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TAGS: Intern Joined: 05 Jan 2016
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S^2 + T^2 < 1-2st [#permalink]
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Question Stats: 63% (01:02) correct 36% (01:14) wrong based on 76 sessions
$$s^2 + t^2 < 1-2st$$

 Quantity A Quantity B 1-S T

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

Last edited by Carcass on 20 Jul 2019, 12:37, edited 3 times in total.
Editing the post and adding the OA and Tags Retired Moderator Joined: 07 Jun 2014
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Re: Manhattan Prep GRE Algebra Question S^2 + T^2 < 1-2st [#permalink]
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Expert's post
GREhelp wrote:
Hi,
Any help on the following question would be greatly appreciated. It is a question from the Manhattan Prep GRE Algebra Book.

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

Quantity A:
1-S

Quantity B:
t

We have

$$s^2 + t^2 < 1-2st$$ addind 2st on both sides

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

$$(s + t)^2< 1$$ or $$s + t <1$$ or $$t< 1-s$$

Hence option Option A is correct.

Now both s and t can be negative quantities. like -10 and -100 respectively. So A is greater.

Or one of them can be negative and other positive such that they satisfy the above in equality s=2 t= -2.5.

Both of them are positive fractions s=99/100 t= 9/1000. Still Quantity A is greater.
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Sandy
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Re: S^2 + T^2 < 1-2st [#permalink]
Expert's post
Please use the rh rules for posting

• a proper title for the question
• using the tags to identify it
• well formatted in its text

Thank you
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Intern Joined: 05 Jan 2016
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Re: S^2 + T^2 < 1-2st [#permalink]
Hi Sandy,
Thanks for your respond. I was a little confused.

In the solution below you said that:

(s + t)^2< 1 or s + t <1 or t< 1-s

How would that be possible since there are two (s+t) and anything plugged into them would result in a positive number.
(s + t)^2< 1 = (s+t) (s+t) < 1

If you to plug in negative quantities for s and t can, like -10 and -100 respectively. you will get a positive number. For example,
S = -10 T = -100
(-10+ -100) ( -10 + -100) = -110 * -110 = 12100 Retired Moderator Joined: 07 Jun 2014
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Re: S^2 + T^2 < 1-2st [#permalink]
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Expert's post
You cannot choose any value for s and t.

|s+t| has to be less than 1.

That is a constraint so if you put s= 10 then t= -10.99999999..... or -9.000000000000001 or something like that.

Note that s and t are not independent. They are constrained. Hence they must follow |s+t| < 1.

If you put any value that just means that there are no constraints.

You cannot substitute "any" value.
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Re: S^2 + T^2 < 1-2st [#permalink]
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Expert's post
GREhelp wrote:
$$s^2 + t^2 < 1-2st$$

 Quantity A Quantity B 1 - s t

First notice that we have a perfect square, (s + t)², "hiding" in the given inequality. The perfect square is in the form s² + 2st + t²
Here's what I mean:

Given: s² + t² < 1 - 2st
Add 2st to both sides to get: s² + 2st + t² < 1
Factor the left side: (s + t)² < 1

IMPORTANT: If (something)² < 1, then it must be the case that -1 < something < 1
Since we now know that (s + t)² < 1, we can say that -1 < s + t < 1

Given:
Quantity A: 1 - s
Quantity B: t

Add s to both sides to get:
Quantity A: 1
Quantity B: s + t

Since -1 < s + t < 1, we can conclude that Quantity A is greater.

Cheers,
Brent
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Re: S^2 + T^2 < 1-2st [#permalink]
Hi Brent, can you explain why " If (something)² < 1, then it must be the case that -1 < something < 1"? Intern Joined: 17 Sep 2017
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Re: S^2 + T^2 < 1-2st [#permalink]
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msawicka wrote:
Hi Brent, can you explain why " If (something)² < 1, then it must be the case that -1 < something < 1"?

Let me try. It is because square, square root and absolute value go hand in hand. Let's say we have A^2=B, A can be any value, but B must be >=0 (i.e. non-negative, or positive inclusion of zero). We can rewrite it as A= + or - square root of B, which means any number under square root is >=0. In turn, if we are given square root of B = A, then B must be non-negative, so to make sure it satisfies this, we absolute value B => if we square both side, we get absolute value of B=A^2. In case of inequality, we have absolute value of B < A^2. And we have our familiar inequality: absolute value of B < A^2 <=> -A^2 < B < A^2.
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Re: S^2 + T^2 < 1-2st [#permalink]
I'm guessing the reasoning behind the interval is that:

square root of a^2 is the same as |a|
and
if |a|< x
then

-x < a < x

So, I'm guessing that's why we can say:

(s + t)² < 1

|s + t| < 1

so: -1 < s + t < 1

or maybe at this point: (s + t)² < 1 one could say we are taking the square root of both sides which leaves us with

s+t < 1 and s+t <-1

We have + and - because the square root of 1 can be +1 or -1, and we have to switch the direction of an inequality sign when we multiply by a negative.

So we end up with: -1 < s + t < 1

I may be wrong.
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Re: S^2 + T^2 < 1-2st [#permalink]
Expert's post
arc601 wrote:
I'm guessing the reasoning behind the interval is that:

square root of a^2 is the same as |a|
and
if |a|< x
then

-x < a < x

So, I'm guessing that's why we can say:

(s + t)² < 1

|s + t| < 1

so: -1 < s + t < 1

or maybe at this point: (s + t)² < 1 one could say we are taking the square root of both sides which leaves us with

s+t < 1 and s+t <-1

We have + and - because the square root of 1 can be +1 or -1, and we have to switch the direction of an inequality sign when we multiply by a negative.

So we end up with: -1 < s + t < 1

I may be wrong.

Your reasoning (in both cases) is perfect.

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com
If you enjoy my solutions, you'll like my GRE prep course.  Re: S^2 + T^2 < 1-2st   [#permalink] 18 Jul 2019, 09:10
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