Question

"Question 3 Find the following squares by suing the identities. (i) (b − 7)^2 (ii) (xy + 3z)^2 (iii) (6x^2 − 5y)^2 (iv) (2m/3 + 3n/2)^2 (v) (0.4p − 0.5q)^2 (vi) (2xy + 5y)^2

Class 8 Algebraic Expressions and Identities Page 151"

Answer 1

Hiiii....

Your answer is attached.

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Answer 2

An identity is true only for certain values of its variables. An equation is not an identity.

The following are the identities

(a + b)² = a² + 2ab + b² 
(a – b)² = a² – 2ab + b² 
(a – b)(a + b) = a² – b²

 Another useful identity is

 (x + a) (x + b) = x² + (a + b) x + ab

If the given expression is the difference of two squares we use the formula

a² –b² = (a+b)(a-b)

 

• The above four identities are useful in carrying out squares and products of algebraic expressions. They also allow easy alternative methods to calculate products of numbers and so on.

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Solution:

i)

(b – 7)²

= (b)²-2×b×7 +(7)²

[(a + b)² = a² + 2ab + b² ]

b² - 14b + 49

 

ii) (xy + 3z)²

( xy)² +2×xy×3z + (3z)²

[(a + b)² = a² + 2ab + b² ]

x²y²+6xyz+9z²

 

iii) (6x²– 5y)²
(6x²)²-2×6x²×5y +(5y)²

[(a – b)² = a² – 2ab + b² ]

36x⁴ - 60x²y + 25y²

 

iv) [(2m/3}) + (3n/2)]²

(2m/3)²+2×2m/3×3n/2+(3n/2)²

[(a + b)² = a² + 2ab + b² ]

(4m²/9) +2mn+(9n²/4)

 

v) (0.4p – 0.5q)²

(0.4p)² -2×0.4p×0.5q +(0.5q)²

[(a – b)² = a² – 2ab + b² ]

0.16p² - 0.4pq + 0.25q²

 

vi) (2xy + 5y)²

(2xy)²+2×2xy×5y+(5y)²

[(a + b)² = a² + 2ab + b² ]

4x²y² + 20xy²+ 25y²

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Hope this will help you.....