Respuestas

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2012-06-04T05:43:53+02:00

1.

\[\begin{gathered} 2{x^2} - 5x - 12 = 0 \hfill \\ = \frac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4 \cdot 2 \cdot \left( { - 12} \right)} }}{{2 \cdot 2}} \hfill \\ = \frac{{5 \pm \sqrt {25 + 96} }}{4} \hfill \\ = \frac{{5 \pm \sqrt {121} }}{4} \hfill \\ = \frac{{5 \pm 11}}{4} \hfill \\ \end{gathered}

 

Entonces: \[x = \dfrac{{5 + 11}}{4} \vee x = \dfrac{{5 - 11}}{4}

 

2. \[\begin{gathered} 14{x^2} = 8x \hfill \\ 14{x^2} - 8x = 0 \hfill \\ 2x\left( {7x - 4} \right) = 0 \hfill \\ \Rightarrow 2x = 0 \vee 7x - 4 = 0 \hfill \\ \Rightarrow x = 0 \vee x = \frac{4}{7} \hfill \\ \end{gathered}

 

3. \[\begin{gathered} {\left( {2x - 3} \right)^2} = 5 \hfill \\ 2x - 3 = \pm \sqrt 5 \hfill \\ 2x = 3 \pm \sqrt 5 \hfill \\ x = \frac{{3 \pm \sqrt 5 }}{2} \hfill \\ \Rightarrow x = \frac{{3 + \sqrt 5 }}{2} \vee x = \frac{{3 - \sqrt 5 }}{2} \hfill \\ \end{gathered}

 

4. \[\begin{gathered} 2x = \frac{2}{x} + 3 \hfill \\ 2x = \frac{{2 + 3x}}{x} \hfill \\ 2{x^2} = 3x + 2 \hfill \\ 2{x^2} - 3x - 2 = 0 \hfill \\ x = 2 \vee x = \frac{{ - 1}}{2} \hfill \\ \end{gathered}

 

5. \[\begin{gathered} 8{x^2} - 22x - 6 = 0 \hfill \\ 4{x^2} - 11x - 3 = 0 \hfill \\ \left( {4x - 12} \right)\left( {x + \frac{1}{4}} \right) = 0 \hfill \\ \Rightarrow 4x - 12 = 0 \vee x + \frac{1}{4} = 0 \hfill \\ \Rightarrow x = 3 \vee x = - \frac{1}{4} \hfill \\ \end{gathered}