[Date Index]
[Thread Index]
[Author Index]
Question about Reduce
*To*: mathgroup at smc.vnet.net
*Subject*: [mg71272] Question about Reduce
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Sun, 12 Nov 2006 06:48:45 -0500 (EST)
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!
Prev by Date:
**Re: [Help Needed] Function to solve polynomial**
Next by Date:
**Re: finite group theory w/mathematica**
Previous by thread:
**Re: Re: Developer`UseFrontEnd + FrontEndExecute + GetBoundingBoxSizePacket**
Next by thread:
**Re: Re: Question about Reduce**
| |