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Question Number 215837 by MATHEMATICSAM last updated on 19/Jan/25

If b^3 + a^2 c + ac^2 = 3abc then prove that one root of ax^2 + bx + c = 0 is the square of the other one.

$$\mathrm{If}\:{b}^{\mathrm{3}} \:+\:{a}^{\mathrm{2}} {c}\:+\:{ac}^{\mathrm{2}} \:=\:\mathrm{3}{abc}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{one}\:\mathrm{root}\:\mathrm{of}\:{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{the}\:\mathrm{square}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one}. \\ $$

Answered by MrGaster last updated on 19/Jan/25

given b^3 +a^2 c+ac^2 =3abc β be the roots of ax^2 +bx+c=0 by Vieta′s formulas:Σα=−(b/a)∧Πα=(c/a) if one root is the□ of the other,then ∃γ:α=γ^2 ∧β=γ∨α=γ∧β=γ^2 without loss of generality,assume α=γ^2 ∧β=γ ∴Σα=γ^2 +γ=−(b/a) Πα=γ^3 =(c/a) from beth sides by a^3 :((b/a))^3 +((c/a))+((c/a))^2 =3((b/a))((c/a)) substitute Σα and Πα:(−Σα)^3 +Πα+(Πα)^2 =3(−Σα)(Πα) (−(γ^2 +γ))^3 +γ^3 +(γ^3 )^2 =3(−(γ^2 +γ))(γ^3 ) −3γ^4 =−3γ^4 bx+c=0 is indeed the □ of the other. ps:“□” denotes squaring.

$$\mathrm{given}\:{b}^{\mathrm{3}} +{a}^{\mathrm{2}} {c}+{ac}^{\mathrm{2}} =\mathrm{3}{abc} \\ $$$$\beta\:{be}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\mathrm{by}\:\mathrm{Vieta}'{s}\:\mathrm{formulas}:\Sigma\alpha=−\frac{{b}}{{a}}\wedge\Pi\alpha=\frac{{c}}{{a}} \\ $$$$\mathrm{if}\:\mathrm{one}\:\mathrm{root}\:\mathrm{is}\:\mathrm{the}\Box\:\mathrm{of}\:\mathrm{the}\:\mathrm{other},\mathrm{then}\:\exists\gamma:\alpha=\gamma^{\mathrm{2}} \wedge\beta=\gamma\vee\alpha=\gamma\wedge\beta=\gamma^{\mathrm{2}} \\ $$$$\mathrm{without}\:\mathrm{loss}\:\mathrm{of}\:\mathrm{generality},\mathrm{assume}\:\alpha=\gamma^{\mathrm{2}} \wedge\beta=\gamma \\ $$$$\therefore\Sigma\alpha=\gamma^{\mathrm{2}} +\gamma=−\frac{{b}}{{a}} \\ $$$$\Pi\alpha=\gamma^{\mathrm{3}} =\frac{{c}}{{a}} \\ $$$$\mathrm{from}\:\mathrm{beth}\:\mathrm{sides}\:\mathrm{by}\:{a}^{\mathrm{3}} :\left(\frac{{b}}{{a}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{c}}{{a}}\right)+\left(\frac{\mathrm{c}}{{a}}\right)^{\mathrm{2}} =\mathrm{3}\left(\frac{{b}}{{a}}\right)\left(\frac{{c}}{{a}}\right) \\ $$$$\mathrm{substitute}\:\Sigma\alpha\:\mathrm{and}\:\Pi\alpha:\left(−\Sigma\alpha\right)^{\mathrm{3}} +\Pi\alpha+\left(\Pi\alpha\right)^{\mathrm{2}} =\mathrm{3}\left(−\Sigma\alpha\right)\left(\Pi\alpha\right) \\ $$$$\left(−\left(\gamma^{\mathrm{2}} +\gamma\right)\right)^{\mathrm{3}} +\gamma^{\mathrm{3}} +\left(\gamma^{\mathrm{3}} \right)^{\mathrm{2}} =\mathrm{3}\left(−\left(\gamma^{\mathrm{2}} +\gamma\right)\right)\left(\gamma^{\mathrm{3}} \right) \\ $$$$−\mathrm{3}\gamma^{\mathrm{4}} =−\mathrm{3}\gamma^{\mathrm{4}} \\ $$$${bx}+{c}=\mathrm{0}\:\mathrm{is}\:\mathrm{indeed}\:\mathrm{the}\:\Box\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}. \\ $$$$\mathrm{ps}:``\Box''\:\:\mathrm{denotes}\:\mathrm{squaring}. \\ $$

Commented by A5T last updated on 19/Jan/25

From the wording of the question, what is required to prove is : b^3 +a^2 c+ac^2 =3abc⇒x_1 =γ ∧ x_2 =γ^2 You showed the converse here : x_1 =γ ∧ x_2 =γ^2 ⇒ b^3 +a^2 c+ac^2 =3abc Both are not quite the same. A⇒B ≢ B⇒A .

