NCERT Solutions for Class 8 Maths<\/a> Chapter 9 Algebraic Expressions and Identities Ex 9.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise 9.5<\/h2>\n
Question 1.
\nUse a suitable identity to get each of the following products.
\n(i) (x + 3) (x + 3)
\n(ii) (2y + 5) (2y + 5)
\n(iii) (2a – 7) (2a – 7)
\n(iv) (3a – \\(\\frac{1}{2}\\))(3a – \\(\\frac{1}{2}\\))
\n(v) (1.1m – 0.4) (1.1m + 0.4)
\n(vi) (a2<\/sup> + b2<\/sup>) (-a2<\/sup> + b2<\/sup>)
\n(vii) (6x – 7) (6x + 7)
\n(viii) (-a + c) (-a + c)
\n(ix) \\(\\left(\\frac{x}{2}+\\frac{3 y}{4}\\right)\\left(\\frac{x}{2}+\\frac{3 y}{5}\\right)\\)
\n(x) (7a – 9b) (7a – 9b)
\nAnswer:
\n(i) (x + 3) (x + 3) = (x + 3)2<\/sup>
\n= x2<\/sup> + 2 \u00d7 x \u00d7 3 + 32<\/sup>
\n= x2<\/sup> + 6x +9
\n[Using the identity (a + b)2<\/sup> = a2<\/sup>+ 2ab + b2<\/sup>]<\/p>\n(ii) (2y + 5) (2y + 5) = (2y + 5)2<\/sup>
\n= (2y)2<\/sup> + 2 \u00d7 2y \u00d7 5 + 52<\/sup>
\n= 4y2<\/sup> + 20y + 25
\n[Using the identity (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n(iii) (2a – 7) (2a – 7) = (2a – 7)2<\/sup>
\n= (2a)2<\/sup> – 2 \u00d7 2a \u00d7 7 + (7)2<\/sup>
\n[using the identity (a – b)2<\/sup> = a2<\/sup> -2ab + b2<\/sup>]
\n= 4a2<\/sup> – 28a + 49<\/p>\n(iv) (3a – \\(\\frac { 1 }{ 2 }\\)) (3a – \\(\\frac { 1 }{ 2 }\\)) = (3a – \\(\\frac { 1 }{ 2 }\\))2<\/sup>
\n= (3a)2<\/sup> – 2 \u00d7 3a \u00d7 \\(\\frac { 1 }{ 2 }\\) + \\(\\frac { 1 }{ 2 }\\)2<\/sup>
\n= 9a2<\/sup> – 3a + \\(\\frac { 1 }{ 4 }\\)
\n[using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n(v) (1.1m – 0.4) (1.1m + 0.4)
\n= (1.1m)2<\/sup> – (0.4)2<\/sup> = 1.21m2<\/sup> – 0.16
\n[using the identity (a \u00d7 b)(a – b) = (a2<\/sup> – b2<\/sup>)]<\/p>\n(vi) (a2<\/sup> + b2<\/sup>) (-a2<\/sup> + b2<\/sup>)
\n= (b2<\/sup> + a2<\/sup>) (b2<\/sup> – a2<\/sup>) = (b2<\/sup>)2<\/sup> – (a2<\/sup>)2<\/sup>
\n= b4<\/sup> – a4<\/sup>
\n[using identity (a + b) (a – b) = a2<\/sup> – b2<\/sup> ]<\/p>\n(vii) (6x – 7) (6x + 7)
\n= (6x)2<\/sup> – 72<\/sup> = 36x2<\/sup> – 49
\n[using the identity (a + b)(a – b) = a2<\/sup> – b2<\/sup>]<\/p>\n(viii) (- a + c) (- a + c) = (-a + c)2<\/sup>
\n= (-a)2<\/sup> -2 \u00d7 a \u00d7 c + c2<\/sup> = a2<\/sup> – 2ac + c2<\/sup>
\n[Using identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n<\/p>\n
(x) (7a – 9b) (7a – 9b) = (7a – 9b)2<\/sup>
\n= (7a)2<\/sup> – 2 \u00d7 7a \u00d7 9b \u00d7 (-9b)2<\/sup>
\n[using identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]
\n= 49a2<\/sup> – 126ab + 81b2<\/sup><\/p>\n<\/p>\n
Question 2.
\nUse the identity (x + a) (x + b) = x2<\/sup> (a + b) x + ab to find the following products.
\n(i) (x + 3) (x + 7)
\n(ii) (4x + 5) (4x + 1)
\n(iii) (4x – 5) (4x – 1)
\n(iv) (4x + 5) (4x – 1)
\n(v) (2x + 5y) (2x + 3y)
\n(vi) (2a2<\/sup> + 9) (2a2<\/sup> + 5)
\n(vii) (xyz – 4) (xyz – 2)
\nAnswer:
\n(i) (x + 3)(x + 7)
\n= x2<\/sup> + (3 + 7) x + 3 \u00d7 7 = x2<\/sup> + 10x + 21<\/p>\n(ii) (4x + 5)(4x + 1)
\n= (4x)2<\/sup> + (5 + 1) 4x + 5 \u00d7 1
\n= 16x2<\/sup> + 6 \u00d7 4x + 5
\n= 16x2<\/sup> + 24x + 5<\/p>\n(iii) (4x – 5) (4x – 1)
\n= (4x)2<\/sup> + (-5 -1) 4x + (-5) (-1)
\n= 16x2<\/sup> + (-6) 4x + 5
\n= 16x2<\/sup> – 24x + 5<\/p>\n(iv) (4x + 5) (4x – 1)
\n= (4x)2<\/sup> + (5 – 1) 4x + 5 x (-1)
\n= 16x2<\/sup> + (4) 4x – 5
\n= 16x2<\/sup> + 16x – 5<\/p>\n(v) (2x + 5y) (2x + 3y)
\n= (2x)2<\/sup> + (5y + 3y) 2x + 5y \u00d7 3y
\n= 4x2<\/sup> + (8y) 2x + 15y2<\/sup>
\n= 4x2<\/sup> + 16xy + 15y2<\/sup><\/p>\n(vi) (2a2<\/sup> + 9) (2a2<\/sup> + 5)
\n= (2a2<\/sup>)2<\/sup> + (9 + 5) 2a2<\/sup> + 9 \u00d7 5
\n= 4a4<\/sup> + (14) 2a2<\/sup> + 45
\n= 4a4<\/sup> + 28a2<\/sup> + 45<\/p>\n(vii) (xyz – 4) (xyz – 2)
\n= (xyz)2<\/sup> + (-4 -2) xyz + (-4) (-2)
\n= x2<\/sup>y2<\/sup>z2<\/sup> + (- 6) xyz + 8
\n= x2<\/sup>y2<\/sup>z2<\/sup> – 6xyz + 8<\/p>\n<\/p>\n
Question 3.
