Solver Problem 'Maybe' - Printable Version +- HP Forums ( https://www.hpmuseum.org/forum)+-- Forum: HP Calculators (and very old HP Computers) ( /forum-3.html)+--- Forum: HP Prime ( /forum-5.html)+--- Thread: Solver Problem 'Maybe' ( /thread-7516.html) |

Solver Problem 'Maybe' - BERNARD MICHAUD - 01-04-2017 05:17 PM
Below is the equation for a Circle. R^2=(X-H)^2+(Y-K)^2. The solver will solve for R=7.2801. X=4, H=2, Y=-6 and K=1 R^2=53. You can solve any of the variables except 'X'. Solving for X returns '0'. Re-arranging the formulae to solve for X. Note: SQRT below instead squareRoot sign on calculator. X=H+SQRT-(K^2-2*Y*K-(R^2-Y^2)). Note: 'H+' Now I can solve for any variables including 'X'. 2+2=4 Of course using X=H-SQRT-(K^2-2*Y*K-(R^2-Y^2)). Note: 'H-' cannot solve for X either, since -SQRT-(K^2-2*Y*K-(R^2-Y^2))=-2. 2-2=0 Which may be a clue????? Its seems that the Solver precedes the SQRT sign with H- instead of H+ and returns '0' instead of '4'. The other option is, maybe I've had too much to drink over the Holidays. Thanks BM RE: Solver Problem 'Maybe' - roadrunner - 01-05-2017 02:36 PM
Solving for X works on my prime. What version softwear are you running? The solver app in one of the earlier versions had issues. -road RE: Solver Problem 'Maybe' - BERNARD MICHAUD - 01-05-2017 05:11 PM
(01-05-2017 02:36 PM)roadrunner Wrote: Solving for X works on my prime. What version softwear are you running? The solver app in one of the earlier versions had issues. The software ver. is 10637. Hardware ver. C. on Does not work on the Mac VC either. with ver. 11226. X Returns 3.599E-5 Same problem on the Windows VC. X returns 3.599E-5. ver. 11226 Thanks for your reply. BM RE: Solver Problem 'Maybe' - Nigel (UK) - 01-05-2017 06:27 PM
Put in a guess for X closer to 4 and it should work. The equation has 2 roots, X=0 and X=4; which one you get depends on the initial value of X. Nigel (UK) RE: Solver Problem 'Maybe' - BERNARD MICHAUD - 01-05-2017 07:36 PM
(01-05-2017 06:27 PM)Nigel (UK) Wrote: Put in a guess for X closer to 4 and it should work. The equation has 2 roots, X=0 and X=4; which one you get depends on the initial value of X. Thanks Nigel Should have thought of that, I'll blame that one on 'Lipitor", Read the book 'Lipitor the thief of Memory' BM |