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Квадратна једначина - растављање на чиниоце

Растављање квадратног тринома на линеарне чиниоце. Извођење и једноставни примери.

Задаци

Текст задатака објашњених у видео лекцији.

Пр.1   Растави на чиниоце:

          а)   ${x^2} - 4x + 3$

          б)   $2{x^2} + 7x + 3$

Пр.2   Скратити разломак:

           $\frac{{3{x^2} + 2x - 8}}{{12{x^2} - 7x - 12}}$

Пр.3   Скратити разломак:

           $\frac{{{x^{ - 1}} + 4{x^{ - 2}} + 4{x^{ - 3}}}}{{3 + 8{x^{ - 1}} + 4{x^{ - 2}}}}$

Пр. 1

а) 

\[a{x^2} + bx + c = a\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\]

\[\begin{gathered}
{x^2} - 4x + 3 = 0 \hfill \\
{x_{1,2}} = \frac{{4 \pm \sqrt {16 - 12} }}{2} \hfill \\
{x_{1,2}} = \frac{{4 \pm 2}}{2} \hfill \\
{x_1} = 1,{x_2} = 3 \hfill \\
{x^2} - 4x + 3 = \left( {x - 1} \right)\left( {x - 3} \right) \hfill \\
\end{gathered} \]

б)

\[\begin{gathered}
2{x^2} + 7x + 3 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 7 \pm \sqrt {49 - 24} }}{4} \hfill \\
{x_{1,2}} = \frac{{ - 7 \pm 5}}{4} \hfill \\
{x_1} = - 3,{x_2} = - \frac{1}{2} \hfill \\
2{x^2} + 7x + 3 = 2\left( {x - \left( { - 3} \right)} \right)\left( {x - \left( { - \frac{1}{2}} \right)} \right) = \hfill \\
= 2\left( {x + 3} \right)\left( {x + \frac{1}{2}} \right) = \left( {x + 3} \right)\left( {2x + 1} \right) \hfill \\
\end{gathered} \]


Пр.2

\[\begin{gathered}
3{x^2} + 2x - 8 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 2 \pm \sqrt {100} }}{6} \hfill \\
{x_{1,2}} = \frac{{ - 2 \pm 10}}{6} \hfill \\
{x_1} = - 2,{x_2} = \frac{4}{3} \hfill \\
\hfill \\
12{x^2} - 7x - 12 = 0 \hfill \\
{x_{1,2}} = \frac{{7 \pm \sqrt {625} }}{{24}} \hfill \\
{x_{1,2}} = \frac{{7 \pm 25}}{{24}} \hfill \\
{x_1} = - \frac{3}{4},{x_2} = \frac{4}{3} \hfill \\
\hfill \\
\frac{{3{x^2} + 2x - 8}}{{12{x^2} - 7x - 12}} = \hfill \\
= \frac{{3\left( {x + 2} \right)\left( {x - \frac{4}{3}} \right)}}{{12\left( {x + \frac{3}{4}} \right)\left( {x - \frac{4}{3}} \right)}} = \frac{{x + 2}}{{4\left( {x + \frac{3}{4}} \right)}} = \frac{{x + 2}}{{4x + 3}} \hfill \\
\end{gathered} \]


Пр.3

\[\begin{gathered}
\frac{{{x^{ - 1}} + 4{x^{ - 2}} + 4{x^{ - 3}}}}{{3 + 8{x^{ - 1}} + 4{x^{ - 2}}}} = \frac{{\frac{1}{x} + \frac{4}{{{x^2}}} + \frac{4}{{{x^3}}}}}{{3 + \frac{8}{x} + \frac{4}{{{x^2}}}}} = \frac{{\frac{{{x^2} + 4x + 4}}{{{x^3}}}}}{{\frac{{3{x^2} + 8x + 4}}{{{x^2}}}}} = \hfill \\
= \frac{{{x^2}\left( {{x^2} + 4x + 4} \right)}}{{{x^3}\left( {3{x^2} + 8x + 4} \right)}} \hfill \\
\hfill \\
{x^2} + 4x + 4 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 4 \pm \sqrt 0 }}{2} \hfill \\
{x_{1,2}} = \frac{{7 \pm 25}}{{24}} \hfill \\
{x_{1,2}} = - 2 \hfill \\
\hfill \\
3{x^2} + 8x + 4 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 8 \pm \sqrt {16} }}{6} \hfill \\
{x_{1,2}} = \frac{{ - 8 \pm 4}}{6} \hfill \\
{x_1} = - 2,{x_2} = - \frac{2}{3} \hfill \\
\hfill \\
\frac{{{x^2}\left( {{x^2} + 4x + 4} \right)}}{{{x^3}\left( {3{x^2} + 8x + 4} \right)}} = \hfill \\
= \frac{{\left( {x + 2} \right)\left( {x + 2} \right)}}{{x\left( {3\left( {x + 2} \right)\left( {x + \frac{2}{3}} \right)} \right)}} = \frac{{\left( {x + 2} \right)}}{{x\left( {3x + 2} \right)}} \hfill \\
\end{gathered} \]