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Math Forum / Mathematics / General Topics / March 2010Satisfying Cardons' impossible quadratic equation.
x^2 - 10x = - 40 ----------------------------------- x^2 - 10x + (10/2)+10 =15 + 40 = 55 When solving for -x and -x^2 in the third negative quadrant of the Cartisan coordinate system where both x and y are negative values you need to derive -x and -x^2 this way -- -x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = - 11.4498322128... -x^2 = ((((11.4498322128...*2)+2)^2) -2)/4 = - 154.49832212... Where the full negative values are -- 10x = - 114.498322128756698596659080230619755701172608310906236690287015059644675373511228185049178598263259828260696982420061133188481211584254257246449621593... and x^2 = - 154.498322128756698596659080230619755701172608310906236690287015059644675373511228185049178598263259828260696982420061133188481211584254257246449621593 I wonder what Cardon would say if he were alive? ;-) Where, BTW the real value (not imaginary) value of the sqrt(-1) is -- ((sqrt(6))-2)/2 Boy, am I going to get flack on this one! This was derived from a ploting algorithm I used years back to plot a curve in the third quadrant where finding -x and -y coordinates by using a similar method as above. Basically this algorithm mirrors x + y = x^2 in the first quadrant but the negative curve plots in the opposite direction in the third negative quadrant. Dan > x^2 - 10x = - 40 > ----------------------------------- [quoted text clipped - 31 lines] > > Dan Rewriting my first quadratic example to conform to the rest I am showing below --- x^2 - 10x = - 40 ----------------------------------- x^2 - 10x = 40 - ((10/2)^2) = 15+ 40 = 55 -x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = - 11.4498322128... -x^2 = ((((11.4498322128...*2)+2)^2) -2)/4 = - 154.49832212... - 11.449832212... * 10 = - 114.49832212.. Therefore - 154.49832212... - ( - 114.49832212..) = - 40 ------------------------------------------------------------------------------------------------------ Other quadratics that work using the same algorithm -- x^2 - 12x = - 60 -------------------------------- x^2 - 12x =( 60 - ((12/2)^2) = 24 + 60 = 84 -x = ((sqrt((84*4)+2))-2)/2 = 8.1923881554... + (12/2) = - 14.1923881554... -x^2 = ((((14.1923881554....*2)+2)^2)-2)/4 = - 230.3086578651... 12*14.1923881554... = - 170.3086578651... Therefore - 230.3086578651... - (- 170.3086578651...) = - 60 ------------------------------------------------------------------------------------------------------- x^2 - 14x = - 84 -------------------------------- x^2 - 14x = ( 84 - ((14/2)^2)) = 35 + 84 = 119 -x = ((sqrt((119*4)+2))-2)/2 = 9.9316055545...+ (14/2) = - 16.9316055545... -x^2 = ((((16.9316055545...*2)+2)^2)-2)/4 = - 321.0424777635 - 16.9316055545..*14 = - 237.0424777635.. Therefore - 321.0424777635 - (- 237.0424777635..) = - 84 ---------------------------------------------------------------------------------------------------------- x^2 - 8x = - 24 ---------------------------------- x^2 - 8x = (24 - ((8/2)^2)) = 8 + 24 = 32 -x = ((sqrt((32*4)+2))-2)/2 = - 4.7008771254.. + (8/2) = - 8.7008771254.. -x^2 = (((8.7008771254.. *2)+2)^2)-2)/4 = - 93.6070170039.. - 8.7008771254.. * 8 = - 69.6070170039 Therefore 93.6070170039.. - ( - 69.6070170039) = - 24 --------------------------------------------------------------------------- And so on --- > x^2 - 10x = - 40 > ----------------------------------- > x^2 - 10x + (10/2)+10 =15 + 40 = 55 No. The second equation does not follow from the first. > When solving for -x and -x^2 in the third negative quadrant of the > Cartisan coordinate system where both x and y are negative values > you need to derive -x and -x^2 this way -- What are you talking about? The solution to the (first) equation is x = (10 +- sqr(100 - 160))/2 = 5 +- 15i > -x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = - > 11.4498322128... [quoted text clipped - 23 lines] > > Dan > > x^2 - 10x = - 40 > > ----------------------------------- [quoted text clipped - 39 lines] > > - Show quoted text - William, I rewrote this in my second post to be consistant with the other quadratics in that second post. x^2 - 10x = - 40 ----------------------------------- x^2 - 10x = 40 - ((10/2)^2) = 15+ 40 = 55 -x = ((sqrt((55*4)+2)) -2)/2 = 6.4498322128... + (10/2) = - 11.4498322128... -x^2 = ((((11.4498322128...