"Question 8 Using (x + a) (x + b) = x^2 + (a + b) x + ab, find (i) 103 × 104 (ii) 5.1 × 5.2 (iii) 103 × 98 (iv) 9.7 × 9.8
Class 8 Algebraic Expressions and Identities Page 152"
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:
Using (x + a) (x + b) = x²+ (a + b) x + ab
1) 103 x 104
= (100 + 3)(100 + 4)
= 100² + (3 + 4)100 + (3×4)
= 10000 + 700 + 12
= 10712
2) 5.1 x 5.2
= (5 + 0.1)(5 + 0.2)
= 5² + (0.1+0.2)5 + (0.1×0.2)
= 25 + 1.5 + 0.02
= 26.52
3) 103 x 98
= (100 + 3)(100 - 2)
= 100² + (3-2)100 – (3×2)
= 10000 + 100 - 6
= 10094
4) 9.7 x 9.8
= (9 + 0.7 )(9 + 0.8)
= 9² + (0.7 + 0.8)9 + (0.7×0.8)
= 81 + 13.5 + 0.56
= 95.06
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Hope this will help you.....
Answer:
103 x 104
= (100 + 3)(100 + 4)
By using the formula ,(x + a) (x + b) = x²+ (a + b) x + ab
= 100² + (3 + 4)100 + (3×4)
= 10000 + 700 + 12
= 10712
2) 5.1 x 5.2
= (5 + 0.1)(5 + 0.2)
By using the formula ,(x + a) (x + b) = x²+ (a + b) x + ab
= 5² + (0.1+0.2)5 + (0.1×0.2)
= 25 + 1.5 + 0.02
= 26.52
3) 103 x 98
= (100 + 3)(100 - 2)
By using the formula ,(x + a) (x + b) = x²+ (a + b) x + ab
= 100² + (3-2)100 – (3×2)
= 10000 + 100 - 6
= 10094
Step-by-step explanation: