Odpowiedź:
zad. 1
[tex]a) {x}^{4} - 5 {x}^{2} + 4 = 0 \\ t = {x}^{2} \\ {t}^{2} - 5t + 4 = 0 \\ ∆ = {( - 5)}^{2} - 4 \times 4 \times 1 = 25 - 16 = 9 \\ \sqrt{∆} = \sqrt{9} = 3 \\ t_{1} = \frac{5 - 3}{2} = \frac{2}{2} = 1 \\ t_{2} = \frac{5 + 3}{2} = \frac{8}{2} = 4 \\ \\ {x}^{2} = 1 \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} = 4 \\ x_{1} = 1 \: \: \: \: \: \: x_{2} = - 1 \: \: \: \: \: \: \: \: x_{3} = 2 \: \: \: \: \: \: \: \: x_{4} = - 2[/tex]
[tex]b) {x}^{4} + 8 {x}^{2} - 9 = 0 \\ t = {x}^{2} \\ {t}^{2} + 8t - 9 = 0 \\ ∆ = {8}^{2} - 4 \times 1 \times ( - 9) = 64 + 36 = 100 \\ \sqrt{∆} = \sqrt{100} = 10 \\ t_{1} = \frac{ - 8 - 10}{2} = \frac{ - 18}{2} = - 9 \\ t_{2} = \frac{ - 8 + 10}{2} = \frac{2}{2} = 1 \\ \\ {x}^{2} = - 9 \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} = 1 \\ x \: \: \: e \: \: \: R \: \: \: \: \: \: \: \: \: \: \: \: \: x_{1} = 1 \: \: \: \: \: x_{2} = - 1[/tex]
[tex]c) - 2 {x}^{4} - 3 {x}^{2} - 5 = 0 \\ {x }^{2} = t \\ - 2 {t}^{2} - 3t - 5 = 0 \\ ∆ = ( - 3 {)}^{2} - 4 \times ( - 2) \times ( - 5) = 9 - 40 \\ ∆ < 0 \\ brak \: rozwiazan[/tex]
zad. 2
[tex]a)5 + ( {x}^{2} - 2 {)}^{2} = 2 {x}^{4} - 12 {x}^{2} \\ 5 + {x}^{4} - 4 {x}^{2} + 4 = 2 {x}^{4} - 12 {x}^{2} \\ 9 + {x}^{4} - 4 {x}^{2} = 2 {x}^{4} - 12 {x}^{2} \\ 9 + {x}^{4} - 4 {x}^{2} - 2 {x}^{4} + 12 {x}^{2} = 0 \\ - {x}^{4} + 8 {x}^{2} + 9 = 0 \\ t = {x}^{2} \\ - {t}^{2} + 8t + 9 = 0 \\ ∆ =- 8 {}^{2} - 4 \times ( - 1) \times 9 = 64+ 36 = 100 \\ \sqrt{∆} = \sqrt{100} = 10 \\ t_{1} = \frac{ - 8 - 10}{ - 2} = \frac{ - 18}{ - 2} = 9 \\ t_{2} = \frac{ - 8 + 10}{ - 2} = \frac{2}{ - 2} = - 1 \\ \\ {x}^{2} = 9 \: \: \: \: \: \: \: \: \: {x}^{2} = - 1 \\ x_{1} = 3 \: \: \: \: \: \: \: x_{2} = - 3 \: \: \: \: \: \: \: \: \: \: \: x \: e \: R \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
[tex]b)( {x}^{2} - 3)( {x}^{2} + 3) + 1 = 3 {x}^{4} - 2 {x}^{2} \\ {x}^{4} + 3 {x}^{2} - 3 {x}^{2} - 9 + 1 = 3 {x}^{4} - 2 {x}^{2} \\ {x}^{4} - 8 = 3 {x}^{4} - 2 {x}^{2} \\ - 2 {x}^{4} + 2 {x}^{2} - 8 = 0 \\ t = {x}^{2} \\- 2 {t}^{2} + 2t - 8 = 0 \\ ∆ = {2}^{2} - 4 \times ( - 2) \times ( - 8) \\ ∆ < 0 \\ brak \: rozwiazan[/tex]
[tex]c) - {x}^{4} - 6 {x}^{2} - 9 = 0 \\ t = {x}^{2} \\ - {t}^{2} - 6t - 9 = 0 \\ ∆ = ( - 6 {)}^{2} - 4 \times ( - 1) \times ( - 9) = 36 - 36 = 0 \\ t_{0} = \frac{ 6}{ - 2} = - 3 \\ {x}^{2} = - 3 \\ brak \: rozwiazan[/tex]
[tex]d)3 {x}^{4} - 5 {x}^{2} - 2 = 0 \\ t = {x}^{2} \\ 3 {t}^{2} - 5t - 2 = 0 \\ ∆ = {( - 5)}^{2} - 4 \times 3 \times ( - 2) = 25 + 24 = 49 \\ \sqrt{∆} = \sqrt{49} = 7 \\ t_{1} = \frac{5 - 7}{6} = - \frac{2}{6} = - \frac{1}{3} \\ t_{2} = \frac{5 + 7}{6} = \frac{12}{6} = 2 \\ \\ {x}^{2} = - \frac{1}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} = 2 \\ x \: \: e \: R \ \: \: \: \: \: \: \: x_{1} = \sqrt{2} \: \: \: \: \: \: x_{2} = - \sqrt{2} [/tex]
[tex]e)x - \sqrt{2 - x} = 4 \\ - \sqrt{2 - x} = 4 - x \\ \sqrt{2 - x} = x - 4 \\ 2 - x = {x}^{2} - 8x + 16 \\ 2 - x - {x}^{2} + 8x - 16 = 0 \\ - {x}^{2} - 7x - 14 = 0 \\ ∆ = ( - 7 {)}^{2} - 4 \times ( - 1) \times ( - 14) = 49 - 56 \\ ∆ < 0 \\ brak \: rozwiazan[/tex]
[tex]f) \sqrt{x - 3} = x - 5 \\ x - 3 = {x}^{2} - 10x + 25 \\ x - 3 - {x}^{2} + 10x - 25 = 0 \\ - {x}^{2} + 11x - 28 = 0 \\ ∆ = {11}^{2} - 4 \times ( - 28) \times ( - 1) = 121 - 112 = 9 \\ \sqrt{∆} = \sqrt{9} = 3 \\ x_{1} = \frac{ - 11 - 3}{ - 2} = \frac{ - 14}{ - 2} = 7 \\ x_{2} = \frac{ - 11 + 3}{ - 2} = \frac{ - 8}{ - 2} = 4 \\ \\ x_{2}≠4 \\ \\ x = 7[/tex]
