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Question-138872




Question Number 138872 by peter frank last updated on 19/Apr/21
Answered by bramlexs22 last updated on 19/Apr/21
p(x)=x^3 +(A−7)x^2 +Ax−8  p(x)=0⇒x^3 +(A−7)x^2 +Ax−8=0  let  { ((x_1 =(a/r))),((x_2 =a)),((x_3 =ar)) :} by Vieta′s rule  ⇒x_1 .x_2 .x_3  = 8 ; a^3 =8 ; a=2  ⇒x_2 =2⇒p(2)=0  ⇒8+(A−7).4+2A−8=0  ⇒6A= 28 ; A=((14)/3)  p(x)=x^3 −(7/3)x^2 +((14)/3)x−8
$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{3}} +\left(\mathrm{A}−\mathrm{7}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{Ax}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{let}\:\begin{cases}{\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{a}}{\mathrm{r}}}\\{\mathrm{x}_{\mathrm{2}} =\mathrm{a}}\\{\mathrm{x}_{\mathrm{3}} =\mathrm{ar}}\end{cases}\:\mathrm{by}\:\mathrm{Vieta}'\mathrm{s}\:\mathrm{rule} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} .\mathrm{x}_{\mathrm{3}} \:=\:\mathrm{8}\:;\:\mathrm{a}^{\mathrm{3}} =\mathrm{8}\:;\:\mathrm{a}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{2}} =\mathrm{2}\Rightarrow\mathrm{p}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{8}+\left(\mathrm{A}−\mathrm{7}\right).\mathrm{4}+\mathrm{2A}−\mathrm{8}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{6A}=\:\mathrm{28}\:;\:\mathrm{A}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{14}}{\mathrm{3}}\mathrm{x}−\mathrm{8}\: \\ $$
Commented by peter frank last updated on 19/Apr/21
thank you
$${thank}\:{you} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Apr/21
AnOther Way      P(x)=x^3 +(A−7)x^2 +Ax−8...(i)  Let a , ar , ar^2  are roots.     P(x)=(x−a)(x−ar)(x−ar^2 )    =x^3 −a(r^2 +r+1)x^2 +a^2 (r^2 +r+1 )x−a^3 ...(ii)  Comparing (i) & (ii)  ^• −a(r^2 +r+1)=A−7...(iii)  ^• a^2 (r^2 +r+1)=A...(iv)  ^• a^3 =8⇒a=2...............(v)  (iv)/(iii): −a=(A/(A−7))=−2             A+2A=14              A=((14)/3)   (iii):  a^2 (r^2 +r+1)=A                   ⇒2^2 (r^2 +r+1)=((14)/3)                        6r^2 +6r−1=0          r=((−6±(√(36+24)))/(12))=((−3±(√(15)))/6)  P(x)=x^3 +(((14)/3)−7)x^2 +((14)/3)x−8   •P(x)=x^3 −(7/3)x^2 +((14)/3)x−8  •Roots are: a,ar,ar^2         2,2(((−3±(√(15)))/6)),2(((−3±(√(15)))/6))^2
$$\boldsymbol{{AnOther}}\:\boldsymbol{{Way}} \\ $$$$\:\:\:\:{P}\left({x}\right)={x}^{\mathrm{3}} +\left({A}−\mathrm{7}\right){x}^{\mathrm{2}} +{Ax}−\mathrm{8}…\left({i}\right) \\ $$$${Let}\:{a}\:,\:{ar}\:,\:{ar}^{\mathrm{2}} \:{are}\:{roots}. \\ $$$$\:\:\:{P}\left({x}\right)=\left({x}−{a}\right)\left({x}−{ar}\right)\left({x}−{ar}^{\mathrm{2}} \right) \\ $$$$\:\:={x}^{\mathrm{3}} −{a}\left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right){x}^{\mathrm{2}} +{a}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{r}+\mathrm{1}\:\right){x}−{a}^{\mathrm{3}} …\left({ii}\right) \\ $$$${Comparing}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:^{\bullet} −{a}\left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right)={A}−\mathrm{7}…\left({iii}\right) \\ $$$$\:^{\bullet} {a}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right)={A}…\left({iv}\right) \\ $$$$\:^{\bullet} {a}^{\mathrm{3}} =\mathrm{8}\Rightarrow{a}=\mathrm{2}……………\left({v}\right) \\ $$$$\left({iv}\right)/\left({iii}\right):\:−{a}=\frac{{A}}{{A}−\mathrm{7}}=−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{A}+\mathrm{2}{A}=\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{A}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\:\left({iii}\right):\:\:{a}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right)={A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}^{\mathrm{2}} \left({r}^{\mathrm{2}} +{r}+\mathrm{1}\right)=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6}{r}^{\mathrm{2}} +\mathrm{6}{r}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{r}=\frac{−\mathrm{6}\pm\sqrt{\mathrm{36}+\mathrm{24}}}{\mathrm{12}}=\frac{−\mathrm{3}\pm\sqrt{\mathrm{15}}}{\mathrm{6}} \\ $$$${P}\left({x}\right)={x}^{\mathrm{3}} +\left(\frac{\mathrm{14}}{\mathrm{3}}−\mathrm{7}\right){x}^{\mathrm{2}} +\frac{\mathrm{14}}{\mathrm{3}}{x}−\mathrm{8} \\ $$$$\:\bullet{P}\left({x}\right)={x}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{14}}{\mathrm{3}}{x}−\mathrm{8} \\ $$$$\bullet{Roots}\:{are}:\:{a},{ar},{ar}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\mathrm{2},\mathrm{2}\left(\frac{−\mathrm{3}\pm\sqrt{\mathrm{15}}}{\mathrm{6}}\right),\mathrm{2}\left(\frac{−\mathrm{3}\pm\sqrt{\mathrm{15}}}{\mathrm{6}}\right)^{\mathrm{2}} \\ $$
Commented by peter frank last updated on 12/May/21
thank you
$${thank}\:{you} \\ $$

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