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Re: S^2 + T^2 < 1-2st [#permalink]
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Please use the rh rules for posting

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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

Please advise, for some reason I keep getting this question wrong.
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Re: S^2 + T^2 < 1-2st [#permalink]
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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]
Hi Brent, can you explain why " If (something)² < 1, then it must be the case that -1 < something < 1"?
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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]
2
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

This is based on this page: https://www.mathsisfun.com/algebra/abso ... lving.html

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]
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

This is based on this page: https://www.mathsisfun.com/algebra/abso ... lving.html

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
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Re: S^2 + T^2 < 1-2st [#permalink]
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