ICSE-VIII-Mathematics
12: Algebraic Identities Class 8 Maths
- #2Use direct method to evaluate : (i) (x + 1) (x - 1) (ii) (2 + a)(2 - a) (iii) (3b - 1)(3b + 1) (iv) (4 + 5x)(4 - 5x) (ix) (z - 2/3)(z + 2/3) (v) (2a + 3)(2a - 3) (vi) (xy + 4)(xy - 4) (vii) (ab + x2)(ab -x2) (viii) (3x2 + 5y2)(3x2 - 5y2) (x) (3/5a + ½)(3/5a - ½) (xi) (0.5 - 2a)(0.5 + 2a) (xii) (a/2 - b/3)(a/2 + b/3)Ans : Note: (a + b)(a - b) = a2 - b2 (i) (x + 1)(x - 1)= (x)2 - (1)2
= x2 - 1 (ii) (2 + a)/(2 - a)= (2)2 -
(a)2
= 4 - a2 (iii) (3b - 1)(3b + 1) = (3b)2 - (1)2
= 9b2 - 1 (iv) (4 + 5x)(4 - 5x)
= (4)2 - (5x)2
= 16 - 25x2 (ix) (z - 2/3)(z + 2/3)
= (z)2 - (2/3)2
= z2 - 4/9 (v) (2a + 3)(2a - 3)
= (2a)2 - (3)2
= 4a2 - 9 (vi) (xy + 4)(xy - 4)
= (xy)2 - (4)2
= x2y2 - 16 (vii) (ab + x2) (ab - x2)
= (ab)2 - (x2)2
= a2b2 - x4 (viii) (3x2 + 5y2)(3x2 - 5y2)= (3x2)2 - (5y2)2
= 9x4 - 25y4 (x) (3/5a + ½) (3/5a - ½)
= (3/5a)2 - (1/2)2
= 9/25a2 - ¼ (xi) (0.5 - 2a)(0.5 + 2a)
= (0.5)2 - (2a)2
= 0.25 - 4a2 (xii) (a/2 - b/3)(a/2 + b/3)
= (a/2)2 - (b/3)2
= a2/4 - b2/9
- #2-i(x + 1) (x - 1)Ans : (x + 1)(x - 1)= (x)2 - (1)2
= x2 - 1
- #2-ii(2 + a)(2 - a)Ans : (2 + a)/(2 - a)= (2)2 -
(a)2
= 4 - a2
- #2-iii(3b - 1)(3b + 1)Ans : (3b - 1)(3b + 1) = (3b)2 - (1)2
= 9b2 - 1
- #2-iv(4 + 5x)(4 - 5x)Ans : (4 + 5x)(4 - 5x)
= (4)2 - (5x)2
= 16 - 25x2
- #2-ix(z - 2/3)(z + 2/3)Ans : (z - 2/3)(z + 2/3)
= (z)2 - (2/3)2
= z2 - 4/9
- #2-v(2a + 3)(2a - 3)Ans : (2a + 3)(2a - 3)
= (2a)2 - (3)2
= 4a2 - 9
- #2-vi(xy + 4)(xy - 4)Ans : (xy + 4)(xy - 4)
= (xy)2 - (4)2
= x2y2 - 16
- #2-vii(ab + x2)(ab -x2)Ans : (ab + x2) (ab - x2)
= (ab)2 - (x2)2
= a2b2 - x4
- #2-viii(3x2 + 5y2)(3x2 - 5y2)Ans : (3x2 + 5y2)(3x2 - 5y2)= (3x2)2 - (5y2)2
= 9x4 - 25y4
- #2-x(3/5a + ½)(3/5a - ½)Ans : (3/5a + ½) (3/5a - ½)
= (3/5a)2 - (1/2)2
= 9/25a2 - ¼
- #2-xi(0.5 - 2a)(0.5 + 2a)Ans : (0.5 - 2a)(0.5 + 2a)
= (0.5)2 - (2a)2
= 0.25 - 4a2
- #2-xii(a/2 - b/3)(a/2 + b/3)Ans : (a/2 - b/3)(a/2 + b/3)
= (a/2)2 - (b/3)2
= a2/4 - b2/9