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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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