Question Number 198968 by necx122 last updated on 26/Oct/23
$${Find}\:{the}\:{polynomial}\:{with}\:{roots}\:{that} \\ $$$${exceed}\:{the}\:{roots}\:{of}\: \\ $$$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{14}{x}^{\mathrm{2}} +{x}+\mathrm{62}=\mathrm{0}\:{by}\:\mathrm{3}.\:{Hence} \\ $$$${determine}\:{the}\:{value}\:{of}\:\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}}, \\ $$$${where}\:{a},{b}\:{and}\:{c}\:{are}\:{roots}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/23
$${Let}\:{a},{b},{c}\:{are}\:{roots}\:{of}\:\:\:\mathrm{3}{x}^{\mathrm{3}} −\mathrm{14}{x}^{\mathrm{2}} +{x}+\mathrm{62}=\mathrm{0} \\ $$$${a}+{b}+{c}=\frac{\mathrm{14}}{\mathrm{3}}\:,\:{ab}+{bc}+{ca}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{abc}=−\frac{\mathrm{62}}{\mathrm{3}} \\ $$$${The}\:{required}\:{equation}\:{have}\:{roots}\:{a}+\mathrm{3}, \\ $$$${b}+\mathrm{3},\:{c}+\mathrm{3} \\ $$$$\bullet\left({a}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)+\left({c}+\mathrm{3}\right)={a}+{b}+{c}+\mathrm{9}=\frac{\mathrm{14}}{\mathrm{3}}+\mathrm{9}=\frac{\mathrm{41}}{\mathrm{3}} \\ $$$$\bullet\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)+\left({c}+\mathrm{3}\right)\left({a}+\mathrm{3}\right) \\ $$$$\:\:\:\:={ab}+\mathrm{3}\left({a}+{b}\right)+\mathrm{9}+{bc}+\mathrm{3}\left({b}+{c}\right)+\mathrm{9}+{ca}+\mathrm{3}\left({c}+{a}\right)+\mathrm{9} \\ $$$$\:\:\:={ab}+{bc}+{ca}+\mathrm{6}\left({a}+{b}+{c}\right)+\mathrm{27} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{6}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)+\mathrm{27}=\frac{\mathrm{1}+\mathrm{84}+\mathrm{81}}{\mathrm{3}}=\frac{\mathrm{166}}{\mathrm{3}} \\ $$$$\bullet\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:={abc}+\mathrm{3}\left({ab}+{bc}+{ca}\right)+\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{27} \\ $$$$\:\:\:\:\:=−\frac{\mathrm{62}}{\mathrm{3}}+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{9}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)=\frac{−\mathrm{62}+\mathrm{3}+\mathrm{126}}{\mathrm{3}}=\frac{\mathrm{67}}{\mathrm{3}} \\ $$$${Hence}\:{the}\:{required}\:{equation}\:{will}\:{be}: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{41}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{166}}{\mathrm{3}}{x}+\frac{\mathrm{67}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{3}} +\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}+\mathrm{67}=\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}} \\ $$$$\:\:=\frac{\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)+\left({a}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)}{\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)} \\ $$$$\:\:=\frac{\:\frac{\mathrm{166}}{\mathrm{3}}}{\frac{\mathrm{67}}{\mathrm{3}}}=\frac{\mathrm{166}}{\mathrm{67}} \\ $$$$ \\ $$
Commented by necx122 last updated on 26/Oct/23
This is great and I'm grateful. well understood. Thank you sir.
Commented by mr W last updated on 26/Oct/23
$${i}\:{think}\:{the}\:{question}\:{suggests}\:{one}\:{to} \\ $$$${shift}\:{the}\:{given}\:{polynomial}\:{f}\left({x}\right)\:{to}\:{the} \\ $$$${right}\:{direction}\:{by}\:\mathrm{3}\:{to}\:{get}\:{the}\:{new} \\ $$$${polynomial}\:{at}\:{first}.\:{this}\:{is}\:{the}\:{easiest} \\ $$$${way}\:{to}\:{determine}\:{the}\:{new}\:{polynomial}. \\ $$$$ \\ $$$${please}\:{recheck}\:{your}\:{calculation}\:{sir}! \\ $$$${i}\:{think}\:{something}\:{is}\:{wrong}\:{in}\:{it}, \\ $$$${see}\:{following}\:{diagram}\:{with} \\ $$$$\left(\mathrm{1}\right)={original}\:{polynomial}\:{f}\left({x}\right) \\ $$$$\left(\mathrm{2}\right)={what}\:{you}\:{got} \\ $$$$\left(\mathrm{3}\right)={what}\:{i}\:{got}\:\left({see}\:{below}\right) \\ $$
