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Question Number 28319 by NECx last updated on 24/Jan/18

If the roots of x^2 +px+q=0, q≠0 are α and β.Find the roots of qx^2 +(2q−p^2 )x+q=0 in terms of α and β.

$${If}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0},\:{q}\neq\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta.{Find}\:{the}\:{roots}\:{of} \\ $$$${qx}^{\mathrm{2}} +\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right){x}+{q}=\mathrm{0}\:{in}\:{terms}\:{of} \\ $$$$\alpha\:{and}\:\beta. \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jan/18

x^2 +px+q=0⇒α+β=−p ,αβ=q ⇒p=−(α+β) , q=αβ qx^2 +(2q−p^2 )x+q=0 x=((−(2q−p^2 )±(√((2q−p^2 )^2 −4q^2 )))/(2q)) x=((−(2q−p^2 )±(√(4q^2 −4p^2 q+p^4 −4q^2 )))/(2q)) x=((−(2q−p^2 )±(√(−4p^2 q+p^4 )))/(2q)) x=((−2q+p^2 ±p(√(p^2 −4q)))/(2q)) x=((−2(αβ)+(α+β)^2 ±(−α−β)(√((α+β)^2 −4αβ)))/(2αβ)) x=((−2αβ+(α+β)^2 ∓(α+β)(√((α^2 −2αβ+β^2 )))/(2αβ)) x=((−2αβ+α^2 +2αβ+β^2 ∓(α+β)(α−β))/(2αβ)) x=((α^2 +β^2 ∓(α^2 −β^2 ))/(2αβ)) x=((α^2 +β^2 −(α^2 −β^2 ))/(2αβ)) , ((α^2 +β^2 +(α^2 −β^2 ))/(2αβ)) x=((α^2 +β^2 −α^2 +β^2 ))/(2αβ)) , ((α^2 +β^2 +α^2 −β^2 )/(2αβ)) x=((2β^2 )/(2αβ)) , ((2α^2 )/(2αβ)) x=(β/α),(α/β)

$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\Rightarrow\alpha+\beta=−{p}\:,\alpha\beta={q} \\ $$$$\:\:\:\:\Rightarrow{p}=−\left(\alpha+\beta\right)\:,\:{q}=\alpha\beta \\ $$$${qx}^{\mathrm{2}} +\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right){x}+{q}=\mathrm{0} \\ $$$$\:\:\:{x}=\frac{−\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right)\pm\sqrt{\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{q}^{\mathrm{2}} }}{\mathrm{2}{q}} \\ $$$$\:\:\:{x}=\frac{−\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right)\pm\sqrt{\mathrm{4}{q}^{\mathrm{2}} −\mathrm{4}{p}^{\mathrm{2}} {q}+{p}^{\mathrm{4}} −\mathrm{4}{q}^{\mathrm{2}} }}{\mathrm{2}{q}} \\ $$$$\:\:\:{x}=\frac{−\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right)\pm\sqrt{−\mathrm{4}{p}^{\mathrm{2}} {q}+{p}^{\mathrm{4}} }}{\mathrm{2}{q}} \\ $$$$\:\:\:{x}=\frac{−\mathrm{2}{q}+{p}^{\mathrm{2}} \pm{p}\sqrt{{p}^{\mathrm{2}} −\mathrm{4}{q}}}{\mathrm{2}{q}} \\ $$$$\:\:\:{x}=\frac{−\mathrm{2}\left(\alpha\beta\right)+\left(\alpha+\beta\right)^{\mathrm{2}} \pm\left(−\alpha−\beta\right)\sqrt{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta}}{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{−\mathrm{2}\alpha\beta+\left(\alpha+\beta\right)^{\mathrm{2}} \mp\left(\alpha+\beta\right)\sqrt{\left(\alpha^{\mathrm{2}} −\mathrm{2}\alpha\beta+\beta^{\mathrm{2}} \right.}}{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{−\mathrm{2}\alpha\beta+\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta+\beta^{\mathrm{2}} \mp\left(\alpha+\beta\right)\left(\alpha−\beta\right)}{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \mp\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}{\mathrm{2}\alpha\beta}\:,\:\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\left(\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} \right)}{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{\left.\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)}{\mathrm{2}\alpha\beta}\:,\:\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }{\mathrm{2}\alpha\beta} \\ $$$$\:\:\:{x}=\frac{\mathrm{2}\beta^{\mathrm{2}} }{\mathrm{2}\alpha\beta}\:,\:\frac{\mathrm{2}\alpha^{\mathrm{2}} }{\mathrm{2}\alpha\beta} \\ $$$${x}=\frac{\beta}{\alpha},\frac{\alpha}{\beta} \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jan/18