$$\mathrm{From}\:\mathrm{the}\:\mathrm{wording}\:\mathrm{of}\:\mathrm{the}\:\mathrm{question},\:\mathrm{what}\:\mathrm{is}\: \\ $$$$\mathrm{required}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{is}\:\::\: \\ $$$$\mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \mathrm{c}+\mathrm{ac}^{\mathrm{2}} =\mathrm{3abc}\Rightarrow\mathrm{x}_{\mathrm{1}} =\gamma\:\wedge\:\mathrm{x}_{\mathrm{2}} =\gamma^{\mathrm{2}} \\ $$$$\mathrm{You}\:\mathrm{showed}\:\mathrm{the}\:\mathrm{converse}\:\mathrm{here}\:: \\ $$$$\mathrm{x}_{\mathrm{1}} =\gamma\:\wedge\:\mathrm{x}_{\mathrm{2}} =\gamma^{\mathrm{2}} \:\Rightarrow\:\mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \mathrm{c}+\mathrm{ac}^{\mathrm{2}} =\mathrm{3abc}\:\: \\ $$$$\mathrm{Both}\:\mathrm{are}\:\mathrm{not}\:\mathrm{quite}\:\mathrm{the}\:\mathrm{same}. \\ $$$$\mathrm{A}\Rightarrow\mathrm{B}\:≢\:\mathrm{B}\Rightarrow\mathrm{A}\:. \\ $$

Answered by A5T last updated on 19/Jan/25

Let the roots of ax^2 +bx+c=0 be x_1 and x_2 x_1 +x_2 =((−b)/a); x_1 x_2 =(c/a) b^3 +a^2 c+ac^2 =3abc⇒^(/a^3 ) ((b/a))^3 +((c/a))+((c/a))^2 =3((b/a))((c/a)) ⇒−(x_1 +x_2 )^3 +x_1 x_2 +x_1 ^2 x_2 ^2 =−3(x_1 +x_2 )(x_1 x_2 ) ⇒−(x_1 ^3 +3x_1 ^2 x_2 +3x_1 x_2 ^2 +x_2 ^3 )+x_1 x_2 +x_1 ^2 x_2 ^2 =−3x_1 ^2 x_2 −3x_1 x_2 ^2 ⇒−x_1 ^3 −x_2 ^3 +x_1 x_2 +x_1 ^2 x_2 ^2 =−x_1 ^2 (x_1 −x_2 ^2 )+x_2 (x_1 −x_2 ^2 )=0 ⇒(x_1 ^2 −x_2 )(x_1 −x_2 ^2 )=0 ⇒x_1 ^2 =x_2 or x_2 ^2 =x_1 ■

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0}\:\mathrm{be}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\frac{−\mathrm{b}}{\mathrm{a}};\:\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{2}} \mathrm{c}+\mathrm{ac}^{\mathrm{2}} =\mathrm{3abc}\overset{/\mathrm{a}^{\mathrm{3}} } {\Rightarrow}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{c}}{\mathrm{a}}\right)+\left(\frac{\mathrm{c}}{\mathrm{a}}\right)^{\mathrm{2}} =\mathrm{3}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)\left(\frac{\mathrm{c}}{\mathrm{a}}\right) \\ $$$$\Rightarrow−\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{3}} +\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{3}\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)\left(\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \right) \\ $$$$\Rightarrow−\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} +\mathrm{3x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{3}} \right)+\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} −\mathrm{3x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\Rightarrow−\mathrm{x}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{x}_{\mathrm{2}} ^{\mathrm{3}} +\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \right)+\mathrm{x}_{\mathrm{2}} \left(\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{x}_{\mathrm{2}} \right)\left(\mathrm{x}_{\mathrm{1}} −\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} =\mathrm{x}_{\mathrm{2}} \:\mathrm{or}\:\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{x}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$

Commented by ArshadS last updated on 19/Jan/25

5th line: ⇒−(x_1 ^3 +3x_1 ^2 x_2 +3x_1 x_2 ^2 +x_2 ^3 )+x_1 x_2 +x_1 ^2 x_2 ^2 =−3x_1 ^2 x_2 −3x_1 x_2 ^2

$$\mathrm{5}{th}\:{line}: \\ $$$$\Rightarrow−\left(\mathrm{x}_{\mathrm{1}} ^{\mathrm{3}} +\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} +\mathrm{3x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{3}} \right)+\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} =−\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} \mathrm{x}_{\mathrm{2}} −\mathrm{3x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$

Commented by A5T last updated on 19/Jan/25

It was a typo, thanks.

$$\mathrm{It}\:\mathrm{was}\:\mathrm{a}\:\mathrm{typo},\:\mathrm{thanks}. \\ $$

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