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Math Assignment Class VIII | Algebraic Identities Ch-7
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Mathematics Assignment Class VIII
Algebraic Identities
Algebraic Identities required to solve this assignment
Students need to learn these algebraic identities before start this assignment
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
a2 - b2 = (a + b)(a - b)
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Ans: 40xy
Ans: (x - 7)(x + 4)
Ans: 40
Ans: (3m - 5n + 8x)(3m - 5n - 8x)
4(x + y)2 – 28(x2 –
y2) + 49(x – y)2
= [2(x
+ y)]2 – 2 x 2(x + y) x 7(x + y) + [7(x - y)]2
Using: a2 - 2ab + b2 = (a - b)2
we get
=
[2(x + y) – 7(x - y)]2.
=
(2x + 2y - 7x + 7y)2.
=
(- 5x + 9y)2 or (5x - 9y)2
x2 – 91 = 2.3 x 2.3 + 2 x 0.7 x 2.3 + 0.7 x 0.7
Ans: 10
Solution Hint
x2 – 91 = 2.3 ✕ 2.3 + 2 ✕ 0.7 ✕ 2.3 + 0.7 ✕ 0.7
= (2.3)2 + 2 ✕ 2.3 ✕ 0.7 + (0.7)2
= (2.3 + 0.7)2 = (3)2 = 9
x2 – 91 = 9
x2 = 9 + 91 = 100
x = 10
Answer: - 60xy + 20x
Solution Hint
(5x - 3y + 1)2 - (5x + 3y - 1)2.
Using a2 – b2
= (a + b)(a - b) we get
= (5x - 3y + 1+5x + 3y - 1)(5x - 3y + 1-5x - 3y + 1)
= (10x)(- 6y + 2) = - 60xy + 20x
Question 25:
If 3p - 4q = 8 and pq = 2, find the value of (3p - 4q)2.
Answer: 160
Solution Hint:
3p - 4q = 8
Squaring on both side we get
(3p – 4q)2 = 82
9p2 + 16q2 – 24pq = 64
9p2 + 16q2 – 24 🇽 2 = 64
9p2 + 16q2 – 48 = 64
9p2 + 16q2 = 64 + 48 = 112
(3p – 4q)2 = 9p2 + 16q2 + 24pq
= 112 + 24 🇽 2 = 160
Factorise :
Answer
Question 27: If 5x - 2y = 7 and xy = 2, find the value of (5x + 2y)2.
Question 28: Factorise: (64m2 – 144mn + 81n2) – 25p2
Answer : (8m - 9n + 5p)(8m - 9n - 5p)
Question 29: Simplify : (a2 - b2)(a2 + b2) - (a2 - b2)2.
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