Question

Using suitable identity, evaluate (i) 104 × 97 (ii) 9.8 × 10.2

Answer 1

(i)(100+4)(100-3)
(X+a)(X+b)=x^2+(a+b)X+ab
=100^2+(1)100+(-12)
=10000+100-12
10088

(ii)(10-0.2)(10+0.2)
(a+b)(a-b)
=a^2-b^2
=10^2-(0.2)^2
=100-0.04
=99.96

Answer 2

Concept-

Algebraic equations are the equations which are valid for all values of variable in them are called algebraic identities. They are also used for the factorization of polynomial. In this way, algebraic identities are used in  computation of algebraic expressions and solving different polynomials.

Given-

It is given that we have to use the suitable identities in 104×97 and 9.8×10.2

Find-

We need to find (i) 104×97  (ii) 9.8×10.2

Solution-

(i)  We have,

104 × 97 = (100+4)(100-3)

⇒(100)² + (4-3)100 + 4× (-3)

⇒ 10000 +100 -12

⇒ 10088

[ using the identity , (x + a)(x + b) = x² + (a + b) x + ab]

Hence, the value of 104 ×97 is 10088.

(ii) We have,

9.8 × 10.2 = (10-0.2)(10+0.2)

⇒ (10)² - (0.2)²

Here, a = 2 , b = 0.2

⇒ 100 - 0.04

⇒ 99.96

[ using the identity, (a + b) ( a - b) = a² - b²]

Hence, the value of 9.8×10.2 is 99.96.

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