199/MR00. példa / Matematika - Megoldott feladatok gyűjteménye

MR-199 / 199. példa

Oldja meg az alábbi egyenletet:

$2 x^{5} + x^{4} - 19 x^{3} + 19 x^{2} - x - 2 = 0$

$2 x^{5} + x^{4} - 19 x^{3} + 19 x^{2} - x - 2 = 0$

2 x^{5} + 1 x^{4} - 19 x^{3} + 19 x^{2} - 1 x - 2 = 0

\begin{matrix} 2 & 1 & - 19 & 19 & - 1 & - 2 \\ - - & - - & - - & - - & - - & - - & - - \\ 1 & 2 & 3 & - 16 & 3 & 2 & 0 \end{matrix}

$x_{1} = 1$

(x - 1) (2 x^{4} + 3 x^{3} - 16 x^{2} + 3 x + 2) = 0

2 x^{4} + 3 x^{3} - 16 x^{2} + 3 x + 2 = 0

2 x^{4} + 3 x^{3} - 16 x^{2} + 3 x + 2 = 0 \cdot \frac{1}{x^{2}}

Ossza el az egyenletet $x^{2}$ -el.

\frac{2 x^{4}}{x^{2}} + \frac{3 x^{3}}{x^{2}} - \frac{16 x^{2}}{x^{2}} + \frac{3 x}{x^{2}} + \frac{2}{x^{2}} = 0

2 x^{2} + 3 x - 16 + \frac{3}{x} + \frac{2}{x^{2}} = 0

2 x^{2} + \frac{2}{x^{2}} + 3 x + \frac{3}{x} - 16 = 0

2 (x^{2} + \frac{1}{x^{2}}) + 3 (x + \frac{1}{x}) - 16 = 0

t = x + \frac{1}{x}

t^{2} = {(x + \frac{1}{x})}^{2} = {(\frac{x^{2} + 1}{x})}^{2} = \frac{x^{4} + 2 x^{2} + 1}{x^{2}} = \frac{x^{4}}{x^{2}} + \frac{2 x^{2}}{x^{2}} + \frac{1}{x^{2}} = x^{2} + 2 + \frac{1}{x^{2}}

t^{2} - 2 = x^{2} - \frac{1}{x^{2}}

2 (t^{2} - 2) + 3 t - 16 = 0

2 t^{2} - 4 + 3 t - 16 = 0

2 t^{2} + 3 t - 20 = 0

2 t^{2} + 3 t - 20 = 0 \Rightarrow a = 2; b = 3; c = - 20

t_{1, 2} = \frac{- 3 \pm \sqrt{3^{3} - 4 \cdot 2 \cdot (- 20)}}{2 \cdot 2} = \frac{- 3 \pm \sqrt{9 + 160}}{4} = \frac{- 3 \pm 13}{4}

t_{1} = \frac{- 3 + 13}{4} = \frac{10}{4} = \frac{5}{2}; t_{2} = \frac{- 3 - 13}{4} = \frac{- 16}{4} = - 4

t_{1} = \frac{5}{2}; t_{2} = - 4

t_{1} = \frac{5}{2} \Rightarrow \frac{5}{2} = x + \frac{1}{x}

\frac{5}{2} = x + \frac{1}{x} \cdot 2 x

\frac{5}{2} \cdot 2 x = x \cdot 2 x + \frac{1}{x} \cdot 2 x

5 x = 2 x^{2} + 2

2 x^{2} - 5 x + 2 = 0

2 x^{2} - 5 x + 2 = 0 \Rightarrow a = 2; b = - 5; c = 2

x_{2, 3} = \frac{5 \pm \sqrt{5^{2} - 4 \cdot 2 \cdot 2}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4}

x_{2} = \frac{5 + 3}{4} = \frac{8}{4}; x_{3} = \frac{5 - 3}{4} = \frac{2}{4}

$x_{2} = 2; x_{3} = \frac{1}{2}$

t_{2} = -4 \Rightarrow -4 = x + \frac{1}{x}

-4 = x + \frac{1}{x} \cdot x

- 4 \cdot x = x \cdot x + \frac{1}{x} \cdot x

- 4 x = x^{2} + 1

x^{2} + 4 x + 1 = 0

x^{2} + 4 x + 1 = 0 \Rightarrow a = 1; b = 4; c = 1

x_{4, 5} = \frac{- 4 \pm \sqrt{4^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{- 4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm \sqrt{4 \cdot 3}}{2} = \frac{4 \pm \sqrt{4} \cdot \sqrt{3}}{2} = \frac{4 \pm 2 \sqrt{3}}{2} = \frac{2 (2 \pm \sqrt{3})}{2} = 2 \pm \sqrt{3}

$x_{4} = 2 + \sqrt{3}; x_{5} = 2 - \sqrt{3}$

x_{1} = 1; x_{2} = 2; x_{3} = - \frac{1}{2}; x_{4} = - 2 + \sqrt{3}; x_{5} = - 2 - \sqrt{3}

$x_{1} = 1; x_{2} = 2; x_{3} = - \frac{1}{2}; x_{4} = - 2 + \sqrt{3}; x_{5} = - 2 - \sqrt{3}$

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