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NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities
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Q8) Using \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab, find
(i) 103 x 104
(ii) 5.1 x 5.2
(iii) 103 x 98
(iv) 9.7 x 9.8
Solution:
(i) 103 x 104 = (100+3) x (100+4)
= \left(100\right)^2+\left(3+4\right)\times100+3\times4
[Using identity \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab]
= 10000 + 7 x 100 + 12
= 10000 + 700 + 12
= 10712
(ii) 5.1 x 5.2 = (5 + 0.1) x (5 + 0.2)
= \left(5\right)^2+\left(0.1+0.2\right)\times5+0.1\times0.2
[Using identity \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab]
= 25 + 0.3 x 5 + 0.02
= 25 + 1.5 + 0.02
= 26.52
(iii) 103 x 98 = (100+3) x (100-2)
= \left(100\right)^2+\left(3-2\right)\times100+3\times\left(-2\right)
[Using identity \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab
= 10000 + (3-2) x 100 - 6
= 10000 + 100 - 6
= 10094
(iv) 9.7 x 9.8 = (10 - 0.3) x (10 - 0.2)
= \left(10\right)^2+\left\{\left(-0.3\right)+\left(-0.2\right)\right\}\times10+\left(-0.3\right)\times\left(-0.2\right)
[Using identity \left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab]
= 100 + {-0.3 -0.2} x 10 + 0.06
= 100 - 0.5 x 10 + 0.06
= 100 - 5 + 0.06
= 95.06
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