Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 79015 by ajfour last updated on 22/Jan/20

Rigorously over one month′s  time, I developed a formula for  general cubic.  x^3 +ax^2 +bx+c=0  let  x=((pt+q)/(t+1))  pq=m, p+q=s  ________________________  m^2 {(a^2 +b)^2 −6a(ab−c)}  +m{2(b^2 +ac)(a^2 +b)−              3(ab−c)(ab+3c)}     +(b^2 +ac)^2 −6bc(ab−c)=0  ________________________  s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))}    +{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3      −8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3)   p,q = (s/2)±(√((s^2 /4)−m))  t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c)))  x=((pt+q)/(t+1)) .  (Please help checking..)  (edited a digit 1 in place of 4)

$${Rigorously}\:{over}\:{one}\:{month}'{s} \\ $$$${time},\:{I}\:{developed}\:{a}\:{formula}\:{for} \\ $$$${general}\:{cubic}. \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$$\boldsymbol{{pq}}=\boldsymbol{{m}},\:\boldsymbol{{p}}+\boldsymbol{{q}}=\boldsymbol{{s}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{m}}^{\mathrm{2}} \left\{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{a}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\right\} \\ $$$$+\boldsymbol{{m}}\left\{\mathrm{2}\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)−\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{ab}}+\mathrm{3}\boldsymbol{{c}}\right)\right\} \\ $$$$\:\:\:+\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{bc}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{s}}=−\frac{\mathrm{2}}{\mathrm{3}}\left\{\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right\} \\ $$$$\:\:+\left\{\frac{\mathrm{8}}{\mathrm{27}}\left[\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]^{\mathrm{3}} \right. \\ $$$$\left.\:\:\:−\mathrm{8}\left[\frac{\boldsymbol{{m}}^{\mathrm{3}} +\boldsymbol{{bm}}^{\mathrm{2}} +\boldsymbol{{acm}}+\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\boldsymbol{{p}},\boldsymbol{{q}}\:=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}\pm\sqrt{\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{m}}} \\ $$$$\boldsymbol{{t}}=−\frac{\left(\mathrm{3}\boldsymbol{{pq}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{apq}}+\boldsymbol{{ap}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{bp}}+\boldsymbol{{bq}}+\mathrm{3}\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}^{\mathrm{3}} +\boldsymbol{{ap}}^{\mathrm{2}} +\boldsymbol{{bp}}+\boldsymbol{{c}}\right)} \\ $$$$\boldsymbol{{x}}=\frac{\boldsymbol{{pt}}+\boldsymbol{{q}}}{\boldsymbol{{t}}+\mathrm{1}}\:. \\ $$$$\left({Please}\:{help}\:{checking}..\right) \\ $$$$\left({edited}\:{a}\:{digit}\:\mathrm{1}\:{in}\:{place}\:{of}\:\mathrm{4}\right) \\ $$

Commented by mr W last updated on 22/Jan/20

great sir! you did it!  just one question before further  studying:  we get generally two values for m and  s, for each pair of m and s we get  a pair of p and q and one value of t,  and finally a root x. totally we get  two roots. but generally we have three  roots. what did i understand wrongly?

$${great}\:{sir}!\:{you}\:{did}\:{it}! \\ $$$${just}\:{one}\:{question}\:{before}\:{further} \\ $$$${studying}: \\ $$$${we}\:{get}\:{generally}\:{two}\:{values}\:{for}\:{m}\:{and} \\ $$$${s},\:{for}\:{each}\:{pair}\:{of}\:{m}\:{and}\:{s}\:{we}\:{get} \\ $$$${a}\:{pair}\:{of}\:{p}\:{and}\:{q}\:{and}\:{one}\:{value}\:{of}\:{t}, \\ $$$${and}\:{finally}\:{a}\:{root}\:{x}.\:{totally}\:{we}\:{get} \\ $$$${two}\:{roots}.\:{but}\:{generally}\:{we}\:{have}\:{three} \\ $$$${roots}.\:{what}\:{did}\:{i}\:{understand}\:{wrongly}? \\ $$

Commented by MJS last updated on 22/Jan/20

doesn′t work for  x^3 +3x^2 −13x−15=0  with x_1 =−5, x_2 =−1, x_3 =3  your method gives depending on the choice  of p and q [t is not symmetric in p, q; so we  have to decide p=(s/2)+... or p=(s/2)−...]  x≈.172±.763 or x≈−1.43±2.53  btw. in this case m∉R

$$\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{for} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{with}\:{x}_{\mathrm{1}} =−\mathrm{5},\:{x}_{\mathrm{2}} =−\mathrm{1},\:{x}_{\mathrm{3}} =\mathrm{3} \\ $$$$\mathrm{your}\:\mathrm{method}\:\mathrm{gives}\:\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{choice} \\ $$$$\mathrm{of}\:{p}\:\mathrm{and}\:{q}\:\left[{t}\:\mathrm{is}\:\mathrm{not}\:\mathrm{symmetric}\:\mathrm{in}\:{p},\:{q};\:\mathrm{so}\:\mathrm{we}\right. \\ $$$$\left.\mathrm{have}\:\mathrm{to}\:\mathrm{decide}\:{p}=\frac{{s}}{\mathrm{2}}+...\:\mathrm{or}\:{p}=\frac{{s}}{\mathrm{2}}−...\right] \\ $$$${x}\approx.\mathrm{172}\pm.\mathrm{763}\:\mathrm{or}\:{x}\approx−\mathrm{1}.\mathrm{43}\pm\mathrm{2}.\mathrm{53} \\ $$$$\mathrm{btw}.\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{m}\notin\mathbb{R} \\ $$

Commented by ajfour last updated on 22/Jan/20

its not at all good! but just now  i had been modifying the  Cardano formula, first let me  confirm and i shall post then..

$${its}\:{not}\:{at}\:{all}\:{good}!\:{but}\:{just}\:{now} \\ $$$${i}\:{had}\:{been}\:{modifying}\:{the} \\ $$$${Cardano}\:{formula},\:{first}\:{let}\:{me} \\ $$$${confirm}\:{and}\:{i}\:{shall}\:{post}\:{then}.. \\ $$

Commented by TawaTawa last updated on 23/Jan/20

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com