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# s^2 + t^2 < 1 – 2st

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s^2 + t^2 < 1 – 2st [#permalink]  21 Jun 2017, 12:50
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Question Stats:

68% (01:37) correct 31% (01:23) wrong based on 66 sessions

$$s^2$$ + $$t^2$$ $$< 1 - 2st$$

 Quantity A Quantity B 1-s t

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.
[Reveal] Spoiler: OA

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GRE Instructor
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Re: s^2 + t^2 < 1 – 2st [#permalink]  23 Jun 2017, 13:05
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Expert's post
Carcass wrote:

$$s^2$$ + $$t^2$$ < 1 – 2st

 Quantity A Quantity B 1-s t

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.

We need to see that the given information contains pieces of a factorable quadratic in the form, x² + 2xy + y², which can be factored to (x + y)²

Given: s² + t² < 1 – 2st
Add 2st to both sides to get: s² + 2st + t² < 1
Factor: (s + t)² < 1
Since (some number)² is always greater than or equal to 0, we can conclude that -1 < s + t < 1
Take: s + t < 1
Subtract s from both sides to get: t < 1 - s

[Reveal] Spoiler:
A

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Brent Hanneson – Creator of greenlighttestprep.com

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Re: s^2 + t^2 < 1 – 2st [#permalink]  30 Nov 2018, 04:14
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@GreenlightTestPrep

Given: s² + t² < 1 – 2st
Add 2st to both sides to get: s² + 2st + t² < 1
Factor: (s + t)² < 1

After this step we can also use the rule that
(s + t)²=0
s+t=0
S=-t

Now plugging s=-t in Quantity A we find out the 1+t will always be greater than t. Hence A.
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Re: s^2 + t^2 < 1 – 2st [#permalink]  02 Dec 2018, 22:43
1
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s^2 + t^2 < 1 - 2st
A: 1-s
B: t

s^2 + t^2 + 2st < 1
(s+t)^2 < 1
(s+t) < 1
t < 1-s
So A is bigger than B.
_________________

Re: s^2 + t^2 < 1 – 2st   [#permalink] 02 Dec 2018, 22:43
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