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Question Number 198968 by necx122 last updated on 26/Oct/23

Find the polynomial with roots that exceed the roots of f(x)=3x^3 −14x^2 +x+62=0 by 3. Hence determine the value of (1/(a+3))+(1/(b+3))+(1/(c+3)), where a,b and c are roots.

$${Find}\:{the}\:{polynomial}\:{with}\:{roots}\:{that} \\ $$$${exceed}\:{the}\:{roots}\:{of}\: \\ $$$${f}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{14}{x}^{\mathrm{2}} +{x}+\mathrm{62}=\mathrm{0}\:{by}\:\mathrm{3}.\:{Hence} \\ $$$${determine}\:{the}\:{value}\:{of}\:\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}}, \\ $$$${where}\:{a},{b}\:{and}\:{c}\:{are}\:{roots}. \\ $$

Answered by Rasheed.Sindhi last updated on 26/Oct/23

Let a,b,c are roots of 3x^3 −14x^2 +x+62=0 a+b+c=((14)/3) , ab+bc+ca=(1/3) , abc=−((62)/3) The required equation have roots a+3, b+3, c+3 •(a+3)+(b+3)+(c+3)=a+b+c+9=((14)/3)+9=((41)/3) •(a+3)(b+3)+(b+3)(c+3)+(c+3)(a+3) =ab+3(a+b)+9+bc+3(b+c)+9+ca+3(c+a)+9 =ab+bc+ca+6(a+b+c)+27 =(1/3)+6(((14)/3))+27=((1+84+81)/3)=((166)/3) •(a+3)(b+3)(c+3) =abc+3(ab+bc+ca)+9(a+b+c)+27 =−((62)/3)+3((1/3))+9(((14)/3))=((−62+3+126)/3)=((67)/3) Hence the required equation will be: x^3 +((41)/3)x^2 +((166)/3)x+((67)/3)=0 ⇒3x^3 +41x^2 +166x+67=0 (1/(a+3))+(1/(b+3))+(1/(c+3)) =(((a+3)(b+3)+(b+3)(c+3)+(a+3)(c+3))/((a+3)(b+3)(c+3))) =(( ((166)/3))/((67)/3))=((166)/(67))

$${Let}\:{a},{b},{c}\:{are}\:{roots}\:{of}\:\:\:\mathrm{3}{x}^{\mathrm{3}} −\mathrm{14}{x}^{\mathrm{2}} +{x}+\mathrm{62}=\mathrm{0} \\ $$$${a}+{b}+{c}=\frac{\mathrm{14}}{\mathrm{3}}\:,\:{ab}+{bc}+{ca}=\frac{\mathrm{1}}{\mathrm{3}}\:,\:{abc}=−\frac{\mathrm{62}}{\mathrm{3}} \\ $$$${The}\:{required}\:{equation}\:{have}\:{roots}\:{a}+\mathrm{3}, \\ $$$${b}+\mathrm{3},\:{c}+\mathrm{3} \\ $$$$\bullet\left({a}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)+\left({c}+\mathrm{3}\right)={a}+{b}+{c}+\mathrm{9}=\frac{\mathrm{14}}{\mathrm{3}}+\mathrm{9}=\frac{\mathrm{41}}{\mathrm{3}} \\ $$$$\bullet\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)+\left({c}+\mathrm{3}\right)\left({a}+\mathrm{3}\right) \\ $$$$\:\:\:\:={ab}+\mathrm{3}\left({a}+{b}\right)+\mathrm{9}+{bc}+\mathrm{3}\left({b}+{c}\right)+\mathrm{9}+{ca}+\mathrm{3}\left({c}+{a}\right)+\mathrm{9} \\ $$$$\:\:\:={ab}+{bc}+{ca}+\mathrm{6}\left({a}+{b}+{c}\right)+\mathrm{27} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{6}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)+\mathrm{27}=\frac{\mathrm{1}+\mathrm{84}+\mathrm{81}}{\mathrm{3}}=\frac{\mathrm{166}}{\mathrm{3}} \\ $$$$\bullet\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:={abc}+\mathrm{3}\left({ab}+{bc}+{ca}\right)+\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{27} \\ $$$$\:\:\:\:\:=−\frac{\mathrm{62}}{\mathrm{3}}+\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{9}\left(\frac{\mathrm{14}}{\mathrm{3}}\right)=\frac{−\mathrm{62}+\mathrm{3}+\mathrm{126}}{\mathrm{3}}=\frac{\mathrm{67}}{\mathrm{3}} \\ $$$${Hence}\:{the}\:{required}\:{equation}\:{will}\:{be}: \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{41}}{\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{166}}{\mathrm{3}}{x}+\frac{\mathrm{67}}{\mathrm{3}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{3}} +\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}+\mathrm{67}=\mathrm{0} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}} \\ $$$$\:\:=\frac{\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)+\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)+\left({a}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)}{\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right)} \\ $$$$\:\:=\frac{\:\frac{\mathrm{166}}{\mathrm{3}}}{\frac{\mathrm{67}}{\mathrm{3}}}=\frac{\mathrm{166}}{\mathrm{67}} \\ $$$$ \\ $$

