Ejercicio 282

octubre 30, 2019abril 16, 2019 por Gestor Baldor

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CAPITULO XXXVI

ECUACIONES BINOMIAS Y TRINOMIAS

Ejercicio 282

Resolver las ecuaciones:

$\begin{matrix} x^{4} - 1 & = 0 \\ (x^{2} + 1) (x^{2} - 1) & = 0 \\ (x^{2} + 1) (x + 1) (x - 1) & = 0 \\ x_{1} + 1 & = 0 \\ x_{1} & = - 1 \\ x_{2} - 1 & = 0 \\ x_{2} & = 1 \\ x^{2} + 1 & = 0 \\ x^{2} & = - 1 \\ x & = \pm \sqrt{- 1} \\ x_{3} & = i \\ x_{4} & = - i \end{matrix}$
$\begin{matrix} x^{3} + 1 & = 0 \\ (x + 1) (x^{2} - x + 1) & = 0 \\ x + 1 & = 0 \\ x_{1} & = - 1 \\ x^{2} - x + 1 & = 0 \\ x & = \frac{- (- 1) \pm \sqrt{{(- 1)}^{2} - 4 (1) (1)}}{2 (1)} \\ x & = \frac{1 \pm \sqrt{1 - 4}}{2} \\ x & = \frac{1 \pm \sqrt{- 1} \sqrt{3}}{2} \\ x & = \frac{1 \pm \sqrt{3} i}{2} \\ x_{1} & = \frac{1 + \sqrt{3} i}{2} \\ x_{2} & = \frac{1 - \sqrt{3} i}{2} \end{matrix}$
$\begin{matrix} x^{4} & = 81 \\ x^{4} - 81 & = 0 \\ (x^{2} + 9) (x^{2} - 9) & = 0 \\ (x^{2} + 9) (x - 3) (x + 3) & = 0 \\ x_{1} - 3 & = 0 \\ x_{1} & = 3 \\ x_{2} + 3 & = 0 \\ x_{2} & = - 3 \\ x^{2} + 9 & = 0 \\ x^{2} & = - 9 \\ x & = \pm \sqrt{- 3^{2}} \\ x & = \pm 3 \sqrt{- 1} \\ x_{3} & = 3 i \\ x_{4} & = - 3 i \end{matrix}$
$\begin{matrix} x^{4} - 256 & = 0 \\ (x^{2} + 16) (x^{2} - 16) & = 0 \\ (x^{2} + 16) (x - 4) (x + 4) & = 0 \\ x_{1} - 4 & = 0 \\ x_{1} & = 4 \\ x_{2} + 4 & = 0 \\ x_{2} & = - 4 \\ x^{2} + 16 & = 0 \\ x^{2} & = - 16 \\ x & = \pm \sqrt{- 4^{2}} \\ x & = \pm 4 \sqrt{- 1} \\ x_{3} & = 4 i \\ x_{4} & = - 4 i \end{matrix}$
$\begin{matrix} x^{3} + 8 & = 0 \\ (x + 2) (x^{2} - 2 x + 4) & = 0 \\ x_{1} + 2 & = 0 \\ x_{1} & = - 2 \\ x^{2} - 2 x + 4 & = 0 \\ x & = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (1) (4)}}{2 (1)} \\ x & = \frac{2 \pm \sqrt{4 - 16}}{2} \\ x & = \frac{2 \pm \sqrt{- 12}}{2} \\ x & = \frac{2 \pm \sqrt{(- 3) (2^{2})}}{2} \\ x & = \frac{2 \pm 2 \sqrt{(3) (- 1)}}{2} \\ x & = \frac{2 (1 \pm \sqrt{3} i)}{2} \\ x_{2} & = 1 + \sqrt{3} i \\ x_{3} & = 1 - \sqrt{3} i \end{matrix}$
$\begin{matrix} x^{4} - 625 & = 0 \\ (x^{2} + 25) (x^{2} - 25) & = 0 \\ (x^{2} + 25) (x - 5) (x + 5) & = 0 \\ x_{1} - 5 & = 0 \\ x_{1} & = 5 \\ x_{2} + 5 & = 0 \\ x_{2} & = - 5 \\ x^{2} + 25 & = 0 \\ x^{2} & = - 25 \\ x & = \pm \sqrt{- 25} \\ x & = \pm \sqrt{(- 1) (5^{2})} \\ x & = \pm 5 i \\ x_{3} & = 5 i \\ x_{4} & = - 5 i \end{matrix}$
