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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

	$△ = {(2 m - 1)}^{2} - 4 (m^{2} - 6) = 25 - 4 m$ $△ > 0 \Rightarrow pour m < \frac{25}{4} 2 solutions$ $x_{1} = \frac{2 m - 1 - \sqrt{25 - 4 m}}{2}; x_{2} = \frac{2 m - 1 + \sqrt{25 - 4 m}}{2}$ $∙ x_{1}^{2} = \frac{1}{4} [{4 m}^{2} + 26 - 8 m - 2 (2 m - 1) \sqrt{25 - 4 m}]$ $= m^{2} - 2 m + \frac{13 - (2 m - 1) \sqrt{24 - 4 m}}{2}$ $∙ x_{1} x_{2} = \frac{{(2 m - 1)}^{2} + 4 m - 25}{4} = m^{2} - 6$ $∙ {6 x}_{2}^{2} = \frac{3}{2} [{4 m}^{2} - 8 m + 26 + 2 (2 m - 1) \sqrt{25 - 4 m}]$ $= {6 m}^{2} - 12 m + 39 + 3 (2 m - 1) \sqrt{25 - 4 m}$ $∙ x_{1} - {3 x}_{2} = \frac{2 m - 1 - \sqrt{25 - 4 m}}{2} - \frac{6 m - 3 + 3 \sqrt{25 - 4 m}}{2}$ $= - 2 (m - 1 + 2 \sqrt{25 - 4 m})$ $x_{1}^{2} - {6 x}_{2}^{2} + (x_{1} - {3 x}_{2}) - x_{1} x_{2} =$ $= m^{2} - 2 m + \frac{13 - (2 m - 1) \sqrt{25 - 4 m}}{2}$ $- {6 m}^{2} + 12 m - 39 - 3 \sqrt{25 - 4 m}$ $- 2 m + 2 - 4 \sqrt{25 - 4 m} + 6 - m^{2}$ $\Rightarrow$ $= - {6 m}^{2} + 8 m - 4 (2 m - 1) \sqrt{25 - 4 m} - \frac{49}{2} = 0$ $6 m^{2} - 8 m + 4 (2 m - 1) \sqrt{25 - 4 m} + \frac{49}{2} = 0$ $(1 - 2 m) \sqrt{25 - 4 m} = 6 (m^{2} - \frac{4}{3} m + \frac{49}{12})$ $\sqrt{25 - 4 m} = \frac{6 (m^{2} - 4 m / 3 + 48 / 12)}{1 - 2 m}$ $25 - 4 m = \frac{36 {(m^{2} - \frac{4 m}{3} + \frac{49}{12})}^{2}}{{(1 - 2 m)}^{2}}$ $25 - 104 m + {100 m}^{2} + {16 m}^{2} - {16 m}^{3} =$ $36 (m^{4} - \frac{{8 m}^{3}}{3} + \frac{{4 m}^{2}}{9} + \frac{49^{2}}{144} + \frac{{49 m}^{2}}{6} - \frac{49 m}{3})$ $\Rightarrow {36 m}^{4} - {80 m}^{3} + {278 m}^{2} - 484 m + 2376 = 0$ $m^{4} - \frac{40}{3} m^{3} + \frac{139}{3} m^{2} - \frac{242}{3} m + 396 = 0$ $\Rightarrow m = 5, 86 et m = 8, 092$ $m < \frac{25}{4} \Rightarrow m = 5, 86 (satisfait l [equation)$
	Answered by manxsol last updated on 19/Apr/23
	$x_1 ^2 −x_1 x_2 −6x_2 ^2 =3x_2 −x_1 (x_1 −3x_2 )(x_1 +2x_2 )=3x_2 −x_1 (x_1 −3x_2 )(x_1 +2x_2 +1)=0 x_1 =3x_2 ⇒x_1 +x_2 =4x_2 x_1 x_2 =3x_2 ^2 4x_2 =2m+1 3x_2 ^2 =m^2 −6 3(2m+1)^2 =16m^2 −96 3(4m^2 +4m+1)=16m^2 −96 12m^2 +12m+3=16m^2 −96 4m^2 −12m−99=0 m_1 =−3.69 m_2 =6.69 x_1 +x_2 =−1−2x_2 +x_2 =−1−x_2 x_1 x_2 =(−1−x_2 )x_2 =−x_2 −x_2 ^2 2m+1=−1−x_2 2m=−2−x_2 x_2 =−2−2m m^2 −6=−x_2 −x_2 ^2 m^2 −6=2+2m−(4m^2 +8m+4) 5m^2 +6m−4=0 m_3 =−1.67 m_4 =0.47 restriction Δ>0 ⇒m<6.25 sol={−3.69,−1.67,0.47,6.69}$
	$x_{1}^{2} - x_{1} x_{2} - 6 x_{2}^{2} = 3 x_{2} - x_{1}$ $(x_{1} - 3 x_{2}) (x_{1} + 2 x_{2}) = 3 x_{2} - x_{1}$ $(x_{1} - 3 x_{2}) (x_{1} + 2 x_{2} + 1) = 0$ $x_{1} = 3 x_{2} \Rightarrow x_{1} + x_{2} = 4 x_{2}$ $x_{1} x_{2} = 3 x_{2}^{2}$ $4 x_{2} = 2 m + 1$ $3 x_{2}^{2} = m^{2} - 6$ $3 {(2 m + 1)}^{2} = 16 m^{2} - 96$ $3 (4 m^{2} + 4 m + 1) = 16 m^{2} - 96$ $12 m^{2} + 12 m + 3 = 16 m^{2} - 96$ $4 m^{2} - 12 m - 99 = 0$ $m_{1} = - 3.69$ $m_{2} = 6.69$ $x_{1} + x_{2} = - 1 - 2 x_{2} + x_{2} = - 1 - x_{2}$ $x_{1} x_{2} = (- 1 - x_{2}) x_{2} = - x_{2} - x_{2}^{2}$ $2 m + 1 = - 1 - x_{2}$ $2 m = - 2 - x_{2}$ $x_{2} = - 2 - 2 m$ $m^{2} - 6 = - x_{2} - x_{2}^{2}$ $m^{2} - 6 = 2 + 2 m - (4 m^{2} + 8 m + 4)$ $5 m^{2} + 6 m - 4 = 0$ $m_{3} = - 1.67$ $m_{4} = 0.47$ $r e s t r i c t i o n$ $Δ > 0 \Rightarrow m < 6.25$ $s o l = {- 3.69, - 1.67, 0.47, 6.69}$
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