x^2 +px+q=0 p=−(α+β) , q=αβ qx^2 +(2q−p^2 )x+q=0⇒ αβx^2 +{2αβ−(α+β)^2 }x+αβ=0 Let the roots are A & B A+B=−((2αβ−(α+β)^2 )/(αβ))=((α^2 +β^2 )/(αβ)) AB=((αβ)/(αβ))=1⇒B=(1/A) A+B=A+(1/A)=((α^2 +β^2 )/(αβ))=(α/β)+(β/α) A^2 −((α/β)+(β/α))A+1=0 A^2 −(α/β)A−(β/α)A+((α/β))((β/α))=0 A(A−(α/β))−(β/α)(A−(α/β))=0 (A−(α/β))(A−(β/α))=0 A=(α/β) , (β/α) B=(1/A)=(β/α) , (α/β) Roots are (α/β) & (β/α)

$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0} \\ $$$${p}=−\left(\alpha+\beta\right)\:,\:\mathrm{q}=\alpha\beta \\ $$$${qx}^{\mathrm{2}} +\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right){x}+{q}=\mathrm{0}\Rightarrow \\ $$$$\alpha\beta{x}^{\mathrm{2}} +\left\{\mathrm{2}\alpha\beta−\left(\alpha+\beta\right)^{\mathrm{2}} \right\}{x}+\alpha\beta=\mathrm{0} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{A}\:\&\:\mathrm{B} \\ $$$$\mathrm{A}+\mathrm{B}=−\frac{\mathrm{2}\alpha\beta−\left(\alpha+\beta\right)^{\mathrm{2}} }{\alpha\beta}=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha\beta} \\ $$$$\mathrm{AB}=\frac{\alpha\beta}{\alpha\beta}=\mathrm{1}\Rightarrow\mathrm{B}=\frac{\mathrm{1}}{\mathrm{A}} \\ $$$$\mathrm{A}+\mathrm{B}=\mathrm{A}+\frac{\mathrm{1}}{\mathrm{A}}=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha\beta}=\frac{\alpha}{\beta}+\frac{\beta}{\alpha} \\ $$$$\mathrm{A}^{\mathrm{2}} −\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)\mathrm{A}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{A}^{\mathrm{2}} −\frac{\alpha}{\beta}\mathrm{A}−\frac{\beta}{\alpha}\mathrm{A}+\left(\frac{\alpha}{\beta}\right)\left(\frac{\beta}{\alpha}\right)=\mathrm{0} \\ $$$$\mathrm{A}\left(\mathrm{A}−\frac{\alpha}{\beta}\right)−\frac{\beta}{\alpha}\left(\mathrm{A}−\frac{\alpha}{\beta}\right)=\mathrm{0} \\ $$$$\left(\mathrm{A}−\frac{\alpha}{\beta}\right)\left(\mathrm{A}−\frac{\beta}{\alpha}\right)=\mathrm{0} \\ $$$$\mathrm{A}=\frac{\alpha}{\beta}\:\:,\:\:\frac{\beta}{\alpha} \\ $$$$\mathrm{B}=\frac{\mathrm{1}}{\mathrm{A}}=\frac{\beta}{\alpha}\:,\:\frac{\alpha}{\beta} \\ $$$$\mathrm{Roots}\:\mathrm{are}\:\:\frac{\alpha}{\beta}\:\&\:\frac{\beta}{\alpha} \\ $$

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