Commented by mr W last updated on 26/Oct/23
Commented by AST last updated on 26/Oct/23
$${The}\:{correct}\:{equation}\:{should}\:{be}:\: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{67}=\mathrm{0} \\ $$
Commented by mr W last updated on 26/Oct/23
$${i}\:{think}\:{the}\:{correct}\:{equation}\:{is} \\ $$$${k}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{148}\right)=\mathrm{0}\:{with}\:{k}\in{R},\:{k}\neq\mathrm{0} \\ $$
Commented by mr W last updated on 26/Oct/23
Commented by AST last updated on 26/Oct/23
$${I}'{m}\:{referring}\:{to}\:{the}\:{solution}\:{above}. \\ $$$${The}\:{solution}\:{would}\:{have}\:{been}\:{correct}\:{if}\:{a},{b}\:{and} \\ $$$${c}\:{were}\:{all}\:{real}\:{numbers}.\: \\ $$$$\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right) \\ $$$$={abc}+\mathrm{3}\left({ab}+{bc}+{ca}\right)+\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{27}\:{is}\:{only}\:{true} \\ $$$${when}\:{a},{b},{c}\in\mathbb{R} \\ $$$${When}\:{two}\left({in}\:{the}\:{case}\:{of}\:{a}\:{cubic}\:{equation}\right)\:{come} \\ $$$${from}\:\mathbb{C}\backslash\mathbb{R},{then}\:{we}\:{are}\:{no}\:{longer}\:{dealing}\:{with} \\ $$$${variables}\left({a},{b},{c}\right)\:{but}\:\left({a},\overset{−} {{b}},\overset{−} {{c}}\right)\:{where}\:\overset{−} {{b}}={x}+{yi}. \\ $$$$\Rightarrow\overset{−} {{b}}+\mathrm{3}=\left({x}+\mathrm{3}\right)+{yi};{so}\:{we}\:{are}\:{only}\:{increasing} \\ $$$${the}\:{real}\:{part}\:{by}\:\mathrm{3} \\ $$
Commented by AST last updated on 27/Oct/23
![Let a=a,b^− =x+yi,c^− =z−yi a+b^− +c^− =((14)/3)⇒c^− =z−yi⇒a+b^− +c^− =a+x+z=((14)/3) ab^− +b^− c^− +c^− a=(1/3)⇒a(x+yi)+(x+yi)(z−yi) +a(z−yi)⇒ax+az+xz+y^2 +i(yz−xy)=(1/3) ⇒yz−xy=0⇒z=x⇒2ax+x^2 +y^2 =(1/3) ab^− c^− =((−62)/3)⇒a(x^2 +y^2 )=((−62)/3) a+b^− +c^− =((14)/3)⇒a+2x=((14)/3) ∗(a+3)+(b^− +3)+(c^− +3)=a+x+z+9=((41)/3) ∗(a+3)(b^− +3)(c^− +3)=(a+3)(x+3+yi)(x+3−yi) =(a+3)[(x+3)^2 −(yi)^2 ]=(a+3)[(x+3)^2 +y^2 ] =(a+3)(x^2 +y^2 +6x+9]=a(x^2 +y^2 )+3(2ax+x^2 +y^2 ) +9(a+2x)+27=−((62)/3)+3((1/3))+9(((14)/3))+((81)/3)=((148)/3) ∗(a+3)(b^− +3)+(b^− +3)(c^− +3)+(c^− +3)(a+3) =(a+3)(x+yi+3)+(x+3+yi)(x+3−yi) +(x−yi+3)(a+3) =(a+3)(2x+6)+(x+3)^2 +y^2 =2ax+6a+12x+27 +x^2 +y^2 =(2ax+x^2 +y^2 )+6(a+2x)+27 =(1/3)+6(((14)/3))+((81)/3)=((166)/3) x^3 −((41)/3)x^2 +((166)/3)x−((148)/3)=0 ⇒3x^3 −41x^2 +166x−148=0⇒(1/(a+3))+(1/(b^− +3))+(1/(c^− +3)) =((166)/(148))=((83)/(74))](https://www.tinkutara.com/question/Q199019.png)
$${Let}\:{a}={a},\overset{−} {{b}}={x}+{yi},\overset{−} {{c}}={z}−{yi} \\ $$$${a}+\overset{−} {{b}}+\overset{−} {{c}}=\frac{\mathrm{14}}{\mathrm{3}}\Rightarrow\overset{−} {{c}}={z}−{y}\boldsymbol{{i}}\Rightarrow{a}+\overset{−} {{b}}+\overset{−} {{c}}={a}+{x}+{z}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$${a}\overset{−} {{b}}+\overset{−} {{b}}\overset{−} {{c}}+\overset{−} {{c}a}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{a}\left({x}+{yi}\right)+\left({x}+{yi}\right)\left({z}−{yi}\right) \\ $$$$+{a}\left({z}−{yi}\right)\Rightarrow{ax}+{az}+{xz}+{y}^{\mathrm{2}} +{i}\left({yz}−{xy}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{yz}−{xy}=\mathrm{0}\Rightarrow{z}={x}\Rightarrow\mathrm{2}{ax}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${a}\overset{−} {{b}}\overset{−} {{c}}=\frac{−\mathrm{62}}{\mathrm{3}}\Rightarrow{a}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\frac{−\mathrm{62}}{\mathrm{3}} \\ $$$${a}+\overset{−} {{b}}+\overset{−} {{c}}=\frac{\mathrm{14}}{\mathrm{3}}\Rightarrow{a}+\mathrm{2}{x}=\frac{\mathrm{14}}{\mathrm{3}} \\ $$$$\ast\left({a}+\mathrm{3}\right)+\left(\overset{−} {{b}}+\mathrm{3}\right)+\left(\overset{−} {{c}}+\mathrm{3}\right)={a}+{x}+{z}+\mathrm{9}=\frac{\mathrm{41}}{\mathrm{3}} \\ $$$$\ast\left({a}+\mathrm{3}\right)\left(\overset{−} {{b}}+\mathrm{3}\right)\left(\overset{−} {{c}}+\mathrm{3}\right)=\left({a}+\mathrm{3}\right)\left({x}+\mathrm{3}+{yi}\right)\left({x}+\mathrm{3}−{yi}\right) \\ $$$$=\left({a}+\mathrm{3}\right)\left[\left({x}+\mathrm{3}\right)^{\mathrm{2}} −\left({yi}\right)^{\mathrm{2}} \right]=\left({a}+\mathrm{3}\right)\left[\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} \right] \\ $$$$=\left({a}+\mathrm{3}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}\right]={a}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{3}\left(\mathrm{2}{ax}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$+\mathrm{9}\left({a}+\mathrm{2}{x}\right)+\mathrm{27}=−\frac{\mathrm{62}}{\mathrm{3}}+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{9}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)+\frac{\mathrm{81}}{\mathrm{3}}=\frac{\mathrm{148}}{\mathrm{3}} \\ $$$$ \\ $$$$\ast\left({a}+\mathrm{3}\right)\left(\overset{−} {{b}}+\mathrm{3}\right)+\left(\overset{−} {{b}}+\mathrm{3}\right)\left(\overset{−} {{c}}+\mathrm{3}\right)+\left(\overset{−} {{c}}+\mathrm{3}\right)\left({a}+\mathrm{3}\right) \\ $$$$=\left({a}+\mathrm{3}\right)\left({x}+{yi}+\mathrm{3}\right)+\left({x}+\mathrm{3}+{yi}\right)\left({x}+\mathrm{3}−{yi}\right) \\ $$$$+\left({x}−{yi}+\mathrm{3}\right)\left({a}+\mathrm{3}\right) \\ $$$$=\left({a}+\mathrm{3}\right)\left(\mathrm{2}{x}+\mathrm{6}\right)+\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}{ax}+\mathrm{6}{a}+\mathrm{12}{x}+\mathrm{27} \\ $$$$+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{2}{ax}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{6}\left({a}+\mathrm{2}{x}\right)+\mathrm{27} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{6}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)+\frac{\mathrm{81}}{\mathrm{3}}=\frac{\mathrm{166}}{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{41}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{166}}{\mathrm{3}}{x}−\frac{\mathrm{148}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{148}=\mathrm{0}\Rightarrow\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{\overset{−} {{b}}+\mathrm{3}}+\frac{\mathrm{1}}{\overset{−} {{c}}+\mathrm{3}} \\ $$$$=\frac{\mathrm{166}}{\mathrm{148}}=\frac{\mathrm{83}}{\mathrm{74}} \\ $$
Answered by mr W last updated on 26/Oct/23
$${say}\:{the}\:{new}\:{polynomial}\:{is}\:{p}\left({x}\right). \\ $$$${roots}\:{of}\:{p}\left({x}\right)\:{exceed}\:{the}\:{roots}\:{of}\:{f}\left({x}\right) \\ $$$${by}\:\mathrm{3},\:{that}\:{means}\:{p}\left({x}\right)={f}\left({x}−\mathrm{3}\right). \\ $$$${p}\left({x}\right)=\mathrm{3}\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{14}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({x}−\mathrm{3}\right)+\mathrm{62} \\ $$$${p}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{148} \\ $$$${say}\:{roots}\:{of}\:{p}\left({x}\right)=\mathrm{0}\:{are}\:{A},{B},{C}, \\ $$$${then}\:{A}={a}+\mathrm{3},\:{B}={b}+\mathrm{3},\:{C}={c}+\mathrm{3} \\ $$$$\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}}=\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}} \\ $$$$=\frac{{AB}+{BC}+{CA}}{{ABC}}=\frac{\frac{\mathrm{166}}{\mathrm{3}}}{\frac{\mathrm{148}}{\mathrm{3}}}=\frac{\mathrm{166}}{\mathrm{148}}=\frac{\mathrm{83}}{\mathrm{74}}\:\checkmark \\ $$$$\blacksquare \\ $$
Commented by necx122 last updated on 26/Oct/23
Wow!! Having to exclaim is all i can do at this point. Funnily, both answers appear in the options.
Commented by Rasheed.Sindhi last updated on 27/Oct/23
$$\mathrm{Most}\:\mathrm{efficient}\:\mathrm{way}! \\ $$