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Question Number 202019 by MATHEMATICSAM last updated on 18/Dec/23
If α and β are the roots of the ax^2 + 2bx + c = 0 and α + δ and β + δ are the roots of Ax^2 + 2Bx + C = 0 for some constant δ then prove that ((b^2 − ac)/a^2 ) = ((B^2 − AC)/A^2 ) .
$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\: \\ $$$${ax}^{\mathrm{2}} \:+\:\mathrm{2}{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha\:+\:\delta\:\mathrm{and}\:\beta\:+\:\delta\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{Ax}^{\mathrm{2}} \:+\:\mathrm{2}{Bx}\:+\:{C}\:=\:\mathrm{0}\:\mathrm{for}\:\mathrm{some}\: \\ $$$$\mathrm{constant}\:\delta\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{b}^{\mathrm{2}} \:−\:{ac}}{{a}^{\mathrm{2}} }\:=\:\frac{{B}^{\mathrm{2}} \:−\:{AC}}{{A}^{\mathrm{2}} }\:. \\ $$
Answered by esmaeil last updated on 18/Dec/23
α−β=((√(b^2 −ac))/(∣a∣))=(α+δ)−(β+δ)= ((√(B^2 −AC))/(∣A∣))→((b^2 −ac)/a^2 )=((B^2 −AC)/A^2 ) α−β=((√δ^′ )/(∣a∣))
$$\alpha−\beta=\frac{\sqrt{{b}^{\mathrm{2}} −{ac}}}{\mid{a}\mid}=\left(\alpha+\delta\right)−\left(\beta+\delta\right)= \\ $$$$\frac{\sqrt{{B}^{\mathrm{2}} −{AC}}}{\mid{A}\mid}\rightarrow\frac{{b}^{\mathrm{2}} −{ac}}{{a}^{\mathrm{2}} }=\frac{{B}^{\mathrm{2}} −{AC}}{{A}^{\mathrm{2}} } \\ $$$$\:\:\alpha−\beta=\frac{\sqrt{\delta^{'} }}{\mid{a}\mid} \\ $$
Commented by MM42 last updated on 18/Dec/23
⋛
$$\:\cancel{\lesseqgtr} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Dec/23
Another way •α & β are the roots of ax^2 + 2bx + c = 0 α+β=−((2b)/a) , αβ=(c/a) •If the roots are α+δ & β+δ (δ is fixed constant) The equation will be: (α+δ)+(β+δ)=α+β+2δ=−((2b)/a)+2δ=−((2B)/A) ⇒(B/A)=((−2b+2aδ)/(−2a)) (α+δ).(β+δ)=αβ+δ^2 +(α+β)δ=(c/a)−((2bδ)/a)+δ^2 =(C/A) ⇒(C/A)=((c−2bδ+aδ^2 )/a) ((B^2 − AC)/A^2 )=((B/A))^2 −(C/A)=(((−2b+2aδ)/(−2a)))^2 −((c−2bδ+aδ^2 )/a) =((4b^2 +4a^2 δ^2 −8abδ)/(4a^2 ))−((c−2bδ+aδ^2 )/a) =((4b^2 +4a^2 δ^2 −8abδ−4ac+8abδ−4a^2 δ^2 )/(4a^2 )) =((4b^2 −4ac)/(4a^2 ))=((b^2 −ac)/a^2 ) QED
$${Another}\:{way} \\ $$$$\bullet\alpha\:\&\:\beta\:{are}\:{the}\:{roots}\:{of}\:{ax}^{\mathrm{2}} \:+\:\mathrm{2}{bx}\:+\:{c}\:=\:\mathrm{0} \\ $$$$\:\:\:\alpha+\beta=−\frac{\mathrm{2}{b}}{{a}}\:,\:\alpha\beta=\frac{{c}}{{a}} \\ $$$$\bullet\mathcal{I}{f}\:{the}\:{roots}\:{are}\:\alpha+\delta\:\&\:\beta+\delta\:\left(\delta\:{is}\:{fixed}\:{constant}\right) \\ $$$$\:\:\:\mathcal{T}{he}\:{equation}\:{will}\:{be}: \\ $$$$\:\:\:\left(\alpha+\delta\right)+\left(\beta+\delta\right)=\alpha+\beta+\mathrm{2}\delta=−\frac{\mathrm{2}{b}}{{a}}+\mathrm{2}\delta=−\frac{\mathrm{2}{B}}{{A}} \\ $$$$\Rightarrow\frac{{B}}{{A}}=\frac{−\mathrm{2}{b}+\mathrm{2}{a}\delta}{−\mathrm{2}{a}} \\ $$$$\:\:\:\left(\alpha+\delta\right).