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Question-174533




Question Number 174533 by mnjuly1970 last updated on 03/Aug/22
Answered by behi834171 last updated on 03/Aug/22
x^2 =2x+1⇒x^2 −2x−1=0⇒ { ((𝛂+𝛃=2)),((𝛂.𝛃=−1)) :}  x^3 =2x^2 +x=5x+2  x^4 =5x^2 +2x=10x+5+2x=12x+5  x^5 =12x^2 +5x=24x+12+5x=29x+12  x^6 =29x^2 +12x=58x+29+12x=70x+29  x^7 =70x^2 +29x=140x+70+29x=169x+70  ⇒A=(2𝛂+1).(169𝛃+70)+15(2𝛂+1)+𝛃+3=  =338αβ+140α+169β+70+30α+15+β+3=  =338αβ+170(α+β)+88=  =−338×1+170×2+88=90   .■
$$\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{2}\boldsymbol{{x}}+\mathrm{1}\Rightarrow\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}−\mathrm{1}=\mathrm{0}\Rightarrow\begin{cases}{\boldsymbol{\alpha}+\boldsymbol{\beta}=\mathrm{2}}\\{\boldsymbol{\alpha}.\boldsymbol{\beta}=−\mathrm{1}}\end{cases} \\ $$$$\boldsymbol{{x}}^{\mathrm{3}} =\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{x}}=\mathrm{5}\boldsymbol{{x}}+\mathrm{2} \\ $$$$\boldsymbol{{x}}^{\mathrm{4}} =\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{x}}=\mathrm{10}\boldsymbol{{x}}+\mathrm{5}+\mathrm{2}\boldsymbol{{x}}=\mathrm{12}\boldsymbol{{x}}+\mathrm{5} \\ $$$$\boldsymbol{{x}}^{\mathrm{5}} =\mathrm{12}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{5}\boldsymbol{{x}}=\mathrm{24}\boldsymbol{{x}}+\mathrm{12}+\mathrm{5}\boldsymbol{{x}}=\mathrm{29}\boldsymbol{{x}}+\mathrm{12} \\ $$$$\boldsymbol{{x}}^{\mathrm{6}} =\mathrm{29}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{12}\boldsymbol{{x}}=\mathrm{58}\boldsymbol{{x}}+\mathrm{29}+\mathrm{12}\boldsymbol{{x}}=\mathrm{70}\boldsymbol{{x}}+\mathrm{29} \\ $$$$\boldsymbol{{x}}^{\mathrm{7}} =\mathrm{70}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{29}\boldsymbol{{x}}=\mathrm{140}\boldsymbol{{x}}+\mathrm{70}+\mathrm{29}\boldsymbol{{x}}=\mathrm{169}\boldsymbol{{x}}+\mathrm{70} \\ $$$$\Rightarrow{A}=\left(\mathrm{2}\boldsymbol{\alpha}+\mathrm{1}\right).\left(\mathrm{169}\boldsymbol{\beta}+\mathrm{70}\right)+\mathrm{15}\left(\mathrm{2}\boldsymbol{\alpha}+\mathrm{1}\right)+\boldsymbol{\beta}+\mathrm{3}= \\ $$$$=\mathrm{338}\alpha\beta+\mathrm{140}\alpha+\mathrm{169}\beta+\mathrm{70}+\mathrm{30}\alpha+\mathrm{15}+\beta+\mathrm{3}= \\ $$$$=\mathrm{338}\alpha\beta+\mathrm{170}\left(\alpha+\beta\right)+\mathrm{88}= \\ $$$$=−\mathrm{338}×\mathrm{1}+\mathrm{170}×\mathrm{2}+\mathrm{88}=\mathrm{90}\:\:\:.\blacksquare \\ $$
Commented by Rasheed.Sindhi last updated on 03/Aug/22
■ Nice sir!
$$\blacksquare\:\mathrm{Nice}\:\mathrm{sir}! \\ $$

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