Текст задатака објашњених у видео лекцији.
Пр.1 Наћи решње следеће квадратне једначине:
${x^2} - 3x - 28 = 0$
Пр.2 Решити следећу квадратну једначину:
$9{x^2} - 12x + 4 = 0$
Пр.3 Решити следећу квадратну једначину:
$25{x^2} - 20x + 7 = 0$
Пр.1
\[\begin{gathered}
{x^2} - 3x - 28 = 0 \hfill \\
{x_{1,2}} = \frac{{ - \left( { - 3} \right) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4 \cdot 1 \cdot \left( { - 28} \right)} }}{{2 \cdot 1}} \hfill \\
{x_{1,2}} = \frac{{3 \pm \sqrt {9 + 112} }}{2} \hfill \\
{x_{1,2}} = \frac{{3 \pm 11}}{2} \hfill \\
{x_1} = \frac{{3 - 11}}{2} = - 4 \hfill \\
{x_2} = \frac{{3 + 11}}{2} = 7 \hfill \\
\end{gathered} \]
Пр. 2
\[\begin{gathered}
9{x^2} - 12x + 4 = 0 \hfill \\
{x_{1,2}} = \frac{{12 \pm \sqrt {144 - 4 \cdot 9 \cdot 4} }}{{2 \cdot 9}} \hfill \\
{x_{1,2}} = \frac{{12 \pm \sqrt {144 - 144} }}{{18}} \hfill \\
{x_{1,2}} = \frac{{12 \pm 0}}{{18}} \hfill \\
{x_{1,2}} = \frac{2}{3} \hfill \\
\end{gathered} \]
Пр. 3
\[\begin{gathered}
25{x^2} - 20x + 7 = 0 \hfill \\
{x_{1,2}} = \frac{{20 \pm \sqrt {400 - 4 \cdot 25 \cdot 7} }}{{2 \cdot 25}} \hfill \\
{x_{1,2}} = \frac{{20 \pm \sqrt {400 - 700} }}{{50}} \hfill \\
{x_{1,2}} = \frac{{20 \pm \sqrt { - 300} }}{{50}} \hfill \\
{x_{1,2}} = \frac{{20 \pm \sqrt { - 1 \cdot 3 \cdot 100} }}{{50}} \hfill \\
{x_{1,2}} = \frac{{20 \pm 10i\sqrt 3 }}{{50}} \hfill \\
{x_{1,2}} = \frac{{10\left( {2 \pm i\sqrt 3 } \right)}}{{50}} \hfill \\
{x_{1,2}} = \frac{{\left( {2 \pm i\sqrt 3 } \right)}}{5} \hfill \\
{x_1} = \frac{{2 + i\sqrt 3 }}{5} \hfill \\
{x_2} = \frac{{2 - i\sqrt 3 }}{5} \hfill \\
\end{gathered} \]
