(i) (x+3)(x+3) = (x+3)² = (x)²+2 X x X 3 +(3)² [Using identity (a+b)² =a²+2ab+b²] = x²+6x+9 (ii) (2y+5) (2y+5) = (2y)²+2X2yx5+(5)² = 4y²+20y+25 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
(i) (x+3)(x+3) = (x+3)²
= (x)²+2 X x X 3 +(3)² [Using identity (a+b)² =a²+2ab+b²]
= x²+6x+9
(ii) (2y+5) (2y+5)
= (2y)²+2X2yx5+(5)²
= 4y²+20y+25
Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video
Use a suitable identity to get each of the following products: (i) (x+3)(x+3) (ii) (2y+5) (2y+5)
(i) (x+3)(x+3) = (x+3)² = (x)²+2 X x X 3 +(3)² [Using identity (a+b)² =a²+2ab+b²] = x²+6x+9 (ii) (2y+5) (2y+5) = (2y)²+2X2yx5+(5)² = 4y²+20y+25 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/
(i) (x+3)(x+3) = (x+3)²
= (x)²+2 X x X 3 +(3)² [Using identity (a+b)² =a²+2ab+b²]
= x²+6x+9
(ii) (2y+5) (2y+5)
= (2y)²+2X2yx5+(5)²
= 4y²+20y+25
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/