Question Number 68422 by ajfour last updated on 10/Sep/19
Answered by ajfour last updated on 10/Sep/19
![C=10m^3 p^2 +b(m^3 +6m^2 p+3mp^2 ) +c(3m^2 +6mp+p^2 )+d(6m+4p) +10e D=10m^2 p^3 +b(3m^2 p+6mp^2 +p^3 ) +c(m^2 +6mp+3p^2 )+d(4m+6p) +10e C−D=0 gives_(−) 10m^2 p^2 (m−p) +b[(m^3 −p^3 )+6mp(m−p)−3mp(m−p)] +c[3(m^2 −p^2 )−(m^2 −p^2 )] +d[6(m−p)−4(m−p)]+0 = 0 If m would be p then z=m=p otherwise dividing by m−p gives 10m^2 p^2 +b(m^2 +4mp+p^2 ) +c[2(m+p)]+2d=0 .....(I) (if m≠−p even) eq.(i)×(m+p) gives 10m^2 p^2 (m+p) +b[m^3 +4m^2 p+mp^2 +m^2 p+4mp^2 +p^3 ]+c[2m^2 +4mp+2p^2 ] +d(2m+2p)=0 ......(i) Now C+D=0 gives_(−) 10m^2 p^2 (m+p)+ b[m^3 +p^3 +6mp(m+p)+3mp(m+p)] +c(4m^2 +4p^2 +12mp) +d(10m+10p)+20e =0 ....(II) (II)−(i) gives_(−) b[4mp(m+p)]+ c[2m^2 +8mp+2p^2 ]+ d[8(m+p)]+20e=0 ⇒ 2bmp(m+p)+c[(m+p)^2 +2mp] +4d(m+p)+10e=0 ....(III) Now lets transform eq. (I) &(III) let m+p=s , and mp=g Eq. (I)_(−) 10g^2 +b(s^2 +2g)+2cs+2d=0 Eq.(III)_(−) 2bgs+c(s^2 +2g)+4ds+10e=0 ⇒ g=−((cs^2 +4ds+10e)/(2(bs+c))) substituting for g in eq.(I)_(−) 10(cs^2 +4ds+10e)^2 − 4b(cs^2 +4ds+10e)(bs+c) +4(bs^2 +2cs+2d)(bs+c)^2 =0 ⇒ 5(cs^2 +4ds+10e)^2 −2b(bs+c)(cs^2 +4ds+10e) +2(bs+c)^2 (bs^2 +2cs+2d)=0 This is a degree 4 eq. in s. So it seems i shall obtain s and g=−((cs^2 +4ds+10e)/(2(bs+c))) . m,p = (s/2)±(√((s^2 /4)−g)) Now we then have At^5 +Bt^4 +Et+F=0 or t^5 +((B/A))t^4 +((E/A))t+(F/A)=0 but if m=p clearly z=m hence all coefficients get zero and we need to change to z=((mt+p)/(t−1)) ; this case i shall take up in another post..](Q68429.png)
$${C}=\mathrm{10}{m}^{\mathrm{3}} {p}^{\mathrm{2}} +{b}\left({m}^{\mathrm{3}} +\mathrm{6}{m}^{\mathrm{2}} {p}+\mathrm{3}{mp}^{\mathrm{2}} \right) \\ $$$$\:+{c}\left(\mathrm{3}{m}^{\mathrm{2}} +\mathrm{6}{mp}+{p}^{\mathrm{2}} \right)+{d}\left(\mathrm{6}{m}+\mathrm{4}{p}\right) \\ $$$$\:+\mathrm{10}{e} \\ $$$${D}=\mathrm{10}{m}^{\mathrm{2}} {p}^{\mathrm{3}} +{b}\left(\mathrm{3}{m}^{\mathrm{2}} {p}+\mathrm{6}{mp}^{\mathrm{2}} +{p}^{\mathrm{3}} \right) \\ $$$$\:+{c}\left({m}^{\mathrm{2}} +\mathrm{6}{mp}+\mathrm{3}{p}^{\mathrm{2}} \right)+{d}\left(\mathrm{4}{m}+\mathrm{6}{p}\right) \\ $$$$\:+\mathrm{10}{e} \\ $$$$\underset{−} {{C}−{D}=\mathrm{0}\:{gives}} \\ $$$$\mathrm{10}{m}^{\mathrm{2}} {p}^{\mathrm{2}} \left({m}−{p}\right) \\ $$$$+{b}\left[\left({m}^{\mathrm{3}} −{p}^{\mathrm{3}} \right)+\mathrm{6}{mp}\left({m}−{p}\right)−\mathrm{3}{mp}\left({m}−{p}\right)\right] \\ $$$$+{c}\left[\mathrm{3}\left({m}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)−\left({m}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)\right] \\ $$$$+{d}\left[\mathrm{6}\left({m}−{p}\right)−\mathrm{4}\left({m}−{p}\right)\right]+\mathrm{0}\:=\:\mathrm{0} \\ $$$${If}\:{m}\:{would}\:{be}\:{p}\:{then}\:{z}={m}={p} \\ $$$${otherwise}\:{dividing}\:{by}\:{m}−{p}\:{gives} \\ $$$$ \\ $$$$\mathrm{10}{m}^{\mathrm{2}} {p}^{\mathrm{2}} +{b}\left({m}^{\mathrm{2}} +\mathrm{4}{mp}+{p}^{\mathrm{2}} \right) \\ $$$$+{c}\left[\mathrm{2}\left({m}+{p}\right)\right]+\mathrm{2}{d}=\mathrm{0}\:\:\:\:.....