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Question Number 172024 by Mikenice last updated on 23/Jun/22
let α and β be the root of the equation ax^2 +bx+c=0. find the equation whose roots are ((1/α)+(1/β)) and ((1/α)−(1/β))
$${let}\:\alpha\:{and}\:\beta\:{be}\:{the}\:{root}\:{of}\:{the}\:{equation} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}.\:{find}\:{the}\:{equation}\:{whose}\:{roots} \\ $$$${are}\:\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right)\:{and}\:\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\beta}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 23/Jun/22
Given equation: ax^2 +bx+c=0 α+β=−(b/a) , αβ=(c/a),α−β=((√(b^2 −4ac))/a) Required equation: Sum of the roots=((1/α)+(1/β))+((1/α)−(1/β)) =((α+β)/(αβ))+((−(α−β))/(αβ)) =(((−b/a)−((√(b^2 −4ac)) )/a)/((c/a)))=((−b−(√(b^2 −4ac)) )/c) Product of roots=((1/α)+(1/β))((1/α)−(1/β)) = (1/α^2 )−(1/β^2 )=((−(α+β)(α−β))/((αβ)^2 )) =((−(−b/a)(−b−(√(b^2 −4ac)) /c))/((c/a)^2 )) =−((ab)/c^3 )(b+(√(b^2 −4ac)) ) x^2 −(Sum of the roots)x+(product of the roots)=0 x^2 −(((−b−(√(b^2 −4ac)) )/c))x−((ab)/c^3 )(b+(√(b^2 −4ac)) )=0 c^3 x^2 +c^2 (b+(√(b^2 −4ac)) )x−ab(b+(√(b^2 −4ac)) )=0
$${Given}\:{equation}: \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\alpha+\beta=−\frac{{b}}{{a}}\:,\:\alpha\beta=\frac{{c}}{{a}},\alpha−\beta=\frac{\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{{a}} \\ $$$${Required}\:{equation}: \\ $$$${Sum}\:{of}\:{the}\:{roots}=\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right)+\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\beta}\right) \\ $$$$\:\:\:\:\:\:\:=\frac{\alpha+\beta}{\alpha\beta}+\frac{−\left(\alpha−\beta\right)}{\alpha\beta} \\ $$$$=\frac{\left(−{b}/{a}\right)−\left(\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\right)/{a}}{\left({c}/{a}\right)}=\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{{c}} \\ $$$${Product}\:{of}\:{roots}=\left(\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\right)\left(\frac{\mathrm{1}}{\alpha}−\frac{\mathrm{1}}{\beta}\right) \\ $$$$\:\:\:\:=\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }−\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{−\left(\alpha+\beta\right)\left(\alpha−\beta\right)}{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{−\left(−{b}/{a}\right)\left(−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:/{c}\right)}{\left({c}/{a}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=−\frac{{ab}}{{c}^{\mathrm{3}} }\left({b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\right) \\ $$$$\:{x}^{\mathrm{2}} −\left({Sum}\:{of}\:{the}\:{roots}\right){x}+\left({product}\:{of}\:{the}\:{roots}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left(\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{{c}}\right){x}−\frac{{ab}}{{c}^{\mathrm{3}} }\left({b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\right)=\mathrm{0} \\ $$$${c}^{\mathrm{3}} {x}^{\mathrm{2}} +{c}^{\mathrm{2}} \left({b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\right){x}−{ab}\left({b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:\right)=\mathrm{0} \\ $$
Commented by Mikenice last updated on 23/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$
Commented by peter frank last updated on 23/Jun/22
thanks
$$\mathrm{thanks} \\ $$

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