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(x) to the right direction by 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 (1)=original polynomial f(x) (2)=what you got (3)=what i got (see below)

$${i}\:{think}\:{the}\:{question}\:{suggests}\:{one}\:{to} \\ $$$${shift}\:{the}\:{given}\:{polynomial}\:{f}\left({x}\right)\:{to}\:{the} \\ $$$${right}\:{direction}\:{by}\:\mathrm{3}\:{to}\:{get}\:{the}\:{new} \\ $$$${polynomial}\:{at}\:{first}.\:{this}\:{is}\:{the}\:{easiest} \\ $$$${way}\:{to}\:{determine}\:{the}\:{new}\:{polynomial}. \\ $$$$ \\ $$$${please}\:{recheck}\:{your}\:{calculation}\:{sir}! \\ $$$${i}\:{think}\:{something}\:{is}\:{wrong}\:{in}\:{it}, \\ $$$${see}\:{following}\:{diagram}\:{with} \\ $$$$\left(\mathrm{1}\right)={original}\:{polynomial}\:{f}\left({x}\right) \\ $$$$\left(\mathrm{2}\right)={what}\:{you}\:{got} \\ $$$$\left(\mathrm{3}\right)={what}\:{i}\:{got}\:\left({see}\:{below}\right) \\ $$

Commented by mr W last updated on 26/Oct/23

Commented by deleteduser1 last updated on 26/Oct/23

The correct equation should be: 3x^2 −41x^2 +166x−67=0

$${The}\:{correct}\:{equation}\:{should}\:{be}:\: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{67}=\mathrm{0} \\ $$

Commented by mr W last updated on 26/Oct/23

i think the correct equation is k(3x^2 −41x^2 +166x−148)=0 with k∈R, k≠0

$${i}\:{think}\:{the}\:{correct}\:{equation}\:{is} \\ $$$${k}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{148}\right)=\mathrm{0}\:{with}\:{k}\in{R},\:{k}\neq\mathrm{0} \\ $$

Commented by mr W last updated on 26/Oct/23

Commented by deleteduser1 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. (a+3)(b+3)(c+3) =abc+3(ab+bc+ca)+9(a+b+c)+27 is only true when a,b,c∈R When two(in the case of a cubic equation) come from C\R,then we are no longer dealing with variables(a,b,c) but (a,b^− ,c^− ) where b^− =x+yi. ⇒b^− +3=(x+3)+yi;so we are only increasing the real part by 3

