (a – b)² + 4ab
(ii) ଦର୍ଶାଅ ଯେ (a + b)² + (a – b)² = 2(a² + b²)
(iii) ଦର୍ଶାଅ ଯେ $$\frac{a+b}{2}^2-\frac{a-b}{2}^2=ab$$
(iv) ଦର୍ଶାଅ ଯେ (2a + b)² – (2a – b)² = 8ab
(v) ଦର୍ଶାଅ ଯେ (3x – 2y)² – 12xy = 9x² + 4y²
ସମାଧାନ :
(i) (a + b)² = (a – b)² + 4ab
ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ = (a – b)² + 4ab
= a² + b² – 2ab + 4ab = a² + b² + 2ab = (a + b)² = ବାମପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)

(ii) (a + b)² + (a – b)² = 2(a² + b²)
ବାମ ପାର୍ଶ୍ଵ = (a + b)² + (a – b)²
= (a² + 2ab + b²) + (a² – 2ab + b²) = a² + 2ab + b² + a² – 2ab +b²
= 2a² + 2b² = 2(a² + b²) = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)

(iii) $$\frac{a+b}{2}^2-\frac{a-b}{2}^2=ab$$
ବାମ ପାର୍ଶ୍ଵ = $$\frac{a+b}{2}^2-\frac{a-b}{2}^2$$
= $$\frac{1}{2}$$
$$\left(\frac{a+b}{2}+\frac{a-b}{2}\right)\left(\frac{a+b}{2}-\frac{a-b}{2}\right)=\left(\frac{a+b+a-b}{2}\right)\left(\frac{a+b-a+b}{2}\right)$$
= $$(\frac{2a}{2})(\frac{2b}{2})$$ = ab ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)

(iv) (2a + b)² – (2a – b)² = 8ab
ବାମ ପାର୍ଶ୍ଵ = (2a + b)² – (2a – b)²
= (4a² + 4ab + b²) – (4a² – 4ab + b²)
= 4a² + 4ab + b² – 4a² + 4ab – b² = 8ab = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)

(v) (3x – 2y)² + 12xy = 9x² + 4.y²
ବାମ ପାର୍ଶ୍ଵ = (3x – 2y)² + 12xy = (3x)² – 2.3x.2y + (2y)² + 12xy
= 9x² – 12xy + 4y² + 12xy = 9x² + 4y² = ଦକ୍ଷିଣ ପାର୍ଶ୍ଵ (ପ୍ରମାଣିତ)