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s^2 + t^2 < 1 – 2st

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

63% (01:23) correct 36% (01:08) wrong based on 44 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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Re: s^2 + t^2 < 1 – 2st [#permalink] New post 23 Jun 2017, 13:05
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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

Answer:
[Reveal] Spoiler:
A

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