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# x ≥ 0 y ≥ 0

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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783 [0], given: 397

x ≥ 0 y ≥ 0 [#permalink]  15 Apr 2018, 04:39
Expert's post
00:00

Question Stats:

73% (00:34) correct 26% (01:20) wrong based on 34 sessions
x ≥ 0
y ≥ 0

 Quantity A Quantity B $$\sqrt{x^{12}}-y$$ $$(x^3-\sqrt{y})(x^3+\sqrt{y})$$

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

Drill 1
Question: 8
Page: 511
[Reveal] Spoiler: OA

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Sandy
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Director
Joined: 07 Jan 2018
Posts: 604
Followers: 7

Kudos [?]: 546 [0], given: 88

Re: x ≥ 0 y ≥ 0 [#permalink]  18 Apr 2018, 01:21
qty B is in the form $$(x -y) (x + y)$$ hence it can be transformed into $$x^2 -y^2$$
qty B will become $$(x^3)^2 - y = x^6 - y$$
qty A becomes $$x^6-y$$ when square root is removed hence both quantities are equal
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This is my response to the question and may be incorrect. Feel free to rectify any mistakes

GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4857
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 105

Kudos [?]: 1783 [0], given: 397

Re: x ≥ 0 y ≥ 0 [#permalink]  22 Apr 2018, 12:20
Expert's post
Explanation

In Quantity A, $$\sqrt{x^{12}}-y=\sqrt{(x^{6})^{2}}-y$$.

In Quantity B, you may recognize one of the common quadratics: $$(a + b)(a - b) = a^2 - b^2$$.

If not, FOIL; either way, Quantity B is $$x^6 - y$$. Thus, the two quantities are equal.
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Sandy
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Re: x ≥ 0 y ≥ 0   [#permalink] 22 Apr 2018, 12:20
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