"Question 1 Use a suitable identity to get each of the following products. (i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a − 7) (2a − 7) (iv) (v) (1.1m − 0.4) (1.1 m + 0.4) (vi) (a^2 + b^2) (− a^2 + b^2) (vii) (6x − 7) (6x + 7) (viii) (− a + c) (− a + c) (ix) (x) (7a − 9b) (7a − 9b)
Class 8 Algebraic Expressions and Identities Page 151"
Answers
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:
1)
(x + 3) (x + 3)
(x+3)²= (x)²+2×x×3+(3)²
(a + b)² = a² + 2ab + b²
=x² + 6x + 9
2)
(2y + 5) (2y + 5)
(2y+5)²= (2y)²+2×2y×5+ (5)²
(a + b)² = a² + 2ab + b²
=4y² + 20y + 25
3)
(2a – 7) (2a – 7)
(2a-7)²=(2a)²-2×2a×7+(7)²
(a – b)² = a² – 2ab + b²
= 4a² – 28a + 49
4)
(3a-1/2)(3a-1/2)
(3a-1/2)²= (3a)²-2×3a×1/2 + (1/2)²
(a – b)² = a² – 2ab + b²
= 9a² -3a+(1/4)
5)
(1.1m – 0.4) (1.1m + 0.4)
= (1.1m)² - (0.4)²
(a – b)(a + b) = a² – b²
= 1.21m² – 0.16
6)
(a²+ b²) (– a²+ b²)
= (b²+ a² ) (b² – a²)
= (b²)² - (a²)²
(a – b)(a + b) = a² – b²
= b⁴ - a⁴
7)
(6x – 7) (6x + 7)
(6x)² -(7)²
(a – b)(a + b) = a² – b²
=36x² – 49
8) (– a + c) (– a + c)
(c-a)(c+a)
= (c)² -(a)²
(a – b)(a + b) = a² – b²
= c² - a²
8) [ (x/2)+(3y/4)] [ (x/2)+(3y/4)]
[ (x/2)+(3y/4)]²
(x/2)²+2×x/2×3y/4)+(3y/4)²
(a + b)² = a² + 2ab + b²
=(x²/4) + (9y²/16) +(3xy/4)
9) (7a – 9b) (7a – 9b)
(7a-9b)²=(7a)²- 2×7a×9b +(9b)²
(a - b)² = a² - 2ab + b²
= 49a² – 126ab + 81b²
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Hope this will help you.....
Answer:
(x)49a²-126ab+81b²
Step-by-step explanation:
thanks