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

$
\begin{array}{l}
x^2 + 4x = - 10 \\
(x + 2)^2 - 4 = - 10 \\
(x + 2)^2 = - 6 \\
{\rm{Geen}}\,\,{\rm{oplossing}} \\
\end{array}
$

$
\begin{array}{l}
2\left( {x^2 + 3} \right) = 2 \\
x^2 + 3 = 1 \\
x^2 = - 2 \\
{\rm{Geen}}\,\,{\rm{oplossing}} \\
\end{array}
$

$
\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}
$

$
\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}
$

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