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If-m-r-1-m-r-r-1-2-3-4-be-four-pairs-of-values-of-x-and-y-satisfy-the-equation-x-2-y-2-2gx-2fy-c-0-then-prove-that-m-1-m-2-m-3-m-4-1-

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Question Number 20296 by Tinkutara last updated on 25/Aug/17
If (m_r , (1/m_r )) ; r = 1, 2, 3, 4 be four pairs of values of x and y satisfy the equation x^2 + y^2 + 2gx + 2fy + c = 0, then prove that m_1 .m_2 .m_3 .m_4 = 1.
$$\mathrm{If}\:\left({m}_{{r}} \:,\:\frac{\mathrm{1}}{{m}_{{r}} }\right)\:;\:{r}\:=\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\:\mathrm{be}\:\mathrm{four}\:\mathrm{pairs} \\ $$$$\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:\mathrm{2}{gx}\:+\:\mathrm{2}{fy}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{m}_{\mathrm{1}} .{m}_{\mathrm{2}} .{m}_{\mathrm{3}} .{m}_{\mathrm{4}} \:=\:\mathrm{1}. \\ $$
Answered by ajfour last updated on 25/Aug/17
(m_r +g)^2 +((1/m_r )+f)^2 =g^2 +f^2 −c m_r ^2 (m_r +g)^2 −m_r ^2 (g^2 +f^2 −c) +(fm_r +1)^2 =0 ⇒ m_r are roots of above equation whose constant term is 1 and coefficint of m_r ^4 is also 1. so m_1 .m_2 .m_3 .m_4 =1
$$\left({m}_{{r}} +{g}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{m}_{{r}} }+{f}\right)^{\mathrm{2}} ={g}^{\mathrm{2}} +{f}^{\mathrm{2}} −{c} \\ $$$${m}_{{r}} ^{\mathrm{2}} \left({m}_{{r}} +{g}\right)^{\mathrm{2}} −{m}_{{r}} ^{\mathrm{2}} \left({g}^{\mathrm{2}} +{f}^{\mathrm{2}} −{c}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({fm}_{{r}} +\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{m}_{{r}} \:{are}\:{roots}\:{of}\:{above}\:\:{equation} \\ $$$${whose}\:{constant}\:{term}\:{is}\:\mathrm{1}\:{and} \\ $$$${coefficint}\:{of}\:{m}_{{r}} ^{\mathrm{4}} \:{is}\:{also}\:\mathrm{1}. \\ $$$${so}\:\:\:{m}_{\mathrm{1}} .{m}_{\mathrm{2}} .{m}_{\mathrm{3}} .{m}_{\mathrm{4}} =\mathrm{1}\: \\ $$
Commented by Tinkutara last updated on 25/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by Tinkutara last updated on 25/Aug/17
Let f(x) = x^4 + 2gx^3 + cx^2 + 2fx + 1 = x^2 (x^2 + (1/x^2 ) + 2gx + ((2f)/x) + c) f(x) has roots m_1 , m_2 , m_3 , m_4 so m_1 m_2 m_3 m_4 = 1 by Vieta.
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{4}} \:+\:\mathrm{2}{gx}^{\mathrm{3}} \:+\:{cx}^{\mathrm{2}} \:+\:\mathrm{2}{fx}\:+\:\mathrm{1} \\ $$$$=\:{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\:\mathrm{2}{gx}\:+\:\frac{\mathrm{2}{f}}{{x}}\:+\:{c}\right) \\ $$$${f}\left({x}\right)\:\mathrm{has}\:\mathrm{roots}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} ,\:{m}_{\mathrm{3}} ,\:{m}_{\mathrm{4}} \:\mathrm{so} \\ $$$${m}_{\mathrm{1}} {m}_{\mathrm{2}} {m}_{\mathrm{3}} {m}_{\mathrm{4}} \:=\:\mathrm{1}\:\mathrm{by}\:\mathrm{Vieta}. \\ $$

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