Квадратна једначина – решења 4

Текст задатака објашњених у видео лекцији.

Пр.3   Решити следећу квадратну једначину.

          $2{x^2} - 5x + 3\left| {x - 2} \right| = 0$

Пр.4   Решити следећу једначину.

          $x\left| {2x - 1} \right| - 4\left| {x - 1} \right| =  - 1$

Пр.3

\[\begin{gathered}
2{x^2} - 5x + 3\left| {x - 2} \right| = 0 \hfill \\
\left| {x - 2} \right| = \left\{ \begin{gathered}
x - 2,x - 2 \geqslant 0 \hfill \\
- \left( {x - 2} \right),x - 2 < 0 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x - 2,x \geqslant 2 \hfill \\
- x + 2,x < 2 \hfill \\
\end{gathered} \right. \hfill \\
2{x^2} - 5x + 3\left| {x - 2} \right| = 0 \Leftrightarrow \left\{ \begin{gathered}
2{x^2} - 5x + 3\left( {x - 2} \right) = 0,x \geqslant 2 \hfill \\
2{x^2} - 5x + 3\left( { - x + 2} \right) = 0,x < 2 \hfill \\
\end{gathered} \right. \hfill \\
2{x^2} - 5x + 3\left( {x - 2} \right) = 0 \hfill \\
2{x^2} - 5x + 3x - 6 = 0 \hfill \\
2{x^2} - 2x - 6 = 0\left| { \div 2} \right. \hfill \\
{x^2} - x - 3 = 0 \hfill \\
{x_{1,2}} = \frac{{1 \pm \sqrt {1 + 12} }}{2} \hfill \\
{x_{1,2}} = \frac{{1 \pm \sqrt {13} }}{2} \hfill \\
\end{gathered} \]

Решење коjе задовољава услове задатака $x \geqslant 2$ jе $x = \frac{{1 + \sqrt {13} }}{2}$

\[\begin{gathered}
2{x^2} - 5x + 3\left( { - x + 2} \right) = 0 \hfill \\
2{x^2} - 5x - 3x + 6 = 0 \hfill \\
2{x^2} - 8x + 6 = 0 \hfill \\
{x_{3,4}} = \frac{{8 \pm \sqrt {64 - 48} }}{4} \hfill \\
{x_{3,4}} = \frac{{8 \pm 4}}{4} \hfill \\
{x_3} = 2,{x_4} = 3 \hfill \\
\end{gathered} \]

Ова решења не задовољаваjу услове задатака $x < 2$.

Пр.4

\[\begin{gathered}
x\left| {2x - 1} \right| - 4\left| {x - 1} \right| = - 1 \hfill \\
\left| {2x - 1} \right| = \left\{ \begin{gathered}
2x - 1,2x - 1 \geqslant 0 \hfill \\
- \left( {2x - 1} \right),2x - 1 < 0 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
2x - 1,x \geqslant \frac{1}{2} \hfill \\
- \left( {2x - 1} \right),x < \frac{1}{2} \hfill \\
\end{gathered} \right. \hfill \\
\left| {x - 1} \right| = \left\{ \begin{gathered}
x - 1,x - 1 \geqslant 0 \hfill \\
- \left( {x - 1} \right),x - 1 < 0 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x - 1,x \geqslant 1 \hfill \\
- \left( {x - 1} \right),x < 1 \hfill \\
\end{gathered} \right. \hfill \\
x\left| {2x - 1} \right| - 4\left| {x - 1} \right| = - 1 \Leftrightarrow \left\{ \begin{gathered}
x\left( { - 2x + 1} \right) - 4\left( { - x + 1} \right) = - 1,x < \frac{1}{2} \hfill \\
x\left( {2x - 1} \right) - 4\left( { - x + 1} \right) = - 1,\frac{1}{2} \leqslant x < 1 \hfill \\
x\left( {2x - 1} \right) - 4\left( {x - 1} \right) = - 1,x \geqslant 1 \hfill \\
\end{gathered} \right. \hfill \\
- 2{x^2} + x + 4x - 4 + 1 = 0 \hfill \\
- 2{x^2} + 5x - 3 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 5 \pm 1}}{{ - 4}} \hfill \\
{x_1} = \frac{3}{2},{x_2} = 1 \hfill \\
\end{gathered} \]

Ови решења нису задовољаваjу услове задатака $x < \frac{1}{2}$.

\[\begin{gathered}
x\left( {2x - 1} \right) - 4\left( { - x + 1} \right) = - 1 \hfill \\
2{x^2} - x + 4x - 4 + 1 = 0 \hfill \\
2{x^2} + 3x - 3 = 0 \hfill \\
{x_{3,4}} = \frac{{ - 3 \pm \sqrt {33} }}{4} \hfill \\
\end{gathered} \]

Решење коjе задовољава услове задатака $\frac{1}{2} \leqslant x < 1$ jе $x = \frac{{ - 3 + \sqrt {33} }}{4}$

\[\begin{gathered}
x\left( {2x - 1} \right) - 4\left( {x - 1} \right) = - 1 \hfill \\
2{x^2} - x - 4x + 4 + 1 = 0 \hfill \\
2{x^2} - 5x + 5 = 0 \hfill \\
{x_{5,6}} = \frac{{5 \pm \sqrt { - 15} }}{4} \hfill \\
\end{gathered} \]

Ова једначина нема решење у скупу реалних бројева.