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MathGroup Archive 2006

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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!


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