Question Number 192855 by pascal889 last updated on 29/May/23
Answered by a.lgnaoui last updated on 29/May/23
![{ ((P^2 −2aP−10P+2a^2 +6a −6=0 (1))),((P^2 −27P +27a−27=0 (2) )) :} (1)−(2)⇒ 2a^2 −21a+21+P(17−2a)=0 ⇒ P=((2a^2 −21a+21)/(2a−17)) (3) ⇒(2)((2(a^2 −((21)/2)a+((21)/2)))/(2a−17))=(((a−((21)/4))^2 −(((63)/(16))))/(a−((17)/2))) P=(((a−((21)/4))^2 −((63)/(16)))/(a−((17)/2))) P^2 =(((a−((21)/4))^4 +((63)/(16^2 ))−((63)/8)(a−((21)/4)))/((a−((17)/2))^2 )) (2) P^2 =27(P−a+1) =27((((a−((21)/4))^2 −((63)/(16)))/(a−((17)/2)))−a+1) =27[(((4a−21)^2 −63)/(8(2a−17)))−(((16a−136)(a+1))/(8(2a−17)))] =27[((21^2 −168a−16a +136a +136−63)/(8(2a−17))] P^2 =27×((38−6a)/(2a−17)) (l)⇔P^2 −2(a+5)P+2(a^2 +3a−3)=0 ((27(38−6a))/(2a−17))−2(a+5)(((4a−21)^2 −63)/(4(2a−17)))+2(a^2 +3a−3)=0 4×27(38−6a)−2(a+5)[4a−21)^2 −63] +8(a^2 +3a−3)=0 998−162a−(2a+10)(16a^2 −168a−163) +8(a^2 +3a−3)(2a−17)=0 (998+1630−408)−16a^3 − (336−160−136−48)a^2 (+326−162 +1680−408−48)a 4a^3 +2a^2 −347a+555=0 ⇒a={−10,226633 ; 1,669079 ;8,097282} ⇒P=f(a)=((2a^2 −21a+21)/(2a−17))](https://www.tinkutara.com/question/Q192860.png)
$$\begin{cases}{\mathrm{P}^{\mathrm{2}} −\mathrm{2aP}−\mathrm{10P}+\mathrm{2a}^{\mathrm{2}} +\mathrm{6a}\:\:−\mathrm{6}=\mathrm{0}\:\:\left(\mathrm{1}\right)}\\{\mathrm{P}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{27P}\:\:\:\:\:\:\:\:\:+\mathrm{27a}−\mathrm{27}=\mathrm{0}\:\:\left(\mathrm{2}\right)\:}\end{cases} \\ $$$$\left(\mathrm{1}\right)−\left(\mathrm{2}\right)\Rightarrow\:\mathrm{2a}^{\mathrm{2}} −\mathrm{21a}+\mathrm{21}+\mathrm{P}\left(\mathrm{17}−\mathrm{2a}\right)=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{\mathrm{P}}=\frac{\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{21}\boldsymbol{\mathrm{a}}+\mathrm{21}}{\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}}\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow\left(\mathrm{2}\right)\frac{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} −\frac{\mathrm{21}}{\mathrm{2}}\mathrm{a}+\frac{\mathrm{21}}{\mathrm{2}}\right)}{\mathrm{2a}−\mathrm{17}}=\frac{\left(\mathrm{a}−\frac{\mathrm{21}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{63}}{\mathrm{16}}\right)}{\mathrm{a}−\frac{\mathrm{17}}{\mathrm{2}}} \\ $$$$\boldsymbol{\mathrm{P}}=\frac{\left(\boldsymbol{\mathrm{a}}−\frac{\mathrm{21}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{63}}{\mathrm{16}}}{\boldsymbol{\mathrm{a}}−\frac{\mathrm{17}}{\mathrm{2}}} \\ $$$$\boldsymbol{\mathrm{P}}^{\mathrm{2}} =\frac{\left(\boldsymbol{\mathrm{a}}−\frac{\mathrm{21}}{\mathrm{4}}\right)^{\mathrm{4}} +\frac{\mathrm{63}}{\mathrm{16}^{\mathrm{2}} }−\frac{\mathrm{63}}{\mathrm{8}}\left(\boldsymbol{\mathrm{a}}−\frac{\mathrm{21}}{\mathrm{4}}\right)}{\left(\boldsymbol{\mathrm{a}}−\frac{\mathrm{17}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:\boldsymbol{\mathrm{P}}^{\mathrm{2}} =\mathrm{27}\left(\boldsymbol{\mathrm{P}}−\boldsymbol{\mathrm{a}}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{27}\left(\frac{\left(\boldsymbol{\mathrm{a}}−\frac{\mathrm{21}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{63}}{\mathrm{16}}}{\boldsymbol{\mathrm{a}}−\frac{\mathrm{17}}{\mathrm{2}}}−\boldsymbol{\mathrm{a}}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{27}\left[\frac{\left(\mathrm{4}\boldsymbol{\mathrm{a}}−\mathrm{21}\right)^{\mathrm{2}} −\mathrm{63}}{\mathrm{8}\left(\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}\right)}−\frac{\left(\mathrm{16}\boldsymbol{\mathrm{a}}−\mathrm{136}\right)\left(\boldsymbol{\mathrm{a}}+\mathrm{1}\right)}{\mathrm{8}\left(\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}\right)}\right] \\ $$$$=\mathrm{27}\left[\frac{\mathrm{21}^{\mathrm{2}} −\mathrm{168}\boldsymbol{\mathrm{a}}−\mathrm{16}\boldsymbol{\mathrm{a}}\:+\mathrm{136}\boldsymbol{\mathrm{a}}\:\:+\mathrm{136}−\mathrm{63}}{\mathrm{8}\left(\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}\right.