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Published on: 13/05/2022
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Questions + Answers key
Take MCQ Maths Test1.
Expand the following squares, using suitable identities.
(mn + 3p)2
2.
Expand the following squares, using suitable identities.
(2x + 5)2
3.
Using the identity (x + a)(x + b) = x2 + x(a + b) + ab ,find the following product.
(4x + 3y) ( 4x + 5y)
4.
Using the identity (x + a)(x + b) = x2 + x(a + b) + ab , find the following product.
(x + 3)(x + 7)
5.
Factorise by using the identity: a2 -b2 = (a +b)(a−b)
(i) a2 -1
(ii) 9k2 -25
(iii) 64x2 - 81y2
1.
\((a+b)^{2}=a^{2}+2 a b+b^{2}\)
put a = mn, b = 3p in the above identity
\( (m n+3 p)^{2} =(m n)^{2}-2 \times m n \times 3 p+(3 p)^{2} \)
\(\\ =m^{2} n^{2}+6 m n p+9 p^{2} \)
2.
\((a+b)^{2}=a^{2}+2 a b+b^{2}\)
put a = 2x, b = 5 in the above formula, we get
\( (2 x+5)^{2} =(2 x)^{2}+(2 \times 2 x \times 5)+5^{2} \)
\(\\ =4 x^{2}+20 x+25 \)
3.
(x + a)(x + b) = x2 + x(a + b) + ab
put x = 4x, a = 3y, b = 5y is the above formula, we get
\( (4 x+3 y)(4 x+5 y) =(4 x)^{2}+(4 x)(3 y+5 y)+(3 y \times 5 y) \)
\(\\ =16 x^{2}+(4 x)(8 y)+15 y^{2} \)
\(\\ =16 x^{2}+32 x y+15 y^{2} \)
4.
put x = x, a = 3, b = 7 in the above formula, we get,
\( (x+3)(x+7) =x^{2}+x(3+7)+3 \times 7\)
\( \\ =x^{2}+x(10)+21 \)
\(\\ =x^{2}+10 x+21 \)
5.
(i) a2 −1 = a2 −12 = (a + 1)(a − 1) [since, 12 = 1 x 1 = 1]
Therefore, the factors of a2 -1 are (a + 1) and (a − 1).
(ii) 9k2 − 25 = (32 x k2 ) − 52 = (3k)2 −52 [since, amxbm =(axb)m ]
= (3k + 5)(3k − 5) |since,a2- b2 = (a + b)(a - b)|
Therefore, the factors of 9k2 -25 are (3k + 5) and (3k − 5).
(iii) 64x2 −81y2 = (82xx2 ) − (92xy2 ) = (8x)2 −(9y)2
= (8x + 9y)(8x − 9y)
Therefore, the factors of 64x2 -81y2 are (8x + 9y) and (8x – 9y).
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Tamilnadu Stateboard 7th Standard Subjects
Tamilnadu Stateboard Standards
