Квадратна једначина – триномна једначина

Текст задатака објашњених у видео лекцији.

пр.1)   Решити следећу триномну једначину:

           ${x^6} - 9{x^3} + 8 = 0$

пр.2)   Решити триномну једначину:

           ${x^8} - 17{x^4} + 16 = 0$

Пр.1

\[\begin{gathered}
{x^6} - 9{x^3} + 8 = 0 \hfill \\
{\left( {{x^3}} \right)^2} - 9{x^3} + 8 = 0 \hfill \\
смена:{x^3} = t \hfill \\
{t^2} - 9t + 8 = 0 \hfill \\
{t_{1,2}} = \frac{{9 \pm \sqrt {81 - 4 \cdot 8 \cdot 1} }}{2} \hfill \\
{t_{1,2}} = \frac{{9 \pm 7}}{2} \hfill \\
{t_1} = 8,{t_2} = 1 \hfill \\
\hfill \\
{x^3} = 8 \vee {x^3} = 1 \hfill \\
{x^3} - 8 = 0 \hfill \\
{x^3} - {2^3} = 0 \hfill \\
\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) = 0 \hfill \\
x - 2 = 0 \vee {x^2} + 2x + 4 = 0 \hfill \\
{x_1} = 2 \hfill \\
\hfill \\
{x^2} + 2x + 4 = 0 \hfill \\
{x_{2,3}} = \frac{{ - 2 \pm \sqrt { - 12} }}{2} \hfill \\
{x_{2,3}} = \frac{{ - 2 \pm 2i\sqrt 3 }}{2} \hfill \\
{x_{2,3}} = \frac{{2\left( { - 1 \pm i\sqrt 3 } \right)}}{2} \hfill \\
{x_{2,3}} = - 1 \pm i\sqrt 3 \hfill \\
\hfill \\
{x^3} = 1 \hfill \\
{x^3} - 1 = 0 \hfill \\
\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) = 0 \hfill \\
x - 1 = 0 \vee {x^2} + x + 1 = 0 \hfill \\
{x_4} = 1 \hfill \\
\hfill \\
{x^2} + x + 1 = 0 \hfill \\
{x_{5,6}} = \frac{{ - 1 \pm \sqrt { - 3} }}{2} \hfill \\
{x_{5,6}} = \frac{{ - 1 \pm i\sqrt 3 }}{2} \hfill \\
\end{gathered} \]

\[{x_1} = 2,{x_{2,3}} =  - 1 \pm i\sqrt 3 ,{x_4} = 1,{x_{5,6}} = \frac{{ - 1 \pm i\sqrt 3 }}{2}\]

Пр.2

\[\begin{gathered}
{x^8} - 17{x^4} + 16 = 0 \hfill \\
{\left( {{x^4}} \right)^2} - 17{x^4} + 16 = 0 \hfill \\
:{x^4} = t \hfill \\
{t^2} - 17t + 16 = 0 \hfill \\
{t_{1,2}} = \frac{{17 \pm \sqrt {289 - 4 \cdot 16 \cdot 1} }}{2} \hfill \\
{t_{1,2}} = \frac{{17 \pm 15}}{2} \hfill \\
{t_1} = 1,{t_2} = 16 \hfill \\
\hfill \\
{x^4} = 1 \vee {x^4} = 16 \hfill \\
{x^4} - 1 = 0 \hfill \\
{\left( {{x^2}} \right)^2} - {1^2} = 0 \hfill \\
\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) = 0 \hfill \\
{x^2} - 1 = 0 \hfill \\
{x^2} = 1 \hfill \\
{x_{1,2}} = \pm 1 \hfill \\
{x^2} + 1 = 0 \hfill \\
{x^2} = - 1 \hfill \\
{x_{3,4}} = \pm \sqrt { - 1} \hfill \\
{x_{3,4}} = \pm i \hfill \\
\hfill \\
{x^4} = 16 \hfill \\
{x^4} - 16 = 0 \hfill \\
{\left( {{x^2}} \right)^2} - {\left( {{2^2}} \right)^2} = 0 \hfill \\
\left( {{x^2} - {2^2}} \right)\left( {{x^2} + {2^2}} \right) = 0 \hfill \\
{x^2} - {2^2} = 0 \hfill \\
{x^2} = 4 \hfill \\
{x_{5,6}} = \pm 2 \hfill \\
{x^2} + {2^2} = 0 \hfill \\
{x^2} = - 4 \hfill \\
{x_{7,8}} = \pm \sqrt { - 4} \hfill \\
{x_{7,8}} = \pm 2i \hfill \\
\end{gathered} \]

\[{x_{1,2}} =  \pm 1,{x_{3,4}} =  \pm i,{x_{5,6}} =  \pm 2,{x_{7,8}} =  \pm 2i\]