$\begin{matrix} x^{3} + 64 & = 0 \\ (x + 4) (x^{2} - 4 x + 16) & = 0 \\ x + 4 & = 0 \\ x_{1} + 4 & = 0 \\ x_{1} & = - 4 \\ x^{2} - 4 x + 16 & = 0 \\ x & = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - 4 (1) (16)}}{2 (1)} \\ x & = \frac{4 \pm \sqrt{16 - 64}}{2} \\ x & = \frac{4 \pm \sqrt{- 48}}{2} \\ x & = \frac{4 \pm \sqrt{(- 3) (4^{2})}}{2} \\ x & = \frac{4 \pm 4 \sqrt{3} \sqrt{- 1}}{2} \\ x & = \frac{2 (2 \pm 2 \sqrt{3} i)}{2} \\ x_{2} & = 2 + 2 \sqrt{3} i \\ x_{3} & = 2 - 2 \sqrt{3} i \end{matrix}$
$\begin{matrix} x^{6} - 729 & = 0 \\ (x^{3} - 27) (x^{3} + 27) & = 0 \\ (x - 3) (x^{2} + 3 x + 9) (x + 3) (x^{2} - 3 x + 9) & = 0 \\ x_{1} - 3 & = 0 \\ x_{1} & = 3 \\ x_{2} + 3 & = 0 \\ x_{2} & = - 3 \\ x^{2} + 3 x + 9 & = 0 \\ x & = \frac{- 3 \pm \sqrt{3^{2} - 4 (1) (9)}}{2 (1)} \\ x & = \frac{- 3 \pm \sqrt{9 - 36}}{2} \\ x & = \frac{- 3 \pm \sqrt{- 27}}{2} \\ x & = \frac{- 3 \pm \sqrt{(- 1) (3^{3})}}{2} \\ x & = \frac{- 3 \pm 3 \sqrt{3} \sqrt{- 1}}{2} \\ x & = \frac{- 3 \pm 3 \sqrt{3} i}{2} \\ x_{3} & = \frac{- 3 - 3 \sqrt{3} i}{2} \\ x_{4} & = \frac{- 3 + 3 \sqrt{3} i}{2} \\ x^{2} - 3 x + 9 & = 0 \\ x & = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (9)}}{2 (1)} \\ x & = \frac{3 \pm \sqrt{9 - 36}}{2} \\ x & = \frac{3 \pm \sqrt{- 27}}{2} \\ x & = \frac{3 \pm \sqrt{(- 1) (3^{3})}}{2} \\ x & = \frac{3 \pm 3 \sqrt{3} \sqrt{- 1}}{2} \\ x & = \frac{3 \pm 3 \sqrt{3} i}{2} \\ x_{5} & = \frac{3 - 3 \sqrt{3} i}{2} \\ x_{6} & = \frac{3 + 3 \sqrt{3} i}{2} \end{matrix}$
Hallar las raíces cúbicas de 8.
$\begin{matrix} x^{3} & = 8 \\ x^{3} - 8 & = 0 \\ (x - 2) (x^{2} + 2 x + 4) & = 0 \\ x_{1} - 2 & = 0 \\ x_{1} & = 2 \\ x^{2} + 2 x + 4 & = 0 \\ x & = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (4)}}{2 (1)} \\ x & = \frac{- 2 \pm \sqrt{4 - 16}}{2} \\ x & = \frac{- 2 \pm \sqrt{- 12}}{2} \\ x & = \frac{- 2 \pm \sqrt{(- 1) (3) (2^{2})}}{2} \\ x & = \frac{- 2 \pm 2 \sqrt{- 1} \sqrt{3}}{2} \\ x & = \frac{2 (- 1 \pm \sqrt{3} i)}{2} \\ x_{2} & = - 1 + \sqrt{3} i \\ x_{3} & = - 1 - \sqrt{3} i \end{matrix}$
Hallar las raíces cuartas de 64.
$\begin{matrix} x^{4} & = 64 \\ x^{4} - 64 & = 0 \\ (x^{2} + 8) (x^{2} - 8) & = 0 \\ x^{2} - 8 & = 0 \\ x^{2} & = 8 \\ x & = \pm \sqrt{8} \\ x & = \pm \sqrt{2^{3}} \\ x & = \pm 2 \sqrt{2} \\ x_{1} & = 2 \sqrt{2} \\ x_{2} & = - 2 \sqrt{2} \\ x^{2} + 8 & = 0 \\ x^{2} & = - 8 \\ x & = \pm \sqrt{- 8} \\ x & = \pm \sqrt{- 2^{3}} \\ x & = \pm 2 \sqrt{2} \sqrt{- 1} \\ x_{1} & = 2 \sqrt{2} i \\ x_{2} & = - 2 \sqrt{2} i \end{matrix}$