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x2 + y2 = 81 x2 - y2 = 0

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x2 + y2 = 81 x2 - y2 = 0 [#permalink] New post 12 Aug 2019, 02:05
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Question Stats:

83% (00:44) correct 16% (00:12) wrong based on 6 sessions
\(x^2 + y^2 = 81\)

\(x^2 - y^2 = 0\)

Quantity A
Quantity B
\(x^4-y^4\)
\(0\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: x2 + y2 = 81 x2 - y2 = 0 [#permalink] New post 12 Aug 2019, 09:11
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Expert's post
Carcass wrote:
\(x^2 + y^2 = 81\)

\(x^2 - y^2 = 0\)

Quantity A
Quantity B
\(x^4-y^4\)
\(0\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Nice question!
I like it because we can ignore the first equation, and just use the second equation.

Take: \(x^2 - y^2 = 0\)
Add \(y^2\) to both sides to get: \(x^2 = y^2\)
Square both sides to get: \((x^2)^2 = (y^2)^2\)
Simplify by applying the Power of a Power law to get: \(x^4 = y^4\)
Rewrite as follows: \(x^4 - y^4 = 0\)

Answer: C

Cheers,
Brent
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Re: x2 + y2 = 81 x2 - y2 = 0 [#permalink] New post 17 Aug 2019, 02:15
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# (A+B) * (A-B) = A^2 - B^2
So (x^2+y^2) * (x^2-y^2) = (x^4-y^4)
=> 81 * 0 => 0
Ans is C.
Re: x2 + y2 = 81 x2 - y2 = 0   [#permalink] 17 Aug 2019, 02:15
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x2 + y2 = 81 x2 - y2 = 0

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