*2)+2)^2) -2)/4 = - 154.49832212... - 11.449832212... * 10 = - 114.49832212.. Therefore - 154.49832212... - ( - 114.49832212..) = - 40 ------------------------------------------------------- This is unconventional I know where many algorithms can do the same plot but in truth not in the negative (-x\-y) third quadrant domain duplicating a mirror of x + y = x^2 in the fist quadrant. This is just an exersize to show it can be done by using these special quadratics, if you want to call them that. Wheather (i) in the complex plain has a roll here I am not sure because this is Cartisan. Thanks for your input. Dan > > > x^2 - 10x = - 40 > > > ----------------------------------- [quoted text clipped - 8 lines] > ----------------------------------- > x^2 - 10x = 40 - ((10/2)^2) = 15+ 40 = 55 That is just digging yourself further in. Where before you had -40 = +40, you now have -40 = +15 = +55. > > > > x^2 - 10x = - 40 > > > > ----------------------------------- [quoted text clipped - 12 lines] > Where before you had -40 = +40, > you now have -40 = +15 = +55. Henry, The whole exersize would not be possible if changing signs were not included. So it is a manipulation of (+\-) to make it all possible. Thanks, Dan >>>>> x^2 - 10x = - 40 >>>>> ----------------------------------- [quoted text clipped - 18 lines] > signs were not included. So it is a manipulation of (+\-) > to make it all possible. Would you be kind enough to state the problem that you believe that you have solved? If it is a simple matter of finding roots of x^2-10x=-40 then your various strange manipulations have not provided you with such roots. >> > > > x^2 - 10x = - 40 >> > > > ----------------------------------- [quoted text clipped - 16 lines] >signs were not included. So it is a manipulation of (+\-) >to make it all possible. What does it tell you when, to make a proof work, you have to do something illegal? Let's see what happens if we only allow legal operations. Assume x is real. If x^2 - 10x = - 40, then 0 = x^2 - 10x + 40 = x^2 - 10x + 25 + 15 = (x-5)^2 + 15 That is, 15 greater than a non-negative number is 0. Since this cannot be, x^2 - 10x = - 40 has no real solutions. Rob Johnson <rob@trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font > In article <cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, > [quoted text clipped - 39 lines] > > - Show quoted text - Rob, Yes you are right it is illegal, but thinking outside of the box and plotting this curve starting in the 4th and then passing into the third quadrant so making it easy I start @ y = - 0.5 and x = 0 the mirror of x = 1 and y =0. The main ploting algorithm after -y <= - 0.5 -x = ((sqrt((- y * 4) -1)) -1)/2 change sign for -y too y before sqrt. then change sign for x too -x after result and changing y back to - y. -y say in the above case is -y = -1 then - y = - 1 - x = - 0.336025403... -x^2 = -y + -x Where its mirrored point in the first quadrant is --- x= 1.336025403 y= 0.5 x^2 = x+y These quadratics that I propose are directly related to any -x -y points that have the actual plot point for the quadratic below .-- -x = - 8.7008771254 -x^2 = - 93.6070170039 Then - y = - x^2 - (- x) = - 84.9061398785 giving both -x and -y coordinates for the quadratic below which reside in the third quadrant. Where -x*8 is -24 > - x^2 Taken from this quadratic below. x^2 - 8x = - 24 ---------------------------------- x^2 - 8x = (24 - ((8/2)^2)) = 8 + 24 = 32 -x = ((sqrt((32*4)+2))-2)/2 = - 4.7008771254.. + (8/2) = - 8.7008771254.. -x^2 = (((8.7008771254.. *2)+2)^2)-2)/4 = - 93.6070170039.. - 8.7008771254.. * 8 = - 69.6070170039 Therefore -93.6070170039.. - ( - 69.6070170039) = - 24 As illegal as it is, it still gives plotting points that are correct in the third quadrant an also illegal quadratics stemming from these plotting points. Dan >> In