$${I}'{m}\:{referring}\:{to}\:{the}\:{solution}\:{above}. \\ $$$${The}\:{solution}\:{would}\:{have}\:{been}\:{correct}\:{if}\:{a},{b}\:{and} \\ $$$${c}\:{were}\:{all}\:{real}\:{numbers}.\: \\ $$$$\left({a}+\mathrm{3}\right)\left({b}+\mathrm{3}\right)\left({c}+\mathrm{3}\right) \\ $$$$={abc}+\mathrm{3}\left({ab}+{bc}+{ca}\right)+\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{27}\:{is}\:{only}\:{true} \\ $$$${when}\:{a},{b},{c}\in\mathbb{R} \\ $$$${When}\:{two}\left({in}\:{the}\:{case}\:{of}\:{a}\:{cubic}\:{equation}\right)\:{come} \\ $$$${from}\:\mathbb{C}\backslash\mathbb{R},{then}\:{we}\:{are}\:{no}\:{longer}\:{dealing}\:{with} \\ $$$${variables}\left({a},{b},{c}\right)\:{but}\:\left({a},\overset{−} {{b}},\overset{−} {{c}}\right)\:{where}\:\overset{−} {{b}}={x}+{yi}. \\ $$$$\Rightarrow\overset{−} {{b}}+\mathrm{3}=\left({x}+\mathrm{3}\right)+{yi};{so}\:{we}\:{are}\:{only}\:{increasing} \\ $$$${the}\:{real}\:{part}\:{by}\:\mathrm{3} \\ $$

Answered by mr W last updated on 26/Oct/23

say the new polynomial is p(x). roots of p(x) exceed the roots of f(x) by 3, that means p(x)=f(x−3). p(x)=3(x−3)^3 −14(x−3)^2 +(x−3)+62 p(x)=3x^3 −41x^2 +166x−148 say roots of p(x)=0 are A,B,C, then A=a+3, B=b+3, C=c+3 (1/(a+3))+(1/(b+3))+(1/(c+3))=(1/A)+(1/B)+(1/C) =((AB+BC+CA)/(ABC))=(((166)/3)/((148)/3))=((166)/(148))=((83)/(74)) ✓ ■

$${say}\:{the}\:{new}\:{polynomial}\:{is}\:{p}\left({x}\right). \\ $$$${roots}\:{of}\:{p}\left({x}\right)\:{exceed}\:{the}\:{roots}\:{of}\:{f}\left({x}\right) \\ $$$${by}\:\mathrm{3},\:{that}\:{means}\:{p}\left({x}\right)={f}\left({x}−\mathrm{3}\right). \\ $$$${p}\left({x}\right)=\mathrm{3}\left({x}−\mathrm{3}\right)^{\mathrm{3}} −\mathrm{14}\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({x}−\mathrm{3}\right)+\mathrm{62} \\ $$$${p}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} −\mathrm{41}{x}^{\mathrm{2}} +\mathrm{166}{x}−\mathrm{148} \\ $$$${say}\:{roots}\:{of}\:{p}\left({x}\right)=\mathrm{0}\:{are}\:{A},{B},{C}, \\ $$$${then}\:{A}={a}+\mathrm{3},\:{B}={b}+\mathrm{3},\:{C}={c}+\mathrm{3} \\ $$$$\frac{\mathrm{1}}{{a}+\mathrm{3}}+\frac{\mathrm{1}}{{b}+\mathrm{3}}+\frac{\mathrm{1}}{{c}+\mathrm{3}}=\frac{\mathrm{1}}{{A}}+\frac{\mathrm{1}}{{B}}+\frac{\mathrm{1}}{{C}} \\ $$$$=\frac{{AB}+{BC}+{CA}}{{ABC}}=\frac{\frac{\mathrm{166}}{\mathrm{3}}}{\frac{\mathrm{148}}{\mathrm{3}}}=\frac{\mathrm{166}}{\mathrm{148}}=\frac{\mathrm{83}}{\mathrm{74}}\:\checkmark \\ $$$$\blacksquare \\ $$

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

Most efficient way!

$$\mathrm{Most}\:\mathrm{efficient}\:\mathrm{way}! \\ $$

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