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  1. Asked: In:

    Use the identity (x+a) (x+b)=x²+(a+b)x+ab to find the following products: (i) (x+3)(x+7) (ii) (4x+5)(4x+1)

    Best Answer
    Taruni

    (i) (x+3)(x+7) = (x)²+(3+7)x+3x7 [Using identity(x+a)(x+b)=x²+(a+b)x+ab x²+10x+21 (ii) (4x+5)(4x+1)=(4x)²+(5x1)4x+5x1 [Using identity(x+a)(x+b)=x²+(a+b)x+ab = 16x²+6x4x+5=16x²+24x+5 Class 8 Maths Chapter 9 Exercise 9.5 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-sRead more

    (i) (x+3)(x+7) = (x)²+(3+7)x+3×7
    [Using identity(x+a)(x+b)=x²+(a+b)x+ab
    x²+10x+21
    (ii) (4x+5)(4x+1)=(4x)²+(5×1)4x+5×1
    [Using identity(x+a)(x+b)=x²+(a+b)x+ab
    = 16x²+6x4x+5=16x²+24x+5

    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/

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  2. Asked: In:

    Use a suitable identity to get each of the following products: (i) (x+3)(x+3) (ii) (2y+5) (2y+5)

    Best Answer
    Taruni

    (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:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-9/

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    • 6