Квадратне једначине – решења 3

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Пр.1   Решити следећу једначину.

          ${\left( {3 - x} \right)^2} + {\left( {20 + 4x} \right)^2} = {\left( {x + 15} \right)^2}$

Пр.2   Решити следећу једначину.

          $\frac{{2x - 3}}{{24{x^2} + 36x}} + \frac{2}{{4{x^2} - 9}} = \frac{{4 - x}}{{12{x^2} - 18x}}$

Пр.1

\[\begin{gathered}
{\left( {3 - x} \right)^2} + {\left( {20 + 4x} \right)^2} = {\left( {x + 15} \right)^2} \hfill \\
9 - 6x + {x^2} + 400 + 160x + 16{x^2} = {x^2} + 30x + 255 \hfill \\
9 - 6x + {x^2} + 400 + 160x + 16{x^2} - {x^2} - 30x - 255 = 0 \hfill \\
\left. {16{x^2} + 124x + 184 = 0} \right| \div 4 \hfill \\
4{x^2} + 31x + 46 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 31 \pm \sqrt {961 - 4 \cdot 4 \cdot 46} }}{{2 \cdot 4}} \hfill \\
{x_{1,2}} = \frac{{ - 31 \pm \sqrt {961 - 736} }}{8} \hfill \\
{x_{1,2}} = \frac{{ - 31 \pm 15}}{8} \hfill \\
{x_1} = - \frac{{46}}{8} = - \frac{{23}}{4} \hfill \\
{x_2} = - \frac{{16}}{8} = - 2 \hfill \\
\end{gathered} \]

Пр.2 

\[\begin{gathered}
\frac{{2x - 3}}{{24{x^2} + 36x}} + \frac{2}{{4{x^2} - 9}} = \frac{{4 - x}}{{12{x^2} - 18x}} \hfill \\
\left. {\frac{{2x - 3}}{{12x\left( {2x + 3} \right)}} + \frac{2}{{\left( {2x - 3} \right)\left( {2x + 3} \right)}} = \frac{{4 - x}}{{6x\left( {2x - 3} \right)}}} \right|x \ne 0,x \ne \frac{3}{2},x \ne \frac{3}{2} \hfill \\
{\left( {2x - 3} \right)^2} + 2 \cdot 12x = \left( {4 - x} \right) \cdot 2 \cdot \left( {2x + 3} \right) \hfill \\
{\left( {2x - 3} \right)^2} + 24x = \left( {4 - x} \right) \cdot 2 \cdot \left( {2x + 3} \right) \hfill \\
4{x^2} - 12x + 9 + 24x = 2\left( {8x + 12 - 2{x^2} - 3x} \right) \hfill \\
4{x^2} - 12x + 9 + 24x = 16x + 24 - 4{x^2} - 6x \hfill \\
4{x^2} - 12x + 9 + 24x - 16x - 24 + 4{x^2} + 6x = 0 \hfill \\
8{x^2} + 2x - 15 = 0 \hfill \\
{x_{1,2}} = \frac{{ - 2 \pm \sqrt {4 - 4 \cdot 8 \cdot \left( { - 15} \right)} }}{{2 \cdot 8}} \hfill \\
{x_{1,2}} = \frac{{ - 2 \pm \sqrt {4 + 480} }}{{16}} \hfill \\
{x_{1,2}} = \frac{{ - 2 \pm 22}}{{16}} \hfill \\
{x_1} = \frac{{ - 2 - 22}}{{16}} = \frac{{ - 24}}{{16}} = - \frac{3}{2} \hfill \\
{x_2} = \frac{{ - 2 + 22}}{{16}} = \frac{{20}}{{16}} = \frac{5}{4} \hfill \\
\end{gathered} \]

Рекли смо да $x \ne \frac{-3}{2},$ онда прво решење не прихватамо, а друго решење прихватамо.