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Question-191131

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Question Number 191131 by TUN last updated on 18/Apr/23
Answered by a.lgnaoui last updated on 19/Apr/23
△=(2m−1)^2 −4(m^2 −6)=25−4m △>0 ⇒pour m<((25)/4) 2 solutions x_1 =((2m−1−(√(25−4m)))/2); x_2 =((2m−1+(√(25−4m)))/2) • x_1 ^2 =(1/4)[4m^2 +26−8m−2(2m−1)(√(25−4m)) ] =m^2 −2m+((13−(2m−1)(√(24−4m)))/2) •x_1 x_2 =(((2m−1)^2 +4m−25)/4)=m^2 −6 •6x_2 ^2 =(3/2)[4m^2 −8m+26+2(2m−1)(√(25−4m)) ] =6m^2 −12m+39+3(2m−1)(√(25−4m)) • x_1 −3x_2 =((2m−1−(√(25−4m)) )/2)−((6m−3+3(√(25−4m)))/2) =−2(m−1+2(√(25−4m)) ) x_1 ^2 −6x_2 ^2 +(x_1 −3x_2 )−x_1 x_2 = =m^2 −2m+((13−(2m−1)(√(25−4m)))/2) −6m^2 +12m−39−3(√(25−4m)) −2m+2−4(√(25−4m)) +6−m^2 ⇒ =−6m^2 +8m−4(2m−1)(√(25−4m)) −((49)/2)=0 6m^2 −8m+4(2m−1)(√(25−4m)) +((49)/2)=0 (1−2m )(√(25−4m)) =6(m^2 −(4/3)m+((49)/(12))) (√(25−4m ))=((6(m^2 −4m/3+48/12))/(1−2m)) 25−4m=((36(m^2 −((4m)/3)+((49)/(12)))^2 )/((1−2m)^2 )) 25−104m+100m^2 +16m^2 −16m^3 = 36(m^4 −((8m^3 )/3)+((4m^2 )/9)+((49^2 )/(144))+((49m^2 )/6)−((49m)/3)) ⇒36m^4 −80m^3 +278m^2 −484m+2376=0 m^4 −((40)/3)m^3 +((139)/3)m^2 −((242)/3)m+396=0 ⇒m=5,86 et m=8,092 m<((25)/4) ⇒m=5,86 (satisfait l[equation)
$$\bigtriangleup=\left(\mathrm{2m}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{6}\right)=\mathrm{25}−\mathrm{4m} \\ $$$$\bigtriangleup>\mathrm{0}\:\:\:\Rightarrow\mathrm{pour}\:\:\:\:\mathrm{m}<\frac{\mathrm{25}}{\mathrm{4}}\:\:\:\:\:\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{2m}−\mathrm{1}−\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}};\:\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2m}−\mathrm{1}+\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\bullet\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{4m}^{\mathrm{2}} +\mathrm{26}−\mathrm{8m}−\mathrm{2}\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4m}}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{m}^{\mathrm{2}} −\mathrm{2m}+\frac{\mathrm{13}−\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{24}−\mathrm{4m}}}{\mathrm{2}} \\ $$$$\:\:\bullet\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} =\frac{\left(\mathrm{2m}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4m}−\mathrm{25}}{\mathrm{4}}=\mathrm{m}^{\mathrm{2}} −\mathrm{6} \\ $$$$\:\:\:\: \\ $$$$\:\bullet\mathrm{6x}_{\mathrm{2}} ^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}\left[\mathrm{4m}^{\mathrm{2}} −\mathrm{8m}+\mathrm{26}+\mathrm{2}\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4m}}\:\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{6m}^{\mathrm{2}} −\mathrm{12m}+\mathrm{39}+\mathrm{3}\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4m}}\: \\ $$$$ \\ $$$$\:\:\bullet\:\:\mathrm{x}_{\mathrm{1}} −\mathrm{3x}_{\mathrm{2}} =\frac{\mathrm{2m}−\mathrm{1}−\sqrt{\mathrm{25}−\mathrm{4m}}\:}{\mathrm{2}}−\frac{\mathrm{6m}−\mathrm{3}+\mathrm{3}\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}\left(\mathrm{m}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{25}−\mathrm{4m}}\:\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{6x}_{\mathrm{2}} ^{\mathrm{2}} +\left(\mathrm{x}_{\mathrm{1}} −\mathrm{3x}_{\mathrm{2}} \right)−\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} = \\ $$$$=\mathrm{m}^{\mathrm{2}} −\mathrm{2m}+\frac{\mathrm{13}−\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}} \\ $$$$−\mathrm{6m}^{\mathrm{2}} +\mathrm{12m}−\mathrm{39}−\mathrm{3}\sqrt{\mathrm{25}−\mathrm{4m}}\: \\ $$$$−\mathrm{2m}+\mathrm{2}−\mathrm{4}\sqrt{\mathrm{25}−\mathrm{4m}}\:\:\:+\mathrm{6}−\mathrm{m}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$=−\mathrm{6m}^{\mathrm{2}} +\mathrm{8m}−\mathrm{4}\left(\mathrm{2m}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4m}}\:\:−\frac{\mathrm{49}}{\mathrm{2}}=\mathrm{0} \\ $$$$\:\mathrm{6}\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\mathrm{8}\boldsymbol{\mathrm{m}}+\mathrm{4}\left(\mathrm{2}\boldsymbol{\mathrm{m}}−\mathrm{1}\right)\sqrt{\mathrm{25}−\mathrm{4}\boldsymbol{\mathrm{m}}}\:+\frac{\mathrm{49}}{\mathrm{2}}=\mathrm{0} \\ $$$$ \\ $$$$\:\left(\mathrm{1}−\mathrm{2}\boldsymbol{\mathrm{m}}\:\right)\sqrt{\mathrm{25}−\mathrm{4}\boldsymbol{\mathrm{m}}}\:=\mathrm{6}\left(\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\mathrm{m}}+\frac{\mathrm{49}}{\mathrm{12}}\right) \\ $$$$\sqrt{\mathrm{25}−\mathrm{4m}\:}=\frac{\mathrm{6}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{4m}/\mathrm{3}+\mathrm{48}/\mathrm{12}\right)}{\mathrm{1}−\mathrm{2m}} \\ $$$$\mathrm{25}−\mathrm{4m}=\frac{\mathrm{36}\left(\mathrm{m}^{\mathrm{2}} −\frac{\mathrm{4m}}{\mathrm{3}}+\frac{\mathrm{49}}{\mathrm{12}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\mathrm{2m}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{25}−\mathrm{104m}+\mathrm{100m}^{\mathrm{2}} +\mathrm{16m}^{\mathrm{2}} −\mathrm{16m}^{\mathrm{3}} = \\ $$$$\mathrm{36}\left(\mathrm{m}^{\mathrm{4}} −\frac{\mathrm{8m}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{4m}^{\mathrm{2}} }{\mathrm{9}}+\frac{\mathrm{49}^{\mathrm{2}} }{\mathrm{144}}+\frac{\mathrm{49m}^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{49m}}{\mathrm{3}}\right) \\ $$$$ \\ $$$$\Rightarrow\mathrm{36m}^{\mathrm{4}} −\mathrm{80m}^{\mathrm{3}} +\mathrm{278m}^{\mathrm{2}} −\mathrm{484m}+\mathrm{2376}=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{m}}^{\mathrm{4}} −\frac{\mathrm{40}}{\mathrm{3}}\boldsymbol{\mathrm{m}}^{\mathrm{3}} +\frac{\mathrm{139}}{\mathrm{3}}\boldsymbol{\mathrm{m}}^{\mathrm{2}} −\frac{\mathrm{242}}{\mathrm{3}}\boldsymbol{\mathrm{m}}+\mathrm{396}=\mathrm{0} \\ $$$$ \\ $$$$\Rightarrow\boldsymbol{\mathrm{m}}=\mathrm{5},\mathrm{86}\:\:\:\mathrm{et}\:\:\mathrm{m}=\mathrm{8},\mathrm{092} \\ $$$$ \\ $$$$\:\:\mathrm{m}<\frac{\mathrm{25}}{\mathrm{4}}\:\:\Rightarrow\boldsymbol{\mathrm{m}}=\mathrm{5},\mathrm{86}\:\left(\boldsymbol{\mathrm{satisfait}}\:\boldsymbol{\mathrm{l}}\left[\boldsymbol{\mathrm{equation}}\right)\right. \\ $$$$ \\ $$
Answered by manxsol last updated on 19/Apr/23
