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Find-the-condition-that-one-root-of-ax-2-bx-c-0-a-0-is-square-of-the-other-

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Question Number 132697 by liberty last updated on 15/Feb/21
Find the condition that one root of ax^2 +bx+c = 0 ,a≠ 0 is square of the other .
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{that}\:\mathrm{one} \\ $$$$\mathrm{root}\:\mathrm{of}\:{ax}^{\mathrm{2}} +{bx}+{c}\:=\:\mathrm{0}\:,{a}\neq\:\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:. \\ $$
Commented by liberty last updated on 16/Feb/21
okay
$$\mathrm{okay} \\ $$
Commented by mr W last updated on 16/Feb/21
α+α^2 =−(b/a) α×α^2 =(c/a) ⇒α=((c/a))^(1/3) ⇒α(1+α)=((c/a))^(1/3) (1+((c/a))^(1/3) )=−(b/a) ⇒(b/a)+((c/a))^(1/3) (1+((c/a))^(1/3) )=0
$$\alpha+\alpha^{\mathrm{2}} =−\frac{{b}}{{a}} \\ $$$$\alpha×\alpha^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\Rightarrow\alpha=\sqrt[{\mathrm{3}}]{\frac{{c}}{{a}}} \\ $$$$\Rightarrow\alpha\left(\mathrm{1}+\alpha\right)=\sqrt[{\mathrm{3}}]{\frac{{c}}{{a}}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{a}}}\right)=−\frac{{b}}{{a}} \\ $$$$\Rightarrow\frac{{b}}{{a}}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{a}}}\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\frac{{c}}{{a}}}\right)=\mathrm{0} \\ $$
Answered by EDWIN88 last updated on 16/Feb/21
Suppose υ is a root and that the other one is υ^2 . By Vietha′s rule { ((υ+υ^2 =−(b/a))),((υ^3 = (c/a))) :} We have υ = 0 if only if b=c=0 let we assume that υ ≠ 0 Therefore ((1+υ)/υ^2 ) = −(b/c) ⇒bυ^2 +cυ+c = 0 on the other hand we find aυ^2 +bυ+c = 0 substract the two relations we get (a−b)υ^2 +(b−c)υ = 0 (a−b)υ +(b−c)=0→υ=((c−b)/(a−b)) but we can the two relations multiplying the top one by a and the bottom by b { ((abυ^2 +acυ + ac = 0)),((abυ^2 +b^2 υ +bc = 0)) :} subtract (ac−b^2 )υ+(ac−bc)= 0→υ=((bc−ac)/(ac−b^2 )) that expands to a^2 c +ac^2 −3abc + b^3 = 0
$$\:\mathrm{Suppose}\:\upsilon\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{and}\:\mathrm{that}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one} \\ $$$$\mathrm{is}\:\upsilon^{\mathrm{2}} \:.\:\mathrm{By}\:\mathrm{Vietha}'\mathrm{s}\:\mathrm{rule}\:\begin{cases}{\upsilon+\upsilon^{\mathrm{2}} =−\frac{{b}}{{a}}}\\{\upsilon^{\mathrm{3}} \:=\:\frac{\mathrm{c}}{{a}}}\end{cases} \\ $$$${W}\mathrm{e}\:\mathrm{have}\:\upsilon\:=\:\mathrm{0}\:\mathrm{if}\:\mathrm{only}\:\mathrm{if}\:\mathrm{b}=\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{we}\:\mathrm{assume}\:\mathrm{that}\:\upsilon\:\neq\:\mathrm{0}\: \\ $$$$\mathrm{Therefore}\:\frac{\mathrm{1}+\upsilon}{\upsilon^{\mathrm{2}} }\:=\:−\frac{{b}}{{c}}\:\Rightarrow{b}\upsilon^{\mathrm{2}} +{c}\upsilon+{c}\:=\:\mathrm{0} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand}\:\mathrm{we}\:\mathrm{find}\:{a}\upsilon^{\mathrm{2}} +\mathrm{b}\upsilon+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\mathrm{substract}\:\mathrm{the}\:\mathrm{two}\:\mathrm{relations}\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\left({a}−{b}\right)\upsilon^{\mathrm{2}} +\left({b}−{c}\right)\upsilon\:\:=\:\mathrm{0}\: \\ $$$$\left({a}−{b}\right)\upsilon\:+\left({b}−{c}\right)=\mathrm{0}\rightarrow\upsilon=\frac{{c}−{b}}{{a}−{b}} \\ $$$${but}\:{we}\:{can}\:{the}\:{two}\:{relations}\:{multiplying}\: \\ $$$${the}\:{top}\:{one}\:{by}\:{a}\:{and}\:{the}\:{bottom}\:{by}\:{b} \\ $$$$\begin{cases}{{ab}\upsilon^{\mathrm{2}} +{ac}\upsilon\:+\:{ac}\:=\:\mathrm{0}}\\{{ab}\upsilon^{\mathrm{2}} \:+{b}^{\mathrm{2}} \upsilon\:+{bc}\:=\:\mathrm{0}}\end{cases} \\ $$$${subtract}\:\left({ac}−{b}^{\mathrm{2}} \right)\upsilon+\left({ac}−{bc}\right)=\:\mathrm{0}\rightarrow\upsilon=\frac{{bc}−{ac}}{{ac}−{b}^{\mathrm{2}} } \\ $$$${that}\:{expands}\:{to}\:{a}^{\mathrm{2}} {c}\:+{ac}^{\mathrm{2}} \:−\mathrm{3}{abc}\:+\:{b}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by liberty last updated on 16/Feb/21
okay
$$\mathrm{okay} \\ $$
Commented by EDWIN88 last updated on 16/Feb/21
let for case { ((a=1 ; b=−2)),((c+c^2 +6c−8=0 ⇒c^2 +7c−8=0)) :} (c+8)(c−1)=0 { ((c=−8)),((c=1)) :} If (a,b,c)=(1,−2,−8)⇒x^2 −2x−8=0 (x−4)(x+2)=0 ⇒x_1 =4 ∧ x_2 =−2 and (−2)^2 = 4
$$\mathrm{let}\:\mathrm{for}\:\mathrm{case}\:\begin{cases}{{a}=\mathrm{1}\:;\:{b}=−\mathrm{2}}\\{{c}+{c}^{\mathrm{2}} +\mathrm{6}{c}−\mathrm{8}=\mathrm{0}\:\Rightarrow{c}^{\mathrm{2}} +\mathrm{7}{c}−\mathrm{8}=\mathrm{0}}\end{cases} \\ $$$$\left({c}+\mathrm{8}\right)\left({c}−\mathrm{1}\right)=\mathrm{0}\:\begin{cases}{{c}=−\mathrm{8}}\\{{c}=\mathrm{1}}\end{cases} \\ $$$$\mathrm{If}\:\left({a},\mathrm{b},\mathrm{c}\right)=\left(\mathrm{1},−\mathrm{2},−\mathrm{8}\right)\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}=\mathrm{0} \\ $$$$\:\left({x}−\mathrm{4}\right)\left({x}+\mathrm{2}\right)=\mathrm{0}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\mathrm{4}\:\wedge\:\mathrm{x}_{\mathrm{2}} =−\mathrm{2} \\ $$$$\mathrm{and}\:\left(−\mathrm{2}\right)^{\mathrm{2}} =\:\mathrm{4}\: \\ $$
Answered by MJS_new last updated on 16/Feb/21
a(x−α^2 )(x−α)=0 ax^2 −aα(α+1)x+aα^3 =0 b=−aα(α+1) c=aα^3 not sure what else you need...
