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# For all numbers x and y, the operationΦ is defined by x?y = (x + y)(x

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Math Expert
Joined: 02 Sep 2009
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For all numbers x and y, the operationΦ is defined by x?y = (x + y)(x  [#permalink]

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21 Aug 2018, 06:24
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Difficulty:

15% (low)

Question Stats:

84% (01:18) correct 16% (01:27) wrong based on 34 sessions

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For all numbers x and y, the operation ? is defined by x?y = (x + y)(x - y) + (y - x)(y + x) + xy. What is the value of $$\sqrt{12}?\sqrt{3}$$?

A. 6
B. 12
C. 18
D. 24
E. 36

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For all numbers x and y, the operationΦ is defined by x?y = (x + y)(x  [#permalink]

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21 Aug 2018, 06:32
Given x?y = (x + y)(x - y) + (y - x)(y + x) + xy
x = $$\sqrt{12}$$
y = $$\sqrt{3}$$
x?y = ($$x^2$$-$$y^2$$) + ($$y^2$$-$$x^2$$) + $$\sqrt{12}$$*$$\sqrt{3}$$
= 12-3 + 3-12 + $$\sqrt{36}$$
= 6

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Re: For all numbers x and y, the operationΦ is defined by x?y = (x + y)(x  [#permalink]

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21 Aug 2018, 06:33
Bunuel wrote:
For all numbers x and y, the operation ? is defined by x?y = (x + y)(x - y) + (y - x)(y + x) + xy. What is the value of $$\sqrt{12}?\sqrt{3}$$?

A. 6
B. 12
C. 18
D. 24
E. 36

$$x?y = (x + y)(x - y) + (y - x)(y + x) + xy$$

$$=>x?y=x^2-y^2+y^2-x^2+xy=xy$$

$$\sqrt{12}?\sqrt{3}=\sqrt{12*3}=6$$

Option A
Re: For all numbers x and y, the operationΦ is defined by x?y = (x + y)(x   [#permalink] 21 Aug 2018, 06:33
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