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# For real numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ?

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Math Expert
Joined: 02 Sep 2009
Posts: 54376
For real numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ?  [#permalink]

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27 Mar 2019, 20:57
00:00

Difficulty:

25% (medium)

Question Stats:

84% (02:03) correct 16% (02:33) wrong based on 25 sessions

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For all numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ? (y - x)^2 ?

(A) 0
(B) x^2 + y^2
(C) 2x^2
(D) 2y^2
(E) 4xy

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Intern
Joined: 05 Feb 2019
Posts: 2
Location: Indonesia
GPA: 3.3
Re: For real numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ?  [#permalink]

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27 Mar 2019, 23:23
a?b = (a - b)^2

(x - y)^2 ? (y - x)^2

Algebra method

= (x - y)^2 ? (y - x)^2
= ((x - y)^2 - (y - x)^2)^2
= (( x^2 -2xy + y^2 ) - ( x^2 -2xy + y^2))^2
= open brackets
= (( x^2 -2xy + y^2 - x^2 + 2xy - y^2))^2
= (0)^2
= 0

Numerical Method
let x & y be an easy number
x = 5
y = 2
= (x - y)^2 ? (y - x)^2
= ((5 - 2)^2 - (2 - 5)^2)^2
= ((3)^2 - (-3)^2)^2
= (9 - 9)^2

Correct me if I'm wrong
Thanks
Manager
Joined: 19 Jan 2019
Posts: 70
Re: For real numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ?  [#permalink]

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27 Mar 2019, 23:27
1
raypohan wrote:
a?b = (a - b)^2

(x - y)^2 ? (y - x)^2

Algebra method

= (x - y)^2 ? (y - x)^2
= ((x - y)^2 - (y - x)^2)^2
= (( x^2 -2xy + y^2 ) - ( x^2 -2xy + y^2))^2
= open brackets
= (( x^2 -2xy + y^2 - x^2 + 2xy - y^2))^2
= (0)^2
= 0

Numerical Method
let x & y be an easy number
x = 5
y = 2
= (x - y)^2 ? (y - x)^2
= ((5 - 2)^2 - (2 - 5)^2)^2
= ((3)^2 - (-3)^2)^2
= (9 - 9)^2

Correct me if I'm wrong
Thanks

There's nothing wrong with the solution, but this can be done within seconds when you realise that (x-y)^2 is same as (y-x)^2....

(2-4)^2 is same as (4-2)^2

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Re: For real numbers a and b, define a?b = (a - b)^2. What is (x - y)^2 ?   [#permalink] 27 Mar 2019, 23:27
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