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Question Number 12543 by @ANTARES_VY last updated on 25/Apr/17
This x^2 −𝛂x+𝛂−1=0. the roots of the equation x_1 and x_2 a what′s the value of x_1 ^2 +x_2 ^2 this collection of smille(minimum) value.
$$\boldsymbol{\mathrm{This}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\alpha\mathrm{x}}+\boldsymbol{\alpha}−\mathrm{1}=\mathrm{0}. \\ $$$$\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{roots}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{equation}}\:\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} \:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{a}}\:\:\boldsymbol{\mathrm{what}}'\boldsymbol{\mathrm{s}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{value}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{x}}_{\mathrm{2}} ^{\mathrm{2}} \:\:\boldsymbol{\mathrm{this}}\:\:\boldsymbol{\mathrm{collection}}\:\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{smille}}\left(\boldsymbol{\mathrm{minimum}}\right)\:\:\boldsymbol{\mathrm{value}}. \\ $$
Answered by mrW1 last updated on 25/Apr/17
x^2 −αx+α−1=(x−x_1 )(x−x_2 ) x^2 −αx+α−1=x^x −(x_1 +x_2 )x+x_1 x_2 x_1 +x_2 =α x_1 x_2 =α−1 x_1 ^2 +x_2 ^2 +2x_1 x_2 =α^2 x_1 ^2 +x_2 ^2 =α^2 −2α+2=(α−1)^2 +1≥1
$${x}^{\mathrm{2}} −\alpha{x}+\alpha−\mathrm{1}=\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right) \\ $$$${x}^{\mathrm{2}} −\alpha{x}+\alpha−\mathrm{1}={x}^{{x}} −\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right){x}+{x}_{\mathrm{1}} {x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\alpha \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\alpha−\mathrm{1} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\alpha^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} =\alpha^{\mathrm{2}} −\mathrm{2}\alpha+\mathrm{2}=\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1} \\ $$
Commented by @ANTARES_VY last updated on 25/Apr/17
(𝛂−1)^2 +1≥1 ?????
$$\left(\boldsymbol{\alpha}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1}\:\:\:\:????? \\ $$
Commented by @ANTARES_VY last updated on 25/Apr/17
I do not understand
$$\boldsymbol{\mathrm{I}}\:\:\boldsymbol{\mathrm{do}}\:\:\boldsymbol{\mathrm{not}}\:\:\boldsymbol{\mathrm{understand}} \\ $$
Commented by mrW1 last updated on 25/Apr/17
since (α−1)^2 ≥0 hence (α−1)^2 +1≥1
$${since}\:\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${hence}\:\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\geqslant\mathrm{1} \\ $$
Answered by ridwan balatif last updated on 25/Apr/17
x^2 −αx+α−1=0 x_1 +x_2 =α x_1 .x_2 =α−1 let p=x_1 ^2 +x_2 ^2 p=(x_1 +x_2 )^2 −2x_1 x_2 p=(α)^2 −2(α−1) p=α^2 −2α+2 p will minimum if (dp/dα)=0 2α−2=0 α=1 so the minimum value of p=x_1 ^2 +x_2 ^2 =α^2 −2α+2 =1^2 −2.1+2 p=1 OR i think this is what MrW1think p=α^2 −2α+2 p=(α−1)^2 −1+2 p=(α−1)^2 +1 p will minimum if (α−1)^2 =0, so p=1
$$\mathrm{x}^{\mathrm{2}} −\alpha\mathrm{x}+\alpha−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =\alpha \\ $$$$\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} =\alpha−\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{p}=\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{p}=\left(\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \\ $$$$\mathrm{p}=\left(\alpha\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha−\mathrm{1}\right) \\ $$$$\mathrm{p}=\alpha^{\mathrm{2}} −\mathrm{2}\alpha+\mathrm{2} \\ $$$$\mathrm{p}\:\mathrm{will}\:\mathrm{minimum}\:\mathrm{if}\:\frac{\mathrm{dp}}{\mathrm{d}\alpha}=\mathrm{0} \\ $$$$\mathrm{2}\alpha−\mathrm{2}=\mathrm{0} \\ $$$$\alpha=\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{p}=\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\:\:\:=\alpha^{\mathrm{2}} −\mathrm{2}\alpha+\mathrm{2} \\ $$$$\:\:\:=\mathrm{1}^{\mathrm{2}} −\mathrm{2}.\mathrm{1}+\mathrm{2} \\ $$$$\mathrm{p}=\mathrm{1} \\ $$$$\: \\ $$$$\mathrm{OR} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{is}\:\mathrm{what}\:\mathrm{MrW1think} \\ $$$$\mathrm{p}=\alpha^{\mathrm{2}} −\mathrm{2}\alpha+\mathrm{2} \\ $$$$\mathrm{p}=\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}+\mathrm{2} \\ $$$$\mathrm{p}=\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{p}\:\mathrm{will}\:\mathrm{minimum}\:\mathrm{if}\:\left(\alpha−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0},\:\mathrm{so} \\ $$$$\mathrm{p}=\mathrm{1} \\ $$

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