article<cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, >> [quoted text clipped - 42 lines] > > Yes you are right it is illegal, but thinking outside of the box Changing an equation to another equation and then solving it and saying that the solution is "correct" is not "thinking outside the box", it is lying. > and > plotting this curve starting in the 4th and then passing into the > third quadrant so making it easy I start @ y = - 0.5 and x = 0 > the mirror of x = 1 and y =0. Why are you plotting "this curve starting in the 4th and then passing into the third quadrant" when the equation you are talking about has no points in either of those quadrants? > The main ploting algorithm after -y<= - 0.5 > -x = ((sqrt((- y * 4) -1)) -1)/2 [quoted text clipped - 44 lines] > third quadrant an also illegal quadratics stemming from these plotting > points. What leads you to believe that these "plotting points" are "correct"? > >> In article<cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, > [quoted text clipped - 115 lines] > > - Show quoted text - Your first question -- Starting in the 4th quadrant is the data point x= 0.5 and y = -0.25 which is the very bottom point of x + y = x^2 parabola that covers the 1st 2nd and 4th quadrant. My plotting algorithm mirrors the right side of this parabola which send the plot from this point down and to the left and crossing the -y axis @ y = -0.5 and x = 0 Starting with -y = -0.25 and incrementing -y for each plotted point with -y = -y + -0.01 with the first value for -y = -0.25 then -- -x = ((sqrt((- y * 4) -1)) -1)/2 = -0.5 changing -y to y before the first -1 and sqrt. This shows x to be negative but it is realy 0.5 because the plot is to the right of the y axis. So the first plot point and following points are-- x= 0.5 and - y = -0.25 x= 0.4 and - y = -0.26 x= 0.358578.. - y = -0.27 x= 0.326794.. - y = -0.28 x= 0.3 ...... - y = -0.29 x= 0.276393..- y = -0.30 x= 0.255051..- y = -0.31 .. continue incrementing -y with + -0.01 until x= 0 and -y = -0.5 Which will mirror x = 1 and y = 0 in the first quadrant. From here on where - y < - 0.5 -x becomes x so that has to change to - x and from this point on incrementing - y + -0.01 -x = ((sqrt((- y * 4) -1)) -1)/2 will reverse mirror image every thing in the first quadrant. A very simple algorithm that is easy to impliment in a plotting application. Do a plot for x + y = x^2 starting @ x=0.5 and -y = -0.25 and compare both plots. Shift one of the plots by 180 degrees and the plots will look the same except for the orintation of the x\y axis between the two different plots. That is why these plot points are correct and not a lie. And again my quadratics are based upon these plot points but with 2 algebraic limits (n,r) where x^2 - x*n = -r. I will show these two with lower limit for (n) and upper limit for -r in a later post. Dan > > >> In article<cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, > [quoted text clipped - 163 lines] > > - Show quoted text - Strange enough as it is with these plot points in the third quadrant and special quadratics tied to these plot points there is also an identity for these two simple algerbraic numbers putting lower and upper limits respectively on (n) and (- r). Where (n) is the multiple of x and (-r) is the result --- x^2 - n*x = - r Showing also that (n) and (- r) do not need to be integers which is consistant with the --->oo number of points to plot and associated quadratics in the negative third quadrant domain. (n) will meet a lower limit where (n) as a multiple of (x) and cannot be < (n) value below. n = 2/(10^(1/2) - 4 = 2.38742588672279311... The lowest possible limit for (n) as a positive number multiple of (x) in the quadratic. -- x^2 -nx = - n_1 (-r) coupled with (n) above as a negative value and as the upper limit < 0 in the quadratic. -- x^2 - n*x = - r -r = 1/9 + (1/9 * 10^(1/2)) * -1 = - 0.46247529557426437022.. Gives the higest possible limit for (n_1) as a negative number in the quadraic. -- x^2 - nx = - r Therefore given Cardons quadratic as a limit which means no < (n) or > (- r) is possible in a third quadrant qaudratic is the proof below. x^2 - (( 2.38742588672279311...)