Odpowiedź:
1.
[tex] {x}^{4} + 2 {x}^{2} = 4 {x}^{3} \\ {x}^{4} - 4 {x}^{3} + 2 {x}^{2} = 0 \\ {x}^{2} ( {x}^{2} - 4x + 2) = 0 \\ {x}^{2} = 0 \\ {x}^{2} - 4x + 2 = 0 \\ ∆ = {( - 4)}^{2} - 4 \times 2 \times 1 = 16 - 8 = 8 \\ \sqrt{∆} = \sqrt{8} = 2 \sqrt{2} \\ x_{1} = \frac{ 4 - 2 \sqrt{2} }{2} = \frac{2(2 - \sqrt{2}) }{2} = 2 - \sqrt{2} \\ x_{2} = \frac{4 + 2 \sqrt{2} }{2} = \frac{2(2 + \sqrt{2}) }{2} = 2 + \sqrt{2} \\ [/tex]
czyli x = ...
[tex]x_{1} = 0 \\ x_{2} = 2 - \sqrt{2} \\ x_{3} = 2 + \sqrt{2} [/tex]
2.
[tex]( {x}^{2} + 6x + 2)(8 {x}^{2} - 4x) = 0 \\[/tex]
obliczymy dwie delty z jednego i z drugiego nawiasu
[tex] {x}^{2} + 6x + 2 = 0 \\ ∆ = {6}^{2} - 4 \times 2 \times 1 = 36 - 8 = 28 \\ \sqrt{∆} = \sqrt{28} = 2 \sqrt{7} \\ x_{1} = \frac{ - 6 - 2 \sqrt{7} }{2} = \frac{2( - 3 - \sqrt{7} )}{2} = - 3 - \sqrt{7} \\ x_{2} = \frac{ - 6 + 2 \sqrt{7} }{2} = \frac{2( - 3 + \sqrt{7}) }{2} = - 3 + \sqrt{7} [/tex]
[tex]8 {x}^{2} - 4x = 0 \\ ∆ = ( - 4 {)}^{2} - 4 \times 0 \times 8 = 16 \\ \sqrt{∆} = \sqrt{16} = 4 \\ x_{1} = \frac{4 - 4}{16} = 0 \\ x_{2} = \frac{4 + 4}{16} = \frac{8}{16} = \frac{1}{2} [/tex]
czyli x = ...
[tex]x_{1} = - 3 - \sqrt{7} \\ x_{2} = - 3 + \sqrt{7} \\ x_{3} = 0 \\ x_{4} = \frac{1}{2} [/tex]
3.
[tex](2 {x}^{2} +x + 1)(3 {x}^{2} + 12) = 0[/tex]
obliczymy dwie delty z jednego i z drugiego nawiasu
[tex]2 {x}^{2} + x + 1 = 0 \\ ∆ = {1}^{2} - 4 \times 2 \times 1 = 1 - 8 = - 9 \\ ∆ < 0[/tex]
brak miejsc zerowych
[tex]3 {x}^{2} + 12 = 0 \\ ∆ = 0 - 4 \times 3 \times 12 = 0 - 144 = - 144 \\ ∆ < 0[/tex]
brak miejsc zerowych
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