(i) 51²-49² =(51+49)(51-49) [Using identity (a-b)(a+b)=a²=b²] = 100 x 2 = 200 (ii) (1.02)² – (0.98)²= (1.02-0.98)(1.02-0.98) [Using identity (a-b)(a+b)=a²=b²] = 2.00 x 0.04 = 0.08 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
(i) 51²-49² =(51+49)(51-49) [Using identity (a-b)(a+b)=a²=b²]
= 100 x 2 = 200
(ii) (1.02)² – (0.98)²= (1.02-0.98)(1.02-0.98) [Using identity (a-b)(a+b)=a²=b²]
= 2.00 x 0.04 = 0.08
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
Using (x+a)(x+b)=x²+(a+b)x+ab, find (i) 103 x 104 (ii) 5.1 x 5.2
(i) 103 x 104 = (100 + 3) x (100 + 4) = (100)² +(3+4)x100+3x4 [Using identity (x+a) (x+b) =x²+(a+b)x+ab] = 10000 + 7 x 100 + 12 = 10000 + 700 + 12 = 10712 (ii) 5.1 x 5.2 = (5 + 0.1) x (5 + 0.2) = (5)²+(0.1+0.2)x5+0.1x0.2 [Using identity (x+a) (x+b) =x²+(a+b)x+ab] = 25 + 0.3 x 5 + 0.02 = 25 + 1.5 + 0Read more
(i) 103 x 104 = (100 + 3) x (100 + 4) = (100)² +(3+4)x100+3×4
[Using identity (x+a) (x+b) =x²+(a+b)x+ab]
= 10000 + 7 x 100 + 12 = 10000 + 700 + 12 = 10712
(ii) 5.1 x 5.2 = (5 + 0.1) x (5 + 0.2) = (5)²+(0.1+0.2)x5+0.1×0.2
[Using identity (x+a) (x+b) =x²+(a+b)x+ab]
= 25 + 0.3 x 5 + 0.02 = 25 + 1.5 + 0.02 = 26.52
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Using a²-b²=(a+b)(a-b), find (i) 51²-49² (ii) (1.02) ² – (0.98) ²
(i) 51²-49² =(51+49)(51-49) [Using identity (a-b)(a+b)=a²=b²] = 100 x 2 = 200 (ii) (1.02)² – (0.98)²= (1.02-0.98)(1.02-0.98) [Using identity (a-b)(a+b)=a²=b²] = 2.00 x 0.04 = 0.08 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solRead more
(i) 51²-49² =(51+49)(51-49) [Using identity (a-b)(a+b)=a²=b²]
= 100 x 2 = 200
(ii) (1.02)² – (0.98)²= (1.02-0.98)(1.02-0.98) [Using identity (a-b)(a+b)=a²=b²]
= 2.00 x 0.04 = 0.08
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Using identities, evaluate: (i) 71² (ii) 99 ²
(i) 71² = (70+1)² = (70)² +2x70x1+(1)² [Using identity(a+b)²=a²-2ab+b² = 4900 + 140 + 1 = 5041 (ii) 99² = (70)²+2x70x1+(1)² [Using identity(a+b)²=a²-2ab+b² = 10000 – 200 + 1 = 9801 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-soRead more
(i) 71² = (70+1)² = (70)² +2x70x1+(1)²
[Using identity(a+b)²=a²-2ab+b²
= 4900 + 140 + 1 = 5041
(ii) 99² = (70)²+2x70x1+(1)²
[Using identity(a+b)²=a²-2ab+b²
= 10000 – 200 + 1 = 9801
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Show that: (i) (3x+7) ² -84x=(3x-7) ² (ii) (9p-5q) ² + 180pq =(9p+5q) ²
(i) L.H.S= (3x+7)²-84x=(3x)²+2x3xX7+(7)²-84x [Using identity (a+b)²=a²+2ab+b²] = 9x²+42x+49-84x = 9x²-42x+49 = (3x-7)² [∵ (a-b)²=a²+2ab+b²] = R.H.S. (ii) LHS = (9p-5q)²+180pq=(9p)²-2x9px5q+(5q)²+180pq [Using identity (a-b)²=a²-2ab+b²] = 81p²-90pq+25q²+180pq = 81p²+90pq²+25q² = (9p+5q)² [∵ (a-b)²=a²-Read more
(i) L.H.S= (3x+7)²-84x=(3x)²+2x3xX7+(7)²-84x
[Using identity (a+b)²=a²+2ab+b²]
= 9x²+42x+49-84x
= 9x²-42x+49
= (3x-7)² [∵ (a-b)²=a²+2ab+b²]
= R.H.S.
(ii) LHS = (9p-5q)²+180pq=(9p)²-2x9px5q+(5q)²+180pq
[Using identity (a-b)²=a²-2ab+b²]
= 81p²-90pq+25q²+180pq
= 81p²+90pq²+25q²
= (9p+5q)² [∵ (a-b)²=a²-2ab+b²]
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
Simplify: (i) (a²-b²)² (ii) (2x+5)²-(2x-5)²
(i) (a²-b²)² = (a²)-2xa²xb²+(b²)+(b²)² [Using identity (a-b)²=a²-2ab+b²] = a⁴-2a²b²+b⁴ (ii) (2x+5)²-(2x-5)² =(2x)²+2X2xX5+(5)²-[(2x)-2X2xX5+(5)² [Using identities (a+b)² = a²+2ab+b² and (a-b)²=a²-2ab+b² = 4x²+20x+25x-[4x²-20x+25] = 4x²+20x+25x-4x²+20x-25 = 40x Class 8 Maths Chapter 9 Exercise 9.5 SoRead more
(i) (a²-b²)² = (a²)-2xa²xb²+(b²)+(b²)² [Using identity (a-b)²=a²-2ab+b²]
= a⁴-2a²b²+b⁴
(ii) (2x+5)²-(2x-5)² =(2x)²+2X2xX5+(5)²-[(2x)-2X2xX5+(5)²
[Using identities (a+b)² = a²+2ab+b² and (a-b)²=a²-2ab+b²
= 4x²+20x+25x-[4x²-20x+25]
= 4x²+20x+25x-4x²+20x-25
= 40x
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/