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# If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an

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Re: If the operation ∆ is defined by a∆b = (b-a)^2/a^2 for all number a an [#permalink]
Bunuel wrote:
If the operation ∆ is defined by $$a∆b = \frac{(b-a)^2}{a^2}$$ for all number a and b, and a ≠ 0, then -1∆(1∆ -1) =

A. -1
B. 0
C. 1
D. 9
E. 25

First, we simplify: a∆b = (b-a)^2/a^2 = (b^2 - 2ab + a^2)/a^2 = (b/a)^2 - 2(b/a) + 1 = (b/a - 1)^2

Working from the inside out: 1∆-1 =( -1/1 - 1)^2 = (-2)^2 = 4.
--> -1∆4 = (4/-1 -1)^2 = (-5)^2 = 25 (E)
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Re: If the operation is defined by ab = (b-a)^2/a^2 for all number a an [#permalink]
Top Contributor
Given that a ∆ b = $$\frac{(b - a)^2}{a^2}$$ and we need to find the value of − 1 ∆ (1 ∆ − 1)

Lets start by finding the value of 1 ∆ − 1

To find 1 ∆ − 1 we need to compare what is before and after ∆ in 1 ∆ − 1 and a ∆ b

=> We need to substitute a with 1 and b with -1 in a ∆ b = $$\frac{(b - a)^2}{a^2}$$ to get the value of 1 ∆ − 1

=> 1 ∆ -1 = $$\frac{(-1 - 1)^2}{1^2}$$ = $$\frac{(-2)^2}{1}$$ = 4

=> − 1 ∆ (1 ∆ − 1) = − 1 ∆ 4

Similarly, -1 ∆ 4 = $$\frac{(4 - (-1))^2}{(-1)^2}$$ = $$\frac{(5)^2}{1}$$ = 25