\left({I}\right) \\ $$$$\left({if}\:{m}\neq−{p}\:\:\:{even}\right) \\ $$$${eq}.\left({i}\right)×\left({m}+{p}\right)\:{gives} \\ $$$$\mathrm{10}{m}^{\mathrm{2}} {p}^{\mathrm{2}} \left({m}+{p}\right) \\ $$$$+{b}\left[{m}^{\mathrm{3}} +\mathrm{4}{m}^{\mathrm{2}} {p}+{mp}^{\mathrm{2}} +{m}^{\mathrm{2}} {p}+\mathrm{4}{mp}^{\mathrm{2}} \right. \\ $$$$\left.\:\:\:+{p}^{\mathrm{3}} \right]+{c}\left[\mathrm{2}{m}^{\mathrm{2}} +\mathrm{4}{mp}+\mathrm{2}{p}^{\mathrm{2}} \right] \\ $$$$\:\:\:+{d}\left(\mathrm{2}{m}+\mathrm{2}{p}\right)=\mathrm{0}\:\:\:\:\:\:\:......\left({i}\right) \\ $$$$\underset{−} {{Now}\:{C}+{D}=\mathrm{0}\:{gives}} \\ $$$$\mathrm{10}{m}^{\mathrm{2}} {p}^{\mathrm{2}} \left({m}+{p}\right)+ \\ $$$${b}\left[{m}^{\mathrm{3}} +{p}^{\mathrm{3}} +\mathrm{6}{mp}\left({m}+{p}\right)+\mathrm{3}{mp}\left({m}+{p}\right)\right] \\ $$$$+{c}\left(\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{2}} +\mathrm{12}{mp}\right) \\ $$$$+{d}\left(\mathrm{10}{m}+\mathrm{10}{p}\right)+\mathrm{20}{e}\:=\mathrm{0}\:\:\:....\left({II}\right) \\ $$$$ \\ $$$$\underset{−} {\left({II}\right)−\left({i}\right)\:{gives}} \\ $$$$\:\:\boldsymbol{{b}}\left[\mathrm{4}\boldsymbol{{mp}}\left(\boldsymbol{{m}}+\boldsymbol{{p}}\right)\right]+ \\ $$$$\:\boldsymbol{{c}}\left[\mathrm{2}\boldsymbol{{m}}^{\mathrm{2}} +\mathrm{8}\boldsymbol{{mp}}+\mathrm{2}\boldsymbol{{p}}^{\mathrm{2}} \right]+ \\ $$$$\:\boldsymbol{{d}}\left[\mathrm{8}\left(\boldsymbol{{m}}+\boldsymbol{{p}}\right)\right]+\mathrm{20}\boldsymbol{{e}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\mathrm{2}{bmp}\left({m}+{p}\right)+{c}\left[\left({m}+{p}\right)^{\mathrm{2}} +\mathrm{2}{mp}\right] \\ $$$$+\mathrm{4}{d}\left({m}+{p}\right)+\mathrm{10}{e}=\mathrm{0}\:\:\:\:....\left({III}\right) \\ $$$$ \\ $$$${Now}\:{lets}\:{transform}\:{eq}.\:\left({I}\right)\:\&\left({III}\right) \\ $$$${let}\:\:\boldsymbol{{m}}+\boldsymbol{{p}}=\boldsymbol{{s}}\:,\:{and}\:\boldsymbol{{mp}}=\boldsymbol{{g}} \\ $$$$\underset{−} {\boldsymbol{{Eq}}.\:\left(\boldsymbol{{I}}\right)} \\ $$$$\:\mathrm{10}\boldsymbol{{g}}^{\mathrm{2}} +\boldsymbol{{b}}\left(\boldsymbol{{s}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{g}}\right)+\mathrm{2}\boldsymbol{{cs}}+\mathrm{2}\boldsymbol{{d}}=\mathrm{0} \\ $$$$ \\ $$$$\underset{−} {\boldsymbol{{Eq}}.