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Evaluate (a2b+2ba)2-(a2b-2ba)2-4 - Mathematics

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Question

Evaluate `(a/[2b] + [2b]/a )^2 - ( a/[2b] - [2b]/a)^2 - 4`.

Sum

Solution 1

Consider the given expression :

Let us expand the first term : `[ a/(2b) + (2b)/a]^2`

We know that,

( a + b )2 = a2 + b2 + 2ab

∴ `[ a/(2b) + (2b)/a]^2 = (a/(2b))^2 + ((2b)/a)^2 + 2 xx a/(2b) xx (2b)/a`

= `a^2/(4b)^2 + (4b)^2/a^2 + 2`       ...(1)

Let us expand the second term : `[ a/[2b] - [2b]/a]^2`

We know that,

( a - b )2 = a2 + b2 - 2ab

∴ `[ a/(2b) - (2b)/a]^2 = (a/(2b))^2 + ((2b)/a)^2 - 2 xx a/(2b) xx (2b)/a`

= `a^2/(4b)^2 + (4b)^2/a^2 - 2`       ...(2)

Thus from (1) and (2), the given expression is

`[ a/(2b) + (2b)/a]^2 - [ a/(2b) - (2b)/a]^2 - 4 `

`= a^2/(4b)^2 + (4b)^2 /a^2 + 2 - a^2/(4b)^2 - (4b)^2/a^2 + 2 - 4`
= 0.

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Solution 2

x2 - y2 = (x - y) (x + y)

So,

`= (a/(2b) + (2b)/a)^2 - (a/(2b) - (2b)/a)^2`

`= [(a/(2b) + (2b)/a) - (a/(2b) - (2b)/a)] [(a/(2b) + (2b)/a)] + (a/ (2b) - (2b)/a)`

`= ((4b)/a) ((2a)/(2b))`

= 4

So, 

`(a/(2b)+ (2b)/a)^2 - (a/(2b) - (2b)/a)^2 - 4`

= 4 - 4

= 0

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Chapter 4: Expansions (Including Substitution) - Exercise 4 (A) [Page 58]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 4 Expansions (Including Substitution)
Exercise 4 (A) | Q 4.1 | Page 58
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