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Question Number 10742 by okhema last updated on 24/Feb/17
let the roots of the equation2x^3 −5x^2 +4x+6=0 be α,β and γ. i)state the values of α+β+γ, αβ+αγ+βγ and αβγ. ii)hence or otherwise determine an equation with integer coefficients which as roots (1/α^(2 ) ), (1/β^2 ) , and (1/γ^2 )
$${let}\:{the}\:{roots}\:{of}\:{the}\:{equation}\mathrm{2}{x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{6}=\mathrm{0} \\ $$$${be}\:\alpha,\beta\:{and}\:\gamma. \\ $$$$\left.{i}\right){state}\:{the}\:{values}\:{of}\:\alpha+\beta+\gamma,\:\alpha\beta+\alpha\gamma+\beta\gamma\:{and}\:\alpha\beta\gamma. \\ $$$$\left.{ii}\right){hence}\:{or}\:{otherwise}\:{determine}\:{an}\:{equation}\:{with}\:{integer}\:{coefficients}\:{which}\:{as}\:{roots}\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}\:} },\:\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:,\:{and}\:\frac{\mathrm{1}}{\gamma^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by sandy_suhendra last updated on 24/Feb/17
(i) a=2, b=−5, c=4 and d=6 α+β+γ = ((−b)/a) = (5/2) αβ+αγ+βγ = (c/a) = 2 αβγ = ((−d)/a) = −3
$$\left(\mathrm{i}\right)\:\mathrm{a}=\mathrm{2},\:\mathrm{b}=−\mathrm{5},\:\mathrm{c}=\mathrm{4}\:\mathrm{and}\:\mathrm{d}=\mathrm{6} \\ $$$$\alpha+\beta+\gamma\:=\:\frac{−\mathrm{b}}{\mathrm{a}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\alpha\beta+\alpha\gamma+\beta\gamma\:=\:\frac{\mathrm{c}}{\mathrm{a}}\:=\:\mathrm{2} \\ $$$$\alpha\beta\gamma\:=\:\frac{−\mathrm{d}}{\mathrm{a}}\:=\:−\mathrm{3} \\ $$
Answered by sandy_suhendra last updated on 24/Feb/17
ii) let x_1 =(1/α^2 ) , x_2 =(1/β^2 ) , x_3 =(1/γ^2 ) x_1 +x_2 +x_3 =(1/α^2 )+(1/β^2 )+(1/γ^2 ) =((β^2 γ^2 +α^2 γ^2 +α^2 β^2 )/((αβγ)^2_ )) =(((αβ+αγ+βγ)^2 −2αβγ(α+β+γ))/((αβγ)^2 )) =((2^2 −2(−3)(2.5))/((−3)^2 )) =((19)/9)=((−b)/a) x_1 x_2 +x_1 x_3 +x_2 x_3 =(1/(α^2 β^2 ))+(1/(α^2 γ^2 ))+(1/(β^2 γ^2 )) =((γ^2 +β^2 +α^2 )/(α^2 β^2 γ^2 )) =(((α+β+γ)^2 −2(αβ+αγ+βγ))/((αβγ)^2 )) =(((2.5)^2 −2(2))/((−3)^2 )) =(1/4)=(c/a) x_1 x_2 x_3 =(1/α^2 )×(1/β^2 )×(1/γ^2 )=(1/((αβγ)^2 )) =(1/((−3)^2 ))=(1/9)=((−d)/a) the new equation : x^3 −(((−b)/a))x^2 +((c/a))x−(((−d)/a))=0 x^3 −((19)/9)x^2 +(1/4)x−(1/9)=0 .... (×36) 36x^3 −76x^2 +9x−4=0
$$\left.\mathrm{ii}\right)\: \\ $$$$\mathrm{let}\:\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:,\:\:\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\:,\:\:\:\mathrm{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\gamma^{\mathrm{2}} } \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }+\frac{\mathrm{1}}{\gamma^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }{\left(\alpha\beta\gamma\right)^{\mathrm{2}_{} } } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\alpha\beta+\alpha\gamma+\beta\gamma\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\gamma\left(\alpha+\beta+\gamma\right)}{\left(\alpha\beta\gamma\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{3}\right)\left(\mathrm{2}.\mathrm{5}\right)}{\left(−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{19}}{\mathrm{9}}=\frac{−\mathrm{b}}{\mathrm{a}} \\ $$$$\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} +\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{3}} +\mathrm{x}_{\mathrm{2}} \mathrm{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \gamma^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} \gamma^{\mathrm{2}} }\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\gamma^{\mathrm{2}} +\beta^{\mathrm{2}} +\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\alpha\gamma+\beta\gamma\right)}{\left(\alpha\beta\gamma\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left(\mathrm{2}.\mathrm{5}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{2}\right)}{\left(−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \mathrm{x}_{\mathrm{3}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }×\frac{\mathrm{1}}{\beta^{\mathrm{2}} }×\frac{\mathrm{1}}{\gamma^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\alpha\beta\gamma\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(−\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{9}}=\frac{−\mathrm{d}}{\mathrm{a}} \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{equation}\:: \\ $$$$\mathrm{x}^{\mathrm{3}} −\left(\frac{−\mathrm{b}}{\mathrm{a}}\right)\mathrm{x}^{\mathrm{2}} +\left(\frac{\mathrm{c}}{\mathrm{a}}\right)\mathrm{x}−\left(\frac{−\mathrm{d}}{\mathrm{a}}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{19}}{\mathrm{9}}\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\mathrm{x}−\frac{\mathrm{1}}{\mathrm{9}}=\mathrm{0}\:….\:\left(×\mathrm{36}\right) \\ $$$$\mathrm{36x}^{\mathrm{3}} −\mathrm{76x}^{\mathrm{2}} +\mathrm{9x}−\mathrm{4}=\mathrm{0} \\ $$

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