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PS: EQUATION MOD2 [#permalink]
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10 Jun 2003, 00:30
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R+9=R+R^2



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PS: EQUATION MOD2 [#permalink]
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26 Jun 2003, 23:59
Stolyar, it'll be 1sqrt(10) and 3.
(sqrt10)1 would not be a root of this equation



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sure it should be negative



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Brainless wrote: R = { +3, 3, 1V10, 1+V10 } , V = SQRT
you are really brainless. Try to check your 3.



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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 = (22*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, 05:42, edited 1 time in total.



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Fig, could you please explain R<0 part?



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Fig's is the way to solve these problems with absolute signs.



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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.



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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.



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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.



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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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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.
FranÃ§ais.... ou parlant le franÃ§ais?
De rien... et excellente annÃ©e pour toi !



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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. 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.



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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. 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



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Never mind. BTW, thanks for the 'other modulus threads.'



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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 Thanks Encore une fois merci Fig.
Sorry about that. Not a big deal, though  the same 'thanks' to Fig.



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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 Thanks Encore une fois merci Fig.
Why not learning some French



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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 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 !!










