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# PS: EQUATION MOD2

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SVP
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PS: EQUATION MOD2 [#permalink]  09 Jun 2003, 23:30
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R+9=|R|+R^2
Manager
Joined: 24 Jun 2003
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PS: EQUATION MOD2 [#permalink]  26 Jun 2003, 22:59
Stolyar, it'll be 1-sqrt(10) and 3.

(sqrt10)-1 would not be a root of this equation
SVP
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sure it should be negative
SVP
Joined: 03 Feb 2003
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Kudos [?]: 76 [0], given: 0

Brainless wrote:
R = { +3, -3, 1-V10, 1+V10 } , V = SQRT

you are really brainless. Try to check your -3.
SVP
Joined: 01 May 2006
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Kudos [?]: 98 [0], given: 0

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.
Senior Manager
Joined: 24 Oct 2006
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Fig, could you please explain R<0 part?
VP
Joined: 25 Jun 2006
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Fig's is the way to solve these problems with absolute signs.
SVP
Joined: 01 May 2006
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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.
Senior Manager
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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.
SVP
Joined: 01 May 2006
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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.
Senior Manager
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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.
SVP
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Sumithra wrote:

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.

FranÃ§ais.... ou parlant le franÃ§ais?

De rien... et excellente annÃ©e pour toi !
Senior Manager
Joined: 24 Oct 2006
Posts: 339
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Kudos [?]: 16 [0], given: 0

Fig wrote:
Sumithra wrote:

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.

FranÃ§ais.... ou parlant le franÃ§ais?

De rien... et excellente annÃ©e pour toi !

Merci, Ã  vous aussi.
Je ne suis pas franÃ§ais, mais je connais un peu le franÃ§ais.
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 98 [0], given: 0

Sumithra wrote:
Fig wrote:
Sumithra wrote:

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.

FranÃ§ais.... ou parlant le franÃ§ais?

De rien... et excellente annÃ©e pour toi !

Merci, Ã  vous aussi.
Je ne suis pas franÃ§ais, mais je connais un peu le franÃ§ais.

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
Senior Manager
Joined: 24 Oct 2006
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Never mind. BTW, thanks for the 'other modulus threads.'
Senior Manager
Joined: 24 Oct 2006
Posts: 339
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Kudos [?]: 16 [0], given: 0

trivikram wrote:
Sumithra wrote:

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
Thanks

Encore une fois merci Fig.

Sorry about that. Not a big deal, though -- the same 'thanks' to Fig.
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 98 [0], given: 0

trivikram wrote:
Sumithra wrote:

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
Thanks

Encore une fois merci Fig.

Why not learning some French
VP
Joined: 28 Mar 2006
Posts: 1384
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Kudos [?]: 19 [0], given: 0

Fig wrote:
trivikram wrote:
Sumithra wrote:

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
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 !!
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