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Question Number 150600 by mathdanisur last updated on 13/Aug/21
xyz = 10 x + y + z = - 7 xy + xz + yz = 2 Find ((xy)/z) + ((xz)/y) + ((yz)/x) = ?
$$\mathrm{xyz}\:=\:\mathrm{10} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:-\:\mathrm{7} \\ $$$$\mathrm{xy}\:+\:\mathrm{xz}\:+\:\mathrm{yz}\:=\:\mathrm{2} \\ $$$$\mathrm{Find}\:\:\frac{\mathrm{xy}}{\mathrm{z}}\:+\:\frac{\mathrm{xz}}{\mathrm{y}}\:+\:\frac{\mathrm{yz}}{\mathrm{x}}\:=\:? \\ $$
Answered by amin96 last updated on 13/Aug/21
x^3 +7x^2 +2x−10=0 { ((x=x_1 )),((y=x_2 )),((z=x_3 )) :} (((x_1 x_2 )^2 +(x_1 x_3 )^2 +(x_2 x_3 )^2 )/(x_1 x_2 x_3 ))= =(((x_1 x_2 +x_1 x_3 +x_2 x_3 )^2 −2x_1 ^2 x_2 x_3 −2x_2 ^2 x_1 x_3 −2x_3 ^2 x_1 x_2 =)/(x_1 x_2 x_3 ))= =((4−2x_1 x_2 x_3 (x_1 +x_2 +x_3 ))/(x_1 x_2 x_3 ))=((4−20∙(−7))/(10))=((144)/(10))=14,4
$${x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{10}=\mathrm{0}\:\:\:\begin{cases}{{x}={x}_{\mathrm{1}} }\\{{y}={x}_{\mathrm{2}} }\\{{z}={x}_{\mathrm{3}} }\end{cases} \\ $$$$\frac{\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({x}_{\mathrm{1}} {x}_{\mathrm{3}} \right)^{\mathrm{2}} +\left({x}_{\mathrm{2}} {x}_{\mathrm{3}} \right)^{\mathrm{2}} }{{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} }= \\ $$$$=\frac{\left({x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} {x}_{\mathrm{3}} +{x}_{\mathrm{2}} {x}_{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{2}{x}_{\mathrm{1}} ^{\mathrm{2}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} −\mathrm{2}{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{1}} {x}_{\mathrm{3}} −\mathrm{2}{x}_{\mathrm{3}} ^{\mathrm{2}} {x}_{\mathrm{1}} {x}_{\mathrm{2}} =}{{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} }= \\ $$$$=\frac{\mathrm{4}−\mathrm{2}{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} \left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} \right)}{{x}_{\mathrm{1}} {x}_{\mathrm{2}} {x}_{\mathrm{3}} }=\frac{\mathrm{4}−\mathrm{20}\centerdot\left(−\mathrm{7}\right)}{\mathrm{10}}=\frac{\mathrm{144}}{\mathrm{10}}=\mathrm{14},\mathrm{4} \\ $$
Commented by mathdanisur last updated on 14/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$
Answered by puissant last updated on 14/Aug/21
let δ_1 =x+y+z ; δ_2 =xy+yz+xz ; δ_3 =xyz P(x)=x^3 −δ_1 x^2 +δ_2 x−δ_3 ⇒P=x^3 +7x^2 +2x−10 P=(x−x_1 )(x−x_2 )(x−x_3 ) P=(x−1)(x^2 +8x+10) ⇒ P=(x−1)(x+4+(√6))(x+4−(√6)) take x=1 , y=−4−(√6) , z=−4+(√6).. ((xy)/z)+((xz)/y)+((yz)/x) = ((−4−(√6))/(−4+(√6))) +((−4+(√6))/(−4−(√6)))+(4^2 −((√6))^2 ) = 14,4...