}\right] \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{P}}^{\mathrm{2}} =\mathrm{27}×\frac{\mathrm{38}−\mathrm{6}\boldsymbol{\mathrm{a}}}{\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}} \\ $$$$\:\:\: \\ $$$$ \\ $$$$\left(\mathrm{l}\right)\Leftrightarrow\mathrm{P}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{a}+\mathrm{5}\right)\mathrm{P}+\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{3a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{27}\left(\mathrm{38}−\mathrm{6a}\right)}{\mathrm{2a}−\mathrm{17}}−\mathrm{2}\left(\mathrm{a}+\mathrm{5}\right)\frac{\left(\mathrm{4a}−\mathrm{21}\right)^{\mathrm{2}} −\mathrm{63}}{\mathrm{4}\left(\mathrm{2a}−\mathrm{17}\right)}+\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{3a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$ \\ $$$$\left.\mathrm{4}×\mathrm{27}\left(\mathrm{38}−\mathrm{6a}\right)−\mathrm{2}\left(\mathrm{a}+\mathrm{5}\right)\left[\mathrm{4a}−\mathrm{21}\right)^{\mathrm{2}} −\mathrm{63}\right] \\ $$$$+\mathrm{8}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{3a}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{998}−\mathrm{162a}−\left(\mathrm{2a}+\mathrm{10}\right)\left(\mathrm{16a}^{\mathrm{2}} −\mathrm{168a}−\mathrm{163}\right) \\ $$$$+\mathrm{8}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{3a}−\mathrm{3}\right)\left(\mathrm{2a}−\mathrm{17}\right)=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{998}+\mathrm{1630}−\mathrm{408}\right)−\mathrm{16a}^{\mathrm{3}} − \\ $$$$\left(\mathrm{336}−\mathrm{160}−\mathrm{136}−\mathrm{48}\right)\mathrm{a}^{\mathrm{2}} \left(+\mathrm{326}−\mathrm{162}\right. \\ $$$$\left.\:+\mathrm{1680}−\mathrm{408}−\mathrm{48}\right)\mathrm{a} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{4}\boldsymbol{\mathrm{a}}^{\mathrm{3}} +\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{347}\boldsymbol{\mathrm{a}}+\mathrm{555}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{a}}=\left\{−\mathrm{10},\mathrm{226633}\:;\:\:\mathrm{1},\mathrm{669079}\:\:\:;\mathrm{8},\mathrm{097282}\right\} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{P}}=\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{a}}\right)=\frac{\mathrm{2}\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{21}\boldsymbol{\mathrm{a}}+\mathrm{21}}{\mathrm{2}\boldsymbol{\mathrm{a}}−\mathrm{17}} \\ $$$$ \\ $$
Answered by York12 last updated on 30/May/23
$${p}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ap}+\mathrm{6}{a}−\mathrm{10}{p}−\mathrm{6}=\mathrm{0}\:……\left({i}\right) \\ $$$${p}^{\mathrm{2}} −\mathrm{27}{p}+\mathrm{27}{a}−\mathrm{27}=\mathrm{0}\:…….\left({ii}\right) \\ $$$${from}\:\left({ii}\right) \\ $$$${p}^{\mathrm{2}} −\mathrm{27}\left({p}−{a}+\mathrm{1}\right)\:\rightarrow\left({I}\right) \\ $$$${then}\:\left({i}\right)\:{can}\:{be}\:{written}\:{as}\: \\ $$$${p}^{\mathrm{2}} \:−\mathrm{27}\left({p}−{a}+\mathrm{1}\right)\:+\mathrm{17}{p}\:−\mathrm{21}{a}\:+\mathrm{21}+\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ap} \\ $$$${from}\:\left({I}\right)\: \\ $$$$\mathrm{17}{p}\:−\mathrm{21}{a}\:+\mathrm{21}\:+\mathrm{2}{a}^{\mathrm{2}} \:−\mathrm{2}{ap}\:=\mathrm{0} \\ $$$$\mathrm{2}{ap}\:−\mathrm{17}{p}\:−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{21}{a}−\mathrm{21}=\mathrm{0} \\ $$$${p}\left(\mathrm{2}{a}−\mathrm{17}\right)=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{21}{a}+\mathrm{21} \\ $$$${p}=\frac{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{21}{a}+\mathrm{21}}{\mathrm{2}{a}−\mathrm{17}} \\ $$$$ \\ $$