Respuesta :

[tex] {x}^{4} - 5 {x}^{3 } + 7 {x}^{2} - 5x + 6 = 0 \\ - 5x( {x}^{2} + 1) + ( {x}^{2} )^{2} + 7 {x}^{2} + 6 = 0 \\ {x}^{2} = u \\ - 5x( {u} + 1) + (u^{2} + 7 u + 6) =0[/tex]

[tex]u^{2} + 7 u + 6=0 \\ u_1 = - 1 \\ u_2= - 6 \\ (u-u_1)(u-u_2) \\ u^{2} + 7 u + 6 = (u + 1)(u + 6)[/tex]

[tex]- 5x( {u} + 1) + u^{2} + 7 u + 6 = 0 \\ - 5x(u + 1) + (u + 1)(u + 6) = 0 \\ u = {x}^{2} \\ - 5x( {x}^{2} + 1) + ( {x}^{2} + 1)( {x}^{2} + 6) =0[/tex]

[tex]( {x}^{2} + 1)( {x}^{2} - 5x + 6) = 0[/tex]

Now we have 2 equations

[tex] {x}^{2} + 1 = 0 \\ {x}^{2} = - 1 \\ x_1 = i \\ x_2 = - i[/tex]

[tex] {x}^{2} - 5x + 6 = 0 \\ D = {( - 5)}^{2} - 4 \times 6 = 1 = {1}^{2} \\ x_3= \frac{5 + 1}{2} = 3 \\ x_4 = \frac{5 - 1}{2} = 2[/tex]

[tex]Answer:x_1=-i, x_2=i, x_3=2, \\ x_4=3[/tex]

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