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Re: Question about Reduce
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71311] Re: Question about Reduce
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Tue, 14 Nov 2006 05:06:35 -0500 (EST)
*References*: <ej735s$j0g$1@smc.vnet.net>
Thanks to anyone that replied to me.
Regards
Dimitris
dimitris wrote:
> Consider the following simple quadratic equation
>
> eq = a*x^2 + b*x + c == 0;
>
> Solve[eq, x]
> {{x -> (-b - Sqrt[b^2 - 4*a*c])/(2*a)}, {x -> (-b + Sqrt[b^2 -
> 4*a*c])/(2*a)}}
>
> For the complete set of solutions we use Reduce
>
> Reduce[eq, x]
> (a != 0 && (x == (-b - Sqrt[b^2 - 4*a*c])/(2*a) || x == (-b + Sqrt[b^2
> - 4*a*c])/(2*a))) || (a == 0 && b != 0 && x == -(c/b)) || (c == 0 && b
> == 0 && a == 0)
>
> The following commnds give the desired results
>
> Reduce[eq && b^2 - 4*a*c < 0, x, Reals]
> False
>
> Reduce[eq && a == 0, x]
> (a == 0 && b != 0 && x == -(c/b)) || (c == 0 && b == 0 && a == 0)
>
> Reduce[eq && c == 0 && a == 0, x]
> (c == 0 && b == 0 && a == 0) || (c == 0 && a == 0 && b != 0 && x == 0)
>
> Howver why the following don't simplify to the double root?
>
> Reduce[a*x^2 + b*x + c == 0 && b^2 - 4*ac == 0 && a != 0, x]
> ac == b^2/4 && a != 0 && (x == (-b - Sqrt[b^2 - 4*a*c])/(2*a) || x ==
> (-b + Sqrt[b^2 - 4*a*c])/(2*a))
>
> Reduce[a*x^2 + b*x + c == 0 && ac == b^2/4 && a != 0, x]
> ac == b^2/4 && a != 0 && (x == (-b - Sqrt[b^2 - 4*a*c])/(2*a) || x ==
> (-b + Sqrt[b^2 - 4*a*c])/(2*a))
>
> Thanks in advance for any help!
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