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PS: EQUATION MOD2

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PS: EQUATION MOD2 [#permalink] New post 09 Jun 2003, 23:30
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R+9=|R|+R^2
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PS: EQUATION MOD2 [#permalink] New post 26 Jun 2003, 22:59
Stolyar, it'll be 1-sqrt(10) and 3.

(sqrt10)-1 would not be a root of this equation
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 [#permalink] New post 26 Jun 2003, 23:02
sure it should be negative :oops:
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 [#permalink] New post 02 Jul 2003, 21:35
Brainless wrote:
R = { +3, -3, 1-V10, 1+V10 } , V = SQRT


you are really brainless. Try to check your -3.
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 [#permalink] New post 28 Dec 2006, 15:02
R+9=|R|+R^2

o If R > 0, the equation is transformed to:

R+9 = R + R^2
<=> R^2 = 9
=> R = 3 as we stated that R is positive.

o If R < 0, the equation is transformed to:

R+9 = -R + R^2
<=> R^2 - 2*R - 9 = 0

Delta = 4 + 36 = 40

R1 = (2+2*sqrt(10)) / 2 = 1 + sqrt(10)
R2 = (2-2*sqrt(10)) / 2 = 1 - sqrt(10)

As we stated R < 0, the only solution to consider is 1 - sqrt(10)

Finally, the anwswers are R = 3 and R = 1 - sqrt(10).

Last edited by Fig on 29 Dec 2006, 04:42, edited 1 time in total.
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 [#permalink] New post 28 Dec 2006, 19:27
Fig, could you please explain R<0 part?
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 [#permalink] New post 28 Dec 2006, 19:53
Fig's is the way to solve these problems with absolute signs.
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 [#permalink] New post 28 Dec 2006, 22:52
Sumithra wrote:
Fig, could you please explain R<0 part?


Ok... I can try :)

The second equation is determined by fixing a domain of validity to the original equation. That domain is R < 0. And on it, the |R| = - R.

So we arrived to an equation that is correct only for values of R < 0. In other words, we consider only the solutions of R that are < 0, the first condition to a domain that helps simplify the original equation.
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 [#permalink] New post 29 Dec 2006, 03:50
Fig wrote:
Sumithra wrote:
Fig, could you please explain R<0 part?


Ok... I can try :)

The second equation is determined by fixing a domain of validity to the original equation. That domain is R < 0. And on it, the |R| = - R.

So we arrived to an equation that is correct only for values of R < 0. In other words, we consider only the solutions of R that are < 0, the first condition to a domain that helps simplify the original equation.


Thank you. Sorry to bother again. I understand why you picked +ve and -ve results from both the conditions. What I do not understand is: Delta = 4 + 36 = 40. Where did you get this from and how did you proceed? Maybe something basic. Would appreciate if you are able to explain.
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 [#permalink] New post 29 Dec 2006, 04:41
Sumithra wrote:
Fig wrote:
Sumithra wrote:
Fig, could you please explain R<0 part?


Ok... I can try :)

The second equation is determined by fixing a domain of validity to the original equation. That domain is R < 0. And on it, the |R| = - R.

So we arrived to an equation that is correct only for values of R < 0. In other words, we consider only the solutions of R that are < 0, the first condition to a domain that helps simplify the original equation.


Thank you. Sorry to bother again. I understand why you picked +ve and -ve results from both the conditions. What I do not understand is: Delta = 4 + 36 = 40. Where did you get this from and how did you proceed? Maybe something basic. Would appreciate if you are able to explain.


All is ok... not bothering at all :) U are welcome :)

It's to solve a binomial expression :
> a*x^2 + b*x + c
> Delta = b^2 - 4*a*c

> Root 1 = (-b + sqrt(Delta)) / (2*a)
> Root 2 = (-b - sqrt(Delta)) / (2*a)

And here, Delta = (-2)^2 - 4*(1)*(-9) = 4 + 36 = 40 = ( 2*sqrt(10) )^2

After that, we know that the root must be included in the domain defined. So here, R1 and R2 must be negative to be a solution of the original equation.

Actually, R1 is positive and so cannot be included.
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 [#permalink] New post 29 Dec 2006, 06:22
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.

Encore une fois merci Fig.
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 [#permalink] New post 29 Dec 2006, 06:27
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.

Encore une fois merci Fig.


Français.... ou parlant le français? :)

De rien... et excellente année pour toi ! :)
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 [#permalink] New post 29 Dec 2006, 06:58
Fig wrote:
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.

Encore une fois merci Fig.


Français.... ou parlant le français? :)

De rien... et excellente année pour toi ! :)


Merci, à vous aussi.
Je ne suis pas français, mais je connais un peu le français.
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 [#permalink] New post 29 Dec 2006, 07:16
Sumithra wrote:
Fig wrote:
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.

Encore une fois merci Fig.


Français.... ou parlant le français? :)

De rien... et excellente année pour toi ! :)


Merci, à vous aussi.
Je ne suis pas français, mais je connais un peu le français.


Merci :)

Ahhhh... Le 'vous', je mettais permis de vous tutoyer :D :)... Sans m'en rendre compte ;)

By using the english without such differences, I had forgetten the traditional way here ;) :)
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 [#permalink] New post 29 Dec 2006, 11:40
Never mind. BTW, thanks for the 'other modulus threads.'
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 [#permalink] New post 29 Dec 2006, 18:21
trivikram wrote:
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.
I understand your mail till here :-D
Thanks

Encore une fois merci Fig.


Sorry about that. Not a big deal, though -- the same 'thanks' to Fig.
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 [#permalink] New post 30 Dec 2006, 01:44
trivikram wrote:
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.
I understand your mail till here :-D
Thanks

Encore une fois merci Fig.


Why not learning some French ;) :)
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 [#permalink] New post 30 Dec 2006, 06:08
Fig wrote:
trivikram wrote:
Sumithra wrote:
Ah..ha! The quadratic formula!

I don't remember applying it in any questions I've come across. It slipped my mind.

Thank you trivikram.
I understand your mail till here :-D
Thanks

Encore une fois merci Fig.


Why not learning some French ;) :)


Sure...if I land up in INSEAD I will surely do it before coming !! :-D
  [#permalink] 30 Dec 2006, 06:08
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