\nFind the following squares by using the identities.
\n(i) (b – 7)2<\/sup>
\n(ii) (xy + 3z)2<\/sup>
\n(iii) (6x2<\/sup> – 5y)2<\/sup>
\n(iv)(\\(\\frac { 2 }{ 3 }\\)m + \\(\\frac { 3 }{ 2 }\\)n)2<\/sup>
\n(v) (0.4p – 0.5q)2<\/sup>
\n(vi) (2xy + 5y)2<\/sup>
\nAnswer:
\n(i) (b – 7)2<\/sup> = b2<\/sup> – 2 b 7 + (-7)2<\/sup>
\n= b2<\/sup> – 14b + 49
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup> ]<\/p>\n(ii) (xy + 3z)2<\/sup> = (xy)2<\/sup> + 2 (xy) 3z + (3z)2<\/sup>
\n= x2<\/sup>y2<\/sup> + 6xyz + 9z2<\/sup>
\n[using the identity (a x b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n(iii) (6x2<\/sup> – 5y)2<\/sup>
\n= (6x2<\/sup>)2<\/sup> – 2 \u00d7 6x2<\/sup> \u00d7 5y + (5y)2<\/sup>
\n= 36x2<\/sup> – 60x2<\/sup>y + 25y2<\/sup>
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup><\/p>\n(iv) ( \\(\\frac { 2 }{ 3 }\\)m + \\(\\frac { 3 }{ 2 }\\)n)2<\/sup>
\n
\n[Using the identity (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]<\/p>\n(v) (0.4p – 0.5q)2<\/sup>
\n= (0.4p)2<\/sup> – 2 \u00d7 0.4p \u00d7 0.5q + (0.5q)2<\/sup>
\n= 0.16p2<\/sup> – 0.4pq + 0.25q2<\/sup>
\n[Using the identity (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n(vi) (2xy + 5y)2<\/sup>
\n= (2xy)2<\/sup> + 2 \u00d7 2xy \u00d7 5y + (5y)2<\/sup>
\n= 4x2<\/sup>y2<\/sup> + 20xy2<\/sup> + 25y2<\/sup><\/p>\n<\/p>\n
Question 4.
\nSimplify:
\n(i) (a2<\/sup> – b2<\/sup>)2<\/sup>
\n(ii) (2x + 5)2<\/sup> – (2x – 5)2<\/sup>
\n(iii) (7m – 8n)2<\/sup> + (7m + 8n)2<\/sup>
\n(iv) (4m + 5n)2<\/sup> + (5m + 4n)2<\/sup>
\n(v) (2.5p – 1.5q)2<\/sup> – (1.5p -2.5q)2<\/sup>
\n(vi) (ab + bc)2<\/sup> – 2ab2<\/sup>c
\n(vii) (m2<\/sup> – n2<\/sup>m2<\/sup>)2<\/sup> + 2m3<\/sup>n2<\/sup>
\nAnswer:
\n(i) (a2<\/sup> – b2<\/sup> )2<\/sup> = (a2<\/sup>)2<\/sup> – 2 \u00d7 a2<\/sup> \u00d7 b2<\/sup> + (b2<\/sup>)2<\/sup>
\n= a4<\/sup> – 2a2<\/sup> b2<\/sup> + b4<\/sup>
\n[using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>]<\/p>\n(ii) (2x + 5)2<\/sup> – (2x – 5)2<\/sup>
\n= (2x)2<\/sup> + 2 \u00d7 2x \u00d7 5 + (5)2<\/sup> – [(2x)2<\/sup> – 2 \u00d7 2x \u00d7 5 + 52<\/sup>]
\n[using (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup>]
\n(a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup>
\n= 4x2<\/sup> + 20x + 25 – (4x2<\/sup> – 20x + 25)
\n= 4x2<\/sup> + 20x + 25 – 4x2<\/sup> + 20x – 25
\n= 4x2<\/sup> – 4x2<\/sup> + 20x + 20x + 25 – 25
\n= 0 + 40x + 0
\n= 40x<\/p>\n(iii) (7m – 8n)2<\/sup> + (7m + 8n)2<\/sup>
\n= (7m)2<\/sup> – 2 \u00d7 7m \u00d7 8n + (8n)2<\/sup> + (7m)2<\/sup> + 2 \u00d7 7m \u00d7 8n + (8n)2<\/sup>
\n[Using (a – b)2<\/sup> = a2<\/sup> – 2ab + b2<\/sup> and (a + b)2<\/sup> = a2<\/sup> + 2ab + b2<\/sup> ]