# NCERT Solutions Class 8 Mathematics Solutions for Exercise 9.5 in Chapter 9 - Algebraic Expressions and Identities

Question 15 Exercise 9.5

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

Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!