$${a}\left({x}−\alpha^{\mathrm{2}} \right)\left({x}−\alpha\right)=\mathrm{0} \\ $$$${ax}^{\mathrm{2}} −{a}\alpha\left(\alpha+\mathrm{1}\right){x}+{a}\alpha^{\mathrm{3}} =\mathrm{0} \\ $$$${b}=−{a}\alpha\left(\alpha+\mathrm{1}\right) \\ $$$${c}={a}\alpha^{\mathrm{3}} \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{what}\:\mathrm{else}\:\mathrm{you}\:\mathrm{need}… \\ $$
Commented by liberty last updated on 16/Feb/21
relation for a,b and c sir
$$\mathrm{relation}\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}\:\mathrm{sir} \\ $$
Commented by MJS_new last updated on 16/Feb/21
a(x^2 +(b/a)x+(c/a))=0 x^2 +px+q=0 (x−α^2 )(x−α)=x^2 +px+q ⇔ p=−α(α+1)∧q=α^3 ⇔ p+q^(2/3) +q^(1/3) =0 ⇔ p=−(1+q^(1/3) )q^(1/3) ⇔ q=((3p−1±(p−1)(√(1−4p)))/2) or p^3 −3pq+q^2 +q=0 with p=(b/a)∧q=(c/a) we get a=((3b−c)/2)±((b−c)/2)(√(1−((4b)/c))) ⇔ b=a^(1/3) (a^(1/3) +c^(1/3) )c^(1/3) ⇔ c=((3b−a)/2)±((b−a)/2)(√(1−((4b)/a))) or a^2 c−3abc+ac^2 +b^3 =0
$${a}\left({x}^{\mathrm{2}} +\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0} \\ $$$$\left({x}−\alpha^{\mathrm{2}} \right)\left({x}−\alpha\right)={x}^{\mathrm{2}} +{px}+{q} \\ $$$$\Leftrightarrow \\ $$$${p}=−\alpha\left(\alpha+\mathrm{1}\right)\wedge{q}=\alpha^{\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$${p}+{q}^{\mathrm{2}/\mathrm{3}} +{q}^{\mathrm{1}/\mathrm{3}} =\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${p}=−\left(\mathrm{1}+{q}^{\mathrm{1}/\mathrm{3}} \right){q}^{\mathrm{1}/\mathrm{3}} \:\Leftrightarrow\:{q}=\frac{\mathrm{3}{p}−\mathrm{1}\pm\left({p}−\mathrm{1}\right)\sqrt{\mathrm{1}−\mathrm{4}{p}}}{\mathrm{2}} \\ $$$$\mathrm{or} \\ $$$${p}^{\mathrm{3}} −\mathrm{3}{pq}+{q}^{\mathrm{2}} +{q}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{with}\:{p}=\frac{{b}}{{a}}\wedge{q}=\frac{{c}}{{a}}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=\frac{\mathrm{3}{b}−{c}}{\mathrm{2}}\pm\frac{{b}−{c}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{4}{b}}{{c}}} \\ $$$$\Leftrightarrow \\ $$$${b}={a}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +{c}^{\mathrm{1}/\mathrm{3}} \right){c}^{\mathrm{1}/\mathrm{3}} \\ $$$$\Leftrightarrow \\ $$$${c}=\frac{\mathrm{3}{b}−{a}}{\mathrm{2}}\pm\frac{{b}−{a}}{\mathrm{2}}\sqrt{\mathrm{1}−\frac{\mathrm{4}{b}}{{a}}} \\ $$$$\mathrm{or} \\ $$$${a}^{\mathrm{2}} {c}−\mathrm{3}{abc}+{ac}^{\mathrm{2}} +{b}^{\mathrm{3}} =\mathrm{0} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Feb/21
{ ((α+α^2 =−(b/a))),((α^3 =(c/a))) :} (α+α^2 )^3 =(−(b/a))^3 α^3 +(α^3 )^2 +3α^3 (α+α^2 )=−(b^3 /a^3 ) (c/a)+((c/a))^2 +3((c/a))(−(b/a))+(b^3 /a^3 )=0 (c/a)+(c^2 /a^2 )−((3bc)/a^2 )+(b^3 /a^3 )=0 a^2 c+ac^2 −3abc+b^3 =0 ac(a+c)+b(b^2 −3ac)=0
$$\begin{cases}{\alpha+\alpha^{\mathrm{2}} =−\frac{{b}}{{a}}}\\{\alpha^{\mathrm{3}} =\frac{{c}}{{a}}}\end{cases} \\ $$$$\left(\alpha+\alpha^{\mathrm{2}} \right)^{\mathrm{3}} =\left(−\frac{{b}}{{a}}\right)^{\mathrm{3}} \\ $$$$\alpha^{\mathrm{3}} +\left(\alpha^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{3}\alpha^{\mathrm{3}} \left(\alpha+\alpha^{\mathrm{2}} \right)=−\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} } \\ $$$$\frac{{c}}{{a}}+\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{{c}}{{a}}\right)\left(−\frac{{b}}{{a}}\right)+\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\frac{{c}}{{a}}+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{3}{bc}}{{a}^{\mathrm{2}} }+\frac{{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} }=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} −\mathrm{3}{abc}+{b}^{\mathrm{3}} =\mathrm{0} \\ $$$${ac}\left({a}+{c}\right)+{b}\left({b}^{\mathrm{2}} −\mathrm{3}{ac}\right)=\mathrm{0} \\ $$

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