*x) = - 0.46247529557426437022... ------------------------------------------------------------ x^2 - 2.38742588672279311..x = ( 0.46247529557426437022..) - (2.38742588672279311../2)^2 = - 0.9624752955742643 + 0.4624752955742643= - 0.5 -x = ((sqrt((-0.5*4)+2))-2)/2 = - 1 + (2.38742588672279311..)/2 = 0.193712943361396555 -x^2 = (((0.193712943361396555... * 2)+2)^2)-2)/4 = 0.9249505911485287 In the case above -0.5 is left negative. -0.193712943361396555 * - 2.38742588672279311 = - 0.4624752955742643 therefore - 0.9249505911485287 - (- 0.4624752955742643) = - 0.4624752955742643 Thus splitting the difference of -x^2/2 and resulting in the limit (- r) The ratio between these two algabraic numbers is the continued fraction of -- 2.3874258867227931 / 0.4624752955742643 = [5,6,6,6,6,6,6,6,6,6,6...] Also -x^2 / - x = ( 2 * 2.3874258867227931...)= low limit for (n)*2 I know, all illegal stuff but interesting to show these two (n) and (- r) limits are algabraic numbers. Has anyone run a plot on -- -x = ((sqrt((- y * 4) -1)) -1)/2 ? Dan >>>>> In article<cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, >> [quoted text clipped - 223 lines] > > -x = ((sqrt((- y * 4) -1)) -1)/2 ? What does ANY of the CRAP have to do with Cardan? >>>> In article<cab9cc13-d98b-462f-8889-ae50c0924...@o3g2000yqb.googlegroups.com>, >> [quoted text clipped - 116 lines] > y = -0.25 which is the very bottom point of x + y = x^2 > parabola that covers the 1st 2nd and 4th quadrant. The original equation was x^2 + 10x = -40. .5^2+10*.5 = .25+5 = 5.25 # -40, so what relevance do you believe this to have to the original equation? > My plotting algorithm mirrors the right side of this > parabola which send the plot from this point down and to > the left and crossing the -y axis @ y = -0.5 and x = 0 What exactly are you trying to accomplish? Is it your intent to mirror y = x^2 +10x about the x axis? > Starting with -y = -0.25 and incrementing -y for each > plotted point with -y = -y + -0.01 with the first value [quoted text clipped - 14 lines] > x= 0 and -y = -0.5 > Which will mirror x = 1 and y = 0 in the first quadrant. 1^2 + 10*1 = 11 # -40, so again what is the relevance to the original equation? > From here on where - y< - 0.5 -x becomes x so that has > to change to - x and from this point on incrementing - y + -0.01 [quoted text clipped - 6 lines] > and the plots will look the same except for the orintation > of the x\y axis between the two different plots. Ok, so you can plot y = x^2 - x. Fine. So can any six year old with a copy of Microsoft Works. What does that have to do with Cardan's equation? > That is why these plot points are correct and not a lie. Then what do they have to do with Cardan's equation? > And again my quadratics are based upon these plot points > but with 2 algebraic limits (n,r) where x^2 - x*n = -r. I will > show these two with lower limit for (n) and upper limit for -r in > a later post. What about it? n is already given as exactly 1, and r as exactly 40. So what do your plot points have to do with anything? Explain, in a sentence containing fewer than 20 words, none of which have more than three syllables, what is is that you are attempting to accomplish. Right now all your comments seem like random meanderings. > Dan > > On Mar 11, 10:35 pm, "J. Clarke"<jclarke.use...@cox.net> wrote: > [quoted text clipped - 195 lines] > > - Show quoted text - Mr J, I thought that Cardons original quadratic -- X^2 - 10x = - 40 had something to do with completing the square?. If that is the case, my crazy quadratics are directly related to the graph of x+y = x^2 in the 1st quadrant but in a reverse mirror image in the third quadrant. So that I believe is the relationship. I could be wrong though and I am sure if I am you will correct me. Thanks for your input. Dan > x^2 - 10x = - 40 > ----------------------------------- > x^2 - 10x + (10/2)+10 =15 + 40 = 55 Wrong. And by the way "Cardon" [sic] solved this correctly 500 years ago. Andrew Usher |
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