Re: Solve / Reduce isolating results.
- To: mathgroup at smc.vnet.net
- Subject: [mg118518] Re: Solve / Reduce isolating results.
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sat, 30 Apr 2011 05:54:17 -0400 (EDT)
There are two real solutions:
Clear[sq1]
Clear[xx]
solutions =
Quiet@Solve[{(1/6)*((sq1*196)*2)^3 - (1/
6)*((((2.75 + xx)*((3.75 + xx)*21^2))*100)*2)^3 ==
1.2902698935932682*^18, (1/6)*((sq1*100)*2)^3 - (1/
6)*((((2.75 + xx)*((3.75 + xx)*21^2))*36)*2)^3 ==
1.883856772422697*^17}, {sq1, xx}];
realOnes = Select[solutions, FreeQ[#, Complex] &]
{{sq1 -> 5292., xx -> -6.75}, {sq1 -> 5292., xx -> 0.25}}
or
realOnes = Select[Chop@solutions, FreeQ[#, Complex] &]
{{sq1 -> 5292., xx -> -6.75}, {sq1 -> 5292., xx -> 0.25}}
Bobby
On Fri, 29 Apr 2011 06:35:15 -0500, Lea Rebanks <lrebanks at netvigator.com>
wrote:
> Dear All,
>
> Given the enclosed code - see below.
> Please could someone help me setup the code to Solve / Reduce for these 2
> missing values only required
>
> {SQ1 -> 5292,
> XX -> 0.25}
>
> (I don't want multiple results.) I want positive numbers & NOT Complex..
>
>
> Many thanks for your help & attention, really appreciated. Please see
> below.
>
> Best regards,
> Lea...
> ...................................................................
>
>
> Clear[SQ1]
> Clear[XX]
> Solve[{(1/6)*((SQ1*196)*2)^3 -
> (1/6)*((((2.75 + XX)*((3.75 + XX)*21^2))*100)*2)^3 ==
> 1.2902698935932682*^18, (1/6)*((SQ1*100)*2)^3 -
> (1/6)*((((2.75 + XX)*((3.75 + XX)*21^2))*36)*2)^3 ==
> 1.883856772422697*^17}, {SQ1, XX}]
--
DrMajorBob at yahoo.com