$${let}\:\:\delta_{\mathrm{1}} ={x}+{y}+{z}\:;\:\delta_{\mathrm{2}} ={xy}+{yz}+{xz}\:;\:\delta_{\mathrm{3}} ={xyz} \\ $$$${P}\left({x}\right)={x}^{\mathrm{3}} −\delta_{\mathrm{1}} {x}^{\mathrm{2}} +\delta_{\mathrm{2}} {x}−\delta_{\mathrm{3}} \\ $$$$\Rightarrow{P}={x}^{\mathrm{3}} +\mathrm{7}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{10} \\ $$$${P}=\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\left({x}−{x}_{\mathrm{3}} \right) \\ $$$${P}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{10}\right) \\ $$$$\Rightarrow\:{P}=\left({x}−\mathrm{1}\right)\left({x}+\mathrm{4}+\sqrt{\mathrm{6}}\right)\left({x}+\mathrm{4}−\sqrt{\mathrm{6}}\right) \\ $$$${take}\:\:{x}=\mathrm{1}\:,\:{y}=−\mathrm{4}−\sqrt{\mathrm{6}}\:,\:{z}=−\mathrm{4}+\sqrt{\mathrm{6}}.. \\ $$$$\frac{{xy}}{{z}}+\frac{{xz}}{{y}}+\frac{{yz}}{{x}}\:=\:\frac{−\mathrm{4}−\sqrt{\mathrm{6}}}{−\mathrm{4}+\sqrt{\mathrm{6}}}\:+\frac{−\mathrm{4}+\sqrt{\mathrm{6}}}{−\mathrm{4}−\sqrt{\mathrm{6}}}+\left(\mathrm{4}^{\mathrm{2}} −\left(\sqrt{\mathrm{6}}\right)^{\mathrm{2}} \right) \\ $$$$\:=\:\mathrm{14},\mathrm{4}… \\ $$
Commented by mathdanisur last updated on 14/Aug/21
Thank you See
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{See} \\ $$
Answered by maged last updated on 14/Aug/21
(xy + xz + yz )^2 = 4 ⇒(xy)^2 + (xz)^2 +( yz)^2 +2xyz(x+y+z) = 4 ⇒(xy)^2 + (xz)^2 +( yz)^2 +2(10)(−7) = 4 ⇒(xy)^2 + (xz)^2 +( yz)^2 = 4+140 ⇒(xy)^2 + (xz)^2 +( yz)^2 = 144.....(1) ∵xyz=10 ,((xy)/z)=((10)/z^2 ) , ((xz)/y)=((10)/y^2 ) , ((yz)/x)=((10)/x^2 ) ⇒((xy)/z)+((xz)/y)+((yz)/x)=10((1/x^2 )+(1/y^2 )+(1/z^2 ))=(((xy)^2 +(xz)^2 +(yz)^2 )/(10))....(2) from eq(1) & eq(2) we get ⇒((xy)/z)+((xz)/y)+((yz)/x)=((144)/(10))=14.4
$$\left(\mathrm{xy}\:+\:\mathrm{xz}\:+\:\mathrm{yz}\:\right)^{\mathrm{2}} =\:\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{xy}\right)^{\mathrm{2}} \:+\:\left(\mathrm{xz}\right)^{\mathrm{2}} \:+\left(\:\mathrm{yz}\right)^{\mathrm{2}} +\mathrm{2xyz}\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\:=\:\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{xy}\right)^{\mathrm{2}} \:+\:\left(\mathrm{xz}\right)^{\mathrm{2}} \:+\left(\:\mathrm{yz}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{10}\right)\left(−\mathrm{7}\right)\:=\:\mathrm{4} \\ $$$$\Rightarrow\left(\mathrm{xy}\right)^{\mathrm{2}} \:+\:\left(\mathrm{xz}\right)^{\mathrm{2}} \:+\left(\:\mathrm{yz}\right)^{\mathrm{2}} \:=\:\mathrm{4}+\mathrm{140} \\ $$$$\Rightarrow\left(\mathrm{xy}\right)^{\mathrm{2}} \:+\:\left(\mathrm{xz}\right)^{\mathrm{2}} \:+\left(\:\mathrm{yz}\right)^{\mathrm{2}} \:=\:\mathrm{144}…..\left(\mathrm{1}\right) \\ $$$$\because\mathrm{xyz}=\mathrm{10}\:\:,\frac{\mathrm{xy}}{\mathrm{z}}=\frac{\mathrm{10}}{\mathrm{z}^{\mathrm{2}} }\:,\:\frac{\mathrm{xz}}{\mathrm{y}}=\frac{\mathrm{10}}{\mathrm{y}^{\mathrm{2}} }\:,\:\frac{\mathrm{yz}}{\mathrm{x}}=\frac{\mathrm{10}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{xy}}{\mathrm{z}}+\frac{\mathrm{xz}}{\mathrm{y}}+\frac{\mathrm{yz}}{\mathrm{x}}=\mathrm{10}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\right)=\frac{\left(\mathrm{xy}\right)^{\mathrm{2}} +\left(\mathrm{xz}\right)^{\mathrm{2}} +\left(\mathrm{yz}\right)^{\mathrm{2}} }{\mathrm{10}}….\left(\mathrm{2}\right) \\ $$$$\mathrm{from}\:\mathrm{eq}\left(\mathrm{1}\right)\:\&\:\mathrm{eq}\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\Rightarrow\frac{\mathrm{xy}}{\mathrm{z}}+\frac{\mathrm{xz}}{\mathrm{y}}+\frac{\mathrm{yz}}{\mathrm{x}}=\frac{\mathrm{144}}{\mathrm{10}}=\mathrm{14}.\mathrm{4} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mathdanisur last updated on 14/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

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