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Question Number 202436 by MATHEMATICSAM last updated on 26/Dec/23

If α, β and γ are the roots of ax^3 + bx + c = 0 then frame an equation whose roots are α^2 , β^2 , γ^2 .

$$\mathrm{If}\:\alpha,\:\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{3}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{frame}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\alpha^{\mathrm{2}} ,\:\beta^{\mathrm{2}} ,\:\gamma^{\mathrm{2}} \:.\: \\ $$

Answered by aleks041103 last updated on 26/Dec/23

ax^3 +bx+c=0, x=α,β,γ ⇒−(c/a)=αβγ (b/a)=αβ+βγ+αγ 0=α+β+γ Ax^3 +Bx^2 +Cx+D=0, x=α^2 ,β^2 ,γ^2 −(D/A)=α^2 β^2 γ^2 =(αβγ)^2 =(c^2 /a^2 ) (C/A)=α^2 β^2 +β^2 γ^2 +α^2 γ^2 =(αβ+βγ+αγ)^2 −2αβγ(α+β+γ)= =(b^2 /a^2 )+2(c/a).0=(b^2 /a^2 ) −(B/A)=α^2 +β^2 +γ^2 =(α+β+γ)^2 −2(αβ+βγ+αγ)= =0^2 −2(b/a)=−((2b)/a) x^3 +((2b)/a)x^2 +(b^2 /a^2 )x−(c^2 /a^2 )=0 ⇒a^2 x^3 +2abx^2 +b^2 x−c^2 =0

$${ax}^{\mathrm{3}} +{bx}+{c}=\mathrm{0},\:{x}=\alpha,\beta,\gamma \\ $$$$\Rightarrow−\frac{{c}}{{a}}=\alpha\beta\gamma \\ $$$$\frac{{b}}{{a}}=\alpha\beta+\beta\gamma+\alpha\gamma \\ $$$$\mathrm{0}=\alpha+\beta+\gamma \\ $$$$ \\ $$$${Ax}^{\mathrm{3}} +{Bx}^{\mathrm{2}} +{Cx}+{D}=\mathrm{0},\:{x}=\alpha^{\mathrm{2}} ,\beta^{\mathrm{2}} ,\gamma^{\mathrm{2}} \\ $$$$−\frac{{D}}{{A}}=\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} =\left(\alpha\beta\gamma\right)^{\mathrm{2}} =\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\frac{{C}}{{A}}=\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} =\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\gamma\left(\alpha+\beta+\gamma\right)= \\ $$$$=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\mathrm{2}\frac{{c}}{{a}}.\mathrm{0}=\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$−\frac{{B}}{{A}}=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)= \\ $$$$=\mathrm{0}^{\mathrm{2}} −\mathrm{2}\frac{{b}}{{a}}=−\frac{\mathrm{2}{b}}{{a}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{2}{b}}{{a}}{x}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{x}−\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} {x}^{\mathrm{3}} +\mathrm{2}{abx}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$

Answered by Frix last updated on 27/Dec/23

x^3 +px^2 +qx+r=0 ((√x^3 )+px+q(√x)+r)((√x^3 )−px+q(√x)−r)=0 x^3 −(p^2 −2q)x^2 −(2pr−q^2 )x−r^2 =0 p=0∧q=(b/a)∧r=(c/a) x^3 +((2bx^2 )/a)+((b^2 x)/a^3 )−(c^2 /a^2 )=0 a^2 x^2 +2abx^2 +b^2 x−c^2 =0

$${x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\left(\sqrt{{x}^{\mathrm{3}} }+{px}+{q}\sqrt{{x}}+{r}\right)\left(\sqrt{{x}^{\mathrm{3}} }−{px}+{q}\sqrt{{x}}−{r}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\left({p}^{\mathrm{2}} −\mathrm{2}{q}\right){x}^{\mathrm{2}} −\left(\mathrm{2}{pr}−{q}^{\mathrm{2}} \right){x}−{r}^{\mathrm{2}} =\mathrm{0} \\ $$$${p}=\mathrm{0}\wedge{q}=\frac{{b}}{{a}}\wedge{r}=\frac{{c}}{{a}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{2}{bx}^{\mathrm{2}} }{{a}}+\frac{{b}^{\mathrm{2}} {x}}{{a}^{\mathrm{3}} }−\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{abx}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}−{c}^{\mathrm{2}} =\mathrm{0} \\ $$

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