\left(\boldsymbol{{III}}\right)} \\ $$$$\:\:\mathrm{2}\boldsymbol{{bgs}}+\boldsymbol{{c}}\left(\boldsymbol{{s}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{g}}\right)+\mathrm{4}\boldsymbol{{ds}}+\mathrm{10}\boldsymbol{{e}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{{g}}=−\frac{\boldsymbol{{cs}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{ds}}+\mathrm{10}\boldsymbol{{e}}}{\mathrm{2}\left(\boldsymbol{{bs}}+\boldsymbol{{c}}\right)} \\ $$$$\underset{−} {{substituting}\:{for}\:\boldsymbol{{g}}\:{in}\:{eq}.\left({I}\right)} \\ $$$$\mathrm{10}\left({cs}^{\mathrm{2}} +\mathrm{4}{ds}+\mathrm{10}{e}\right)^{\mathrm{2}} − \\ $$$$\mathrm{4}{b}\left({cs}^{\mathrm{2}} +\mathrm{4}{ds}+\mathrm{10}{e}\right)\left({bs}+{c}\right) \\ $$$$+\mathrm{4}\left({bs}^{\mathrm{2}} +\mathrm{2}{cs}+\mathrm{2}{d}\right)\left({bs}+{c}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{5}\left(\boldsymbol{{cs}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{ds}}+\mathrm{10}\boldsymbol{{e}}\right)^{\mathrm{2}} \\ $$$$\:\:\:−\mathrm{2}\boldsymbol{{b}}\left(\boldsymbol{{bs}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{cs}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{ds}}+\mathrm{10}\boldsymbol{{e}}\right) \\ $$$$\:\:\:+\mathrm{2}\left(\boldsymbol{{bs}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \left(\boldsymbol{{bs}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{cs}}+\mathrm{2}\boldsymbol{{d}}\right)=\mathrm{0} \\ $$$${This}\:{is}\:{a}\:{degree}\:\mathrm{4}\:{eq}.\:{in}\:\boldsymbol{{s}}. \\ $$$${So}\:{it}\:{seems}\:{i}\:{shall}\:{obtain}\:\boldsymbol{{s}} \\ $$$${and}\:\boldsymbol{{g}}=−\frac{\boldsymbol{{cs}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{ds}}+\mathrm{10}\boldsymbol{{e}}}{\mathrm{2}\left(\boldsymbol{{bs}}+\boldsymbol{{c}}\right)}\:. \\ $$$$\boldsymbol{{m}},\boldsymbol{{p}}\:=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}\pm\sqrt{\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{g}}} \\ $$$${Now}\:{we}\:{then}\:{have} \\ $$$$\:\:\:\boldsymbol{{At}}^{\mathrm{5}} +\boldsymbol{{Bt}}^{\mathrm{4}} +\boldsymbol{{Et}}+\boldsymbol{{F}}=\mathrm{0} \\ $$$${or}\:\:{t}^{\mathrm{5}} +\left(\frac{{B}}{{A}}\right){t}^{\mathrm{4}} +\left(\frac{{E}}{{A}}\right){t}+\frac{{F}}{{A}}=\mathrm{0} \\ $$$${but}\:{if}\:{m}={p}\:\:{clearly}\:{z}={m} \\ $$$${hence}\:{all}\:{coefficients}\:{get}\:{zero} \\ $$$${and}\:{we}\:{need}\:\:{to}\:{change}\: \\ $$$${to}\:{z}=\frac{{mt}+{p}}{{t}−\mathrm{1}}\:;\:{this}\:{case}\:{i}\:{shall}\:{take} \\ $$$${up}\:{in}\:{another}\:{post}.. \\ $$$$ \\ $$
Commented by TawaTawa last updated on 10/Sep/19
$$\mathrm{wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{and}\:\mathrm{increase}\:\mathrm{your}\:\mathrm{knowledge}.\:\mathrm{I}\:\mathrm{will}\:\mathrm{learn} \\ $$$$\mathrm{this}\:\mathrm{with}\:\mathrm{an}\:\mathrm{example}. \\ $$