1

Linkerkolom
${(x - 3)}^{2} = 100$
$x - 3 = 10$ of $x - 3 = ‐ 10$
$x = 13$ of $x = ‐ 7$

${(x + \frac{1}{2})}^{2} = 25$
$x + \frac{1}{2} = 5$ of $x + \frac{1}{2} = ‐ 5$
$x = 4 \frac{1}{2}$ of $x = ‐ 5 \frac{1}{2}$

${(2 x + 1)}^{2} = 36$
$2 x + 1 = 6$ of $2 x + 1 = ‐ 6$
$2 x = 5$ of $2 x = ‐ 7$
$x = 2 \frac{1}{2}$ of $x = ‐ 3 \frac{1}{2}$

Rechterkolom:

${(x - 3)}^{2} = 13$
$x - 3 = \sqrt{13}$ of $x - 3 = ‐ \sqrt{13}$
$x = 3 + \sqrt{13}$ of $x = 3 - \sqrt{13}$

${(x + \frac{1}{2})}^{2} = 48$
$x + \frac{1}{2} = \sqrt{48} = 4 \sqrt{3}$ of $x + \frac{1}{2} = ‐ \sqrt{48} = ‐ 4 \sqrt{3}$
$x = ‐ \frac{1}{2} + 4 \sqrt{3}$ of $x = ‐ \frac{1}{2} - 4 \sqrt{3}$

${(2 x + 1)}^{2} = 68$
$2 x + 1 = \sqrt{68} = 2 \sqrt{17}$ of $2 x + 1 = ‐ \sqrt{68} = ‐ 2 \sqrt{17}$
$2 x = ‐ 1 + 2 \sqrt{17}$ of $2 x = ‐ 1 - 2 \sqrt{17}$
$x = ‐ \frac{1}{2} + \sqrt{17}$ of $x = ‐ \frac{1}{2} - \sqrt{17}$

2

25 ; 5	7 ; $x$ ; $3 \frac{1}{2}$
12 ; 36 ; $x$	4 ; 16
16 ; $x$ ; 8	$5 \frac{1}{2}$ ; $30 \frac{1}{4}$
$20 \frac{1}{4}$ ; $4 \frac{1}{2}$	3 ; 9
11 ; $30 \frac{1}{4}$ ; $x$

3

Linkerkolom:

$x^{2} + 10 x = 90$
$x^{2} + 10 x + 25 = 90 + 25$
${(x + 5)}^{2} = 115$
$x + 5 = \sqrt{115}$ of $x + 5 = ‐ \sqrt{115}$
$x = ‐ 5 + \sqrt{115}$ of $x = ‐ 5 - \sqrt{115}$

$x^{2} + 7 x + 1 = 0$
$x^{2} + 7 x + 11 \frac{1}{4} = 11 \frac{1}{4}$
${(x + 3 \frac{1}{2})}^{2} = 11 \frac{1}{4} = \frac{45}{4}$
$x + 3 \frac{1}{2} = \sqrt{\frac{45}{4}} = \frac{1}{2} \sqrt{45} = 1 \frac{1}{2} \sqrt{5}$ of $x + 3 \frac{1}{2} = ‐ 1 \frac{1}{2} \sqrt{5}$
$x = ‐ 3 \frac{1}{2} + 1 \frac{1}{2} \sqrt{5}$ of $x = ‐ 3 \frac{1}{2} - 1 \frac{1}{2} \sqrt{5}$

$x^{2} + 10 x + 22 = 0$
$x^{2} + 10 x + 22 + 3 = 3$
$x^{2} + 10 x + 25 = 3$
${(x + 5)}^{2} = 3$
$x + 5 = \sqrt{3}$ of $x + 5 = ‐ \sqrt{3}$
$x = ‐ 5 + \sqrt{3}$ of $x = ‐ 5 - \sqrt{3}$

$x^{2} - 11 x = ‐ 7$
$x^{2} - 11 x + 30 \frac{1}{4} = ‐ 7 + 30 \frac{1}{4}$
${(x - 5 \frac{1}{2})}^{2} = 23 \frac{1}{4} = \frac{93}{4}$
$x - 5 \frac{1}{2} = \sqrt{\frac{93}{4}} = \frac{1}{2} \sqrt{93}$ of $x - 5 \frac{1}{2} = ‐ \sqrt{\frac{93}{4}} = ‐ \frac{1}{2} \sqrt{93}$
$x = 5 \frac{1}{2} + \frac{1}{2} \sqrt{93}$ of $x = 5 \frac{1}{2} - \frac{1}{2} \sqrt{93}$

Rechterkolom:
$x^{2} - 12 x = ‐ 23$
$x^{2} - 12 x + 36 = ‐ 23 + 36$
${(x - 6)}^{2} = 13$
$x - 6 = \sqrt{13}$ of $x - 6 = ‐ \sqrt{13}$
$x = 6 + \sqrt{13}$ of $x = 6 - \sqrt{13}$

$x^{2} - 5 x - 1 = 0$
$x^{2} - 5 x - 1 + 7 \frac{1}{4} = 7 \frac{1}{4}$
$x^{2} - 5 x + 6 \frac{1}{4} = 7 \frac{1}{4}$
${(x - 2 \frac{1}{2})}^{2} = 7 \frac{1}{4} = \frac{29}{4}$
$x - 2 \frac{1}{2} = \sqrt{\frac{29}{4}} = \frac{1}{2} \sqrt{29}$ of $x - 2 \frac{1}{2} = ‐ \sqrt{\frac{29}{4}} = ‐ \frac{1}{2} \sqrt{29}$
$x = 2 \frac{1}{2} + \frac{1}{2} \sqrt{29}$ of $x = 2 \frac{1}{2} - \frac{1}{2} \sqrt{29}$

