Wiskundeleraar

Uitwerkingen van de oefeningen

Opgave 1

$ \begin{array}{*{20}c} \begin{array}{l} a. \\ \\ \\ \\ \\ \\ \end{array} & \begin{array}{l} 3x^2 = 5x \\ 3x^2 - 5x = 0 \\ x(3x - 5) = 0 \\ x = 0 \vee 3x - 5 = 0 \\ x = 0 \vee 3x = 5 \\ x = 0 \vee x = 1\frac{2}{3} \\ \end{array} \\ \end{array} $	$ \begin{array}{*{20}c} \begin{array}{l} b. \\ \\ \\ \\ \end{array} & \begin{array}{l} (3x + 3)(2x - 5) = 0 \\ 3x + 3 = 0 \vee 2x - 5 = 0 \\ 3x = - 3 \vee 2x = 5 \\ x = - 1 \vee x = 2\frac{1}{2} \\ \end{array} \\ \end{array} $	$ \begin{array}{*{20}c} \begin{array}{l} c. \\ \\ \\ \\ \end{array} & \begin{array}{l} (3x - 1)^2 = 16 \\ 3x - 1 = - 4 \vee 3x - 1 = 4 \\ 3x = - 3 \vee 3x = 5 \\ x = - 1 \vee x = 1\frac{2}{3} \\ \end{array} \\ \end{array} $
$ \begin{array}{*{20}c} \begin{array}{l} d. \\ \\ \\ \\ \\ \\ \\ \end{array} & \begin{array}{l} (x + 2)^2 + (x + 3)^2 = 1 \\ x^2 + 4x + 4 + x^2 + 6x + 9 = 1 \\ 2x^2 + 10x + 13 = 1 \\ 2x^2 + 10x + 12 = 0 \\ x^2 + 5x + 6 = 0 \\ (x + 2)(x + 3) = 0 \\ x = - 2 \vee x = - 3 \\ \end{array} \\ \end{array} $	$ \begin{array}{*{20}c} \begin{array}{l} e. \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array} & \begin{array}{l} 8x^2 + 2x - 3 = 0 \\ 8\left( {x^2 + \frac{1}{4}x} \right) - 3 = 0 \\ 8\left( {\left( {x + \frac{1}{8}} \right)^2 - \frac{1}{{64}}} \right) - 3 = 0 \\ 8\left( {x + \frac{1}{8}} \right)^2 - \frac{1}{8} - 3 = 0 \\ 8\left( {x + \frac{1}{8}} \right)^2 - 3\frac{1}{8} = 0 \\ 8\left( {x + \frac{1}{8}} \right)^2 = 3\frac{1}{8} \\ \left( {x + \frac{1}{8}} \right)^2 = \frac{{25}}{{64}} \\ x + \frac{1}{8} = - \frac{5}{8} \vee x + \frac{1}{8} = \frac{5}{8} \\ x = - \frac{3}{4} \vee x = \frac{1}{2} \\ \end{array} \\ \end{array} $	$ \begin{array}{*{20}c} \begin{array}{l} f. \\ \\ \\ \end{array} & \begin{array}{l} 12x^2 = 144 \\ x^2 = 12 \\ x = - \sqrt {12} \vee x = \sqrt {12} \\ \end{array} \\ \end{array} $

Opgave 2

$
\begin{array}{l}
f(x) = x^2 + 2x - 1 = \left( {x + 1} \right)^2 - 2 \to Top( - 1, - 2) \\
g(x) = - 2x^2 + 6 \to Top(0,6) \\
h(x) = 3x^2 - 30x + 50 = 3(x - 5)^2 - 25 \to Top(5, - 25) \\
k(x) = - 4x^2 - 16x - 28 = - 4(x + 2)^2 - 12 \to Top( - 2, - 12) \\
\end{array}
$

Opgave 3

a.	$ \begin{array}{l} x^2 + 4x = - 10 \\ (x + 2)^2 - 4 = - 10 \\ (x + 2)^2 = - 6 \\ {\rm{Geen}}\,\,{\rm{oplossing}} \\ \end{array} $
b.	$ \begin{array}{l} 2\left( {x^2 + 3} \right) = 2 \\ x^2 + 3 = 1 \\ x^2 = - 2 \\ {\rm{Geen}}\,\,{\rm{oplossing}} \\ \end{array} $
c.	$ \begin{array}{l} 4x^2 - 4x + 1 = 25 \\ 4x^2 - 4x - 24 = 0 \\ x^2 - x - 6 = 0 \\ (x - 3)(x + 2) = 0 \\ x = 3 \vee x = - 2 \\ \end{array} $
d.	$ \begin{array}{l} (2x + 2)^2 = (3x - 3)^2 \\ 2x + 2 = 3x - 3 \vee 2x + 2 = - 3x + 3 \\ - x = - 5 \vee 5x = 1 \\ x = 5 \vee x = \frac{1}{5} \\ \end{array} $