x_1 ^2 −x_1 x_2 −6x_2 ^2 =3x_2 −x_1 (x_1 −3x_2 )(x_1 +2x_2 )=3x_2 −x_1 (x_1 −3x_2 )(x_1 +2x_2 +1)=0 x_1 =3x_2 ⇒x_1 +x_2 =4x_2 x_1 x_2 =3x_2 ^2 4x_2 =2m+1 3x_2 ^2 =m^2 −6 3(2m+1)^2 =16m^2 −96 3(4m^2 +4m+1)=16m^2 −96 12m^2 +12m+3=16m^2 −96 4m^2 −12m−99=0 m_1 =−3.69 m_2 =6.69 x_1 +x_2 =−1−2x_2 +x_2 =−1−x_2 x_1 x_2 =(−1−x_2 )x_2 =−x_2 −x_2 ^2 2m+1=−1−x_2 2m=−2−x_2 x_2 =−2−2m m^2 −6=−x_2 −x_2 ^2 m^2 −6=2+2m−(4m^2 +8m+4) 5m^2 +6m−4=0 m_3 =−1.67 m_4 =0.47 restriction Δ>0 ⇒m<6.25 sol={−3.69,−1.67,0.47,6.69}
$${x}_{\mathrm{1}} ^{\mathrm{2}} −{x}_{\mathrm{1}} {x}_{\mathrm{2}} −\mathrm{6}{x}_{\mathrm{2}} ^{\mathrm{2}} =\mathrm{3}{x}_{\mathrm{2}} −{x}_{\mathrm{1}} \\ $$$$\left({x}_{\mathrm{1}} \:−\mathrm{3}{x}_{\mathrm{2}} \right)\left({x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} \right)=\mathrm{3}{x}_{\mathrm{2}} −{x}_{\mathrm{1}} \\ $$$$\left({x}_{\mathrm{1}} −\mathrm{3}{x}_{\mathrm{2}} \right)\left({x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{3}{x}_{\mathrm{2}} \Rightarrow{x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\mathrm{4}{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\mathrm{3}{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{4}{x}_{\mathrm{2}} =\mathrm{2}{m}+\mathrm{1} \\ $$$$\mathrm{3}{x}_{\mathrm{2}} ^{\mathrm{2}} ={m}^{\mathrm{2}} −\mathrm{6} \\ $$$$\mathrm{3}\left(\mathrm{2}{m}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{16}{m}^{\mathrm{2}} −\mathrm{96} \\ $$$$\mathrm{3}\left(\mathrm{4}{m}^{\mathrm{2}} +\mathrm{4}{m}+\mathrm{1}\right)=\mathrm{16}{m}^{\mathrm{2}} −\mathrm{96} \\ $$$$\mathrm{12}{m}^{\mathrm{2}} +\mathrm{12}{m}+\mathrm{3}=\mathrm{16}{m}^{\mathrm{2}} −\mathrm{96} \\ $$$$\mathrm{4}{m}^{\mathrm{2}} −\mathrm{12}{m}−\mathrm{99}=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =−\mathrm{3}.\mathrm{69} \\ $$$${m}_{\mathrm{2}} =\mathrm{6}.\mathrm{69} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\mathrm{1}−\mathrm{2}{x}_{\mathrm{2}} +{x}_{\mathrm{2}} =−\mathrm{1}−{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\left(−\mathrm{1}−{x}_{\mathrm{2}} \right){x}_{\mathrm{2}} =−{x}_{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$$\mathrm{2}{m}+\mathrm{1}=−\mathrm{1}−{x}_{\mathrm{2}} \\ $$$$\mathrm{2}{m}=−\mathrm{2}−{x}_{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}−\mathrm{2}{m} \\ $$$${m}^{\mathrm{2}} −\mathrm{6}=−{x}_{\mathrm{2}} −{x}_{\mathrm{2}} ^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −\mathrm{6}=\mathrm{2}+\mathrm{2}{m}−\left(\mathrm{4}{m}^{\mathrm{2}} +\mathrm{8}{m}+\mathrm{4}\right) \\ $$$$\mathrm{5}{m}^{\mathrm{2}} +\mathrm{6}{m}−\mathrm{4}=\mathrm{0} \\ $$$${m}_{\mathrm{3}} =−\mathrm{1}.\mathrm{67} \\ $$$${m}_{\mathrm{4}} =\mathrm{0}.\mathrm{47} \\ $$$${restriction} \\ $$$$\Delta>\mathrm{0}\:\Rightarrow{m}<\mathrm{6}.\mathrm{25} \\ $$$${sol}=\left\{−\mathrm{3}.\mathrm{69},−\mathrm{1}.\mathrm{67},\mathrm{0}.\mathrm{47},\cancel{\mathrm{6}.\mathrm{69}}\right\} \\ $$$$ \\ $$

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