$x^{2} = x + 3$
$x^{2} - x = 3$
$x^{2} - x + \frac{1}{4} = 3 + \frac{1}{4}$
${(x - \frac{1}{2})}^{2} = 3 \frac{1}{4} = \frac{13}{4}$
$x - \frac{1}{2} = \sqrt{\frac{13}{4}} = \frac{1}{2} \sqrt{13}$ of $x - \frac{1}{2} = ‐ \sqrt{\frac{13}{4}} = ‐ \frac{1}{2} \sqrt{13}$
$x = \frac{1}{2} + \frac{1}{2} \sqrt{13}$ of $x = \frac{1}{2} - \frac{1}{2} \sqrt{13}$

$x^{2} - 9 x - 3 = 0$
$x^{2} - 9 x - 3 + 23 \frac{1}{4} = 23 \frac{1}{4}$
$x^{2} - 9 x + 20 \frac{1}{4} = 23 \frac{1}{4}$
${(x - 4 \frac{1}{2})}^{2} = 23 \frac{1}{4} = \frac{93}{4}$
$x - 4 \frac{1}{2} = \sqrt{\frac{93}{4}} = \frac{1}{2} \sqrt{93}$ of $x - 4 \frac{1}{2} = ‐ \sqrt{\frac{93}{4}} = ‐ \frac{1}{2} \sqrt{93}$
$x = 4 \frac{1}{2} + \frac{1}{2} \sqrt{93}$ of $x = 4 \frac{1}{2} - \frac{1}{2} \sqrt{93}$

4

a

$a = 2$ , $b = 12$ en $c = 6$

b

$x = \frac{‐ 12 + \sqrt{12^{2} - 4 \cdot 2 \cdot 6}}{2 \cdot 2} = \frac{‐ 12 + \sqrt{96}}{4} = \frac{‐ 12 + 4 \sqrt{6}}{4} = ‐ 3 + \sqrt{6}$
$x = \frac{‐ 12 - \sqrt{12^{2} - 4 \cdot 2 \cdot 6}}{2 \cdot 2} = \frac{‐ 12 - \sqrt{96}}{4} = \frac{‐ 12 - 4 \sqrt{6}}{4} = ‐ 3 - \sqrt{6}$

5

Dan staat er een lineaire vergelijking.

6

$2 x {}^{2}- 3 x - 35 = 0$ $\begin{array}{l} a = 2 \\ b = ‐ 3 \\ c = ‐ 35 \end{array}} \begin{matrix} D = 9 - 4 \cdot 2 \cdot ‐ 35 = 289 \\ \sqrt{D} = \sqrt{289} = 17 \end{matrix}$ $x = \frac{3 + 17}{4} = 5$ of $x = \frac{3 - 17}{4} = ‐ 3 \frac{1}{2}$	$4 x = 1 + 4 x^{2}$ $4 x^{2} - 4 x + 1 = 0$ $\begin{array}{l} a = 4 \\ b = ‐ 4 \\ c = 1 \end{array}} D = 16 - 4 \cdot 4 \cdot 1 = 0$ $x = ‐ \frac{‐ 4}{8} = \frac{1}{2}$

$2 x^{2} + 4 x - 1 = 0$ $\begin{array}{l} a = 2 \\ b = 4 \\ c = ‐ 1 \end{array}} \begin{matrix} D = 16 - 4 \cdot 2 \cdot ‐ 1 = 24 \\ \sqrt{D} = \sqrt{24} = 2 \sqrt{6} \end{matrix}$ $x = \frac{‐ 4 + 2 \sqrt{6}}{4} = ‐ 1 + \frac{1}{2} \sqrt{6}$ of $x = \frac{‐ 4 - 2 \sqrt{6}}{4} = ‐ 1 - \frac{1}{2} \sqrt{6}$	${(x - 3)}^{2} = 5 - 3 x$ $x^{2} - 6 x + 9 = 5 - 3 x$ $x^{2} - 3 x + 4 = 0$ $\begin{array}{l} a = 1 \\ b = ‐ 3 \\ c = 4 \end{array}} D = 9 - 4 \cdot 1 \cdot 4 = ‐ 7$ $D < 0$ , dus géén oplossingen

$7 x^{2} - 6 x + 2 = 0$ $\begin{array}{l} a = 7 \\ b = ‐ 6 \\ c = 2 \end{array}} D = 36 - 4 \cdot 7 \cdot 2 = ‐ 20$ $D < 0$ , dus géén oplossingen	$5 x - 3 x^{2} = 0$ $\begin{array}{l} a = ‐ 3 \\ b = 5 \\ c = 0 \end{array}} \begin{matrix} D = 25 - 4 \cdot ‐ 3 \cdot 0 = 25 \\ \sqrt{D} = \sqrt{25} = 5 \end{matrix}$ $x = \frac{‐ 5 + 5}{‐ 6} = 0$ of $x = \frac{‐ 5 - 5}{‐ 6} = \frac{‐ 10}{‐6} = 1 \frac{2}{3}$

$\frac{1}{2} x^{2} - 3 x - 4 \frac{1}{2} = 0$ $\begin{array}{l} a = \frac{1}{2} \\ b = ‐ 3 \\ c = ‐ 4 \frac{1}{2} \end{array}} \begin{matrix} D = 9 - 4 \cdot \frac{1}{2} \cdot ‐ 4 \frac{1}{2} = 18 \\ \sqrt{D} = \sqrt{18} = 3 \sqrt{2} \end{matrix}$ $x = \frac{3 + 3 \sqrt{2}}{1} = 3 + 3 \sqrt{2}$ of $x = \frac{3 - 3 \sqrt{2}}{1} = 3 - 3 \sqrt{2}$