\left(\beta+\delta\right)=\alpha\beta+\delta^{\mathrm{2}} +\left(\alpha+\beta\right)\delta=\frac{{c}}{{a}}−\frac{\mathrm{2}{b}\delta}{{a}}+\delta^{\mathrm{2}} =\frac{{C}}{{A}} \\ $$$$\Rightarrow\frac{{C}}{{A}}=\frac{{c}−\mathrm{2}{b}\delta+{a}\delta^{\mathrm{2}} }{{a}} \\ $$$$\:\frac{{B}^{\mathrm{2}} \:−\:{AC}}{{A}^{\mathrm{2}} }=\left(\frac{{B}}{{A}}\right)^{\mathrm{2}} −\frac{{C}}{{A}}=\left(\frac{−\mathrm{2}{b}+\mathrm{2}{a}\delta}{−\mathrm{2}{a}}\right)^{\mathrm{2}} −\frac{{c}−\mathrm{2}{b}\delta+{a}\delta^{\mathrm{2}} }{{a}} \\ $$$$\:\:\:\:=\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} \delta^{\mathrm{2}} −\mathrm{8}{ab}\delta}{\mathrm{4}{a}^{\mathrm{2}} }−\frac{{c}−\mathrm{2}{b}\delta+{a}\delta^{\mathrm{2}} }{{a}} \\ $$$$\:\:\:\:=\frac{\mathrm{4}{b}^{\mathrm{2}} +\cancel{\mathrm{4}{a}^{\mathrm{2}} \delta^{\mathrm{2}} }−\cancel{\mathrm{8}{ab}\delta}−\mathrm{4}{ac}+\cancel{\mathrm{8}{ab}\delta}−\cancel{\mathrm{4}{a}^{\mathrm{2}} \delta^{\mathrm{2}} }}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}{ac}}{\mathrm{4}{a}^{\mathrm{2}} }=\frac{{b}^{\mathrm{2}} −{ac}}{{a}^{\mathrm{2}} } \\ $$$${QED} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Dec/23
((b^2 − ac)/a^2 ) = ((B^2 − AC)/A^2 ) . ((b/a))^2 −(c/a)=((B/A))^2 −(C/A) (1/4)(((−2b)/a))^2 −(c/a)=(1/4)(((−2B)/A))^2 −(C/A) (1/4)(α+β)^2 −αβ=(1/4)( (α+δ)+(β+δ) )^2 −(α+δ)(β+δ) (1/4)(α+β)^2 −αβ=(1/4)(α+β+2δ )^2 −(αβ+(α+β)δ+δ^2 ) (α+β)^2 −4αβ=(α+β+2δ)^2 −4(αβ+(α+β)δ+δ^2 ) =(α+β)^2 +4(α+β)δ+4δ^2 −4αβ−4(α+β)δ−4δ^2 =(α+β)^2 −4αβ lhs=rhs proved
$$\frac{{b}^{\mathrm{2}} \:−\:{ac}}{{a}^{\mathrm{2}} }\:=\:\frac{{B}^{\mathrm{2}} \:−\:{AC}}{{A}^{\mathrm{2}} }\:. \\ $$$$\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} −\frac{{c}}{{a}}=\left(\frac{{B}}{{A}}\right)^{\mathrm{2}} −\frac{{C}}{{A}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{−\mathrm{2}{b}}{{a}}\right)^{\mathrm{2}} −\frac{{c}}{{a}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{−\mathrm{2}{B}}{{A}}\right)^{\mathrm{2}} −\frac{{C}}{{A}} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\alpha+\beta\right)^{\mathrm{2}} −\alpha\beta=\frac{\mathrm{1}}{\mathrm{4}}\left(\:\left(\alpha+\delta\right)+\left(\beta+\delta\right)\:\right)^{\mathrm{2}} −\left(\alpha+\delta\right)\left(\beta+\delta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\alpha+\beta\right)^{\mathrm{2}} −\alpha\beta=\frac{\mathrm{1}}{\mathrm{4}}\left(\alpha+\beta+\mathrm{2}\delta\:\right)^{\mathrm{2}} −\left(\alpha\beta+\left(\alpha+\beta\right)\delta+\delta^{\mathrm{2}} \right) \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta=\left(\alpha+\beta+\mathrm{2}\delta\right)^{\mathrm{2}} −\mathrm{4}\left(\alpha\beta+\left(\alpha+\beta\right)\delta+\delta^{\mathrm{2}} \right) \\ $$$$\:\:\:=\left(\alpha+\beta\right)^{\mathrm{2}} +\cancel{\mathrm{4}\left(\alpha+\beta\right)\delta}+\cancel{\mathrm{4}\delta^{\mathrm{2}} }−\mathrm{4}\alpha\beta−\cancel{\mathrm{4}\left(\alpha+\beta\right)\delta}−\cancel{\mathrm{4}\delta^{\mathrm{2}} } \\ $$$$\:\:\:=\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta \\ $$$${lhs}={rhs} \\ $$$${proved} \\ $$

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