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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.
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.
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
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.