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Re: Solve / Reduce isolating results.

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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@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@yahoo.com

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Re: Solve / Reduce isolating results. DrMajorBob <btreat1@austin.rr.com> - 2011-04-30 09:54 +0000

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