1

$x = 2$ , $x = ‐ 2$	$x = \sqrt{5}$ , $x = ‐ \sqrt{5}$
$x = 1 \pm 2$ , dus $x = ‐ 1$ of $x = 3$	$x = 1 \pm \sqrt{5}$
${(x - 1)}^{2} = 2$ , dus $x = 1 \pm \sqrt{2}$	${(x - 1)}^{2} = 3$ , dus $x = 1 \pm \sqrt{3}$
${(x + 1)}^{2} = 0$ , dus $x = ‐ 1$	${(x + 1)}^{2}$ moet negatief zijn en dat kan niet.

2

-

3

a

$x^{2} + 3 x + 3 x = x^{2} + 6 x$

b

zijde = $x + 3$ ; oppervlakte = ${(x + 3)}^{2}$

c

$x^{2} + 6 x = {(x + 3)}^{2} - 9$

d

$x^{2} + 6 x = 11$
${(x + 3)}^{2} - 9 = 11$
${(x + 3)}^{2} = 20$
$x + 3 = \sqrt{20}$ of $x + 3 = ‐ \sqrt{20}$
$x = ‐ 3 + 2 \sqrt{5}$ of $x = ‐ 3 - 2 \sqrt{5}$

4

a

$x^{2} + 12 x = {(x + 6)}^{2} - 36$

b

$x^{2} + 12 x = 4$
${(x + 6)}^{2} - 36 = 4$
${(x + 6)}^{2} = 40$
$x + 6 = \sqrt{40}$ of $x + 6 = ‐ \sqrt{40}$
$x = ‐ 6 + 2 \sqrt{10}$ of $x = ‐ 6 - 2 \sqrt{10}$

c

$x^{2} + 12 x + 4 = 0$
${(x + 6)}^{2} - 36 + 4 = 0$
${(x + 6)}^{2} = 32$
$x + 6 = \sqrt{32}$ of $x + 6 = ‐ \sqrt{32}$
$x = ‐ 6 + 4 \sqrt{2}$ of $x = ‐ 6 - 4 \sqrt{2}$

5

a

$x^{2} - 20 x = {(x - 10)}^{2} - 100$

b

$x^{2} - 7 x = {(x - 3 \frac{1}{2})}^{2} - 12 \frac{1}{4}$

c

$x^{2} - 8 x = {(x - 4)}^{2} - 16$

d

$x^{2} + 11 x = {(x + 5 \frac{1}{2})}^{2} - 30 \frac{1}{4}$

e

$x^{2} - 21 x = {(x - 10 \frac{1}{2})}^{2} - 110 \frac{1}{4}$

f

$x^{2} + x = {(x + \frac{1}{2})}^{2} - \frac{1}{4}$

g

$x^{2} - x = {(x - \frac{1}{2})}^{2} - \frac{1}{4}$

6

a

$f (x) = {(x - 2)}^{2} - 7$

b

${(x - 2)}^{2}$ neemt alle waarden $\geq 0$ aan en is alleen $0$ als $x = 2$ .

c

$f (x)$ is minimaal $‐ 7$ voor $x = 2$ .

7

a

$‐ 9$ voor $x = 1$

b

$‐ {(x - 2)}^{2}$ is $\leq 0$ voor alle $x$ en alleen $0$ voor $x = 2$ .
Dus $y$ is maximaal $‐ 3$ voor $x = 2$ .

c

Onderdeel a
$y^{'} = 4 x + 4$ , dus de grafiek heeft een horizontale raaklijn in het punt met eerste coordinaat $‐ 1$ .
$y (‐ 1) = ‐ 9$
Onderdeel b
$y^{'} = ‐ 2 x + 4$ , dus dus de grafiek heeft een horizontale raaklijn in het punt met eerste coordinaat $2$ .
$y (2) = ‐ 3$

8

$3 x^{2} + 3 x + 4 = 3 {(x + \frac{1}{2})}^{2} + 3 \frac{1}{4}$
$‐ x^{2} + 5 x + 4 = ‐ {(x - 2 \frac{1}{2})}^{2} + 10 \frac{1}{4}$
$\frac{1}{2} x^{2} - 3 x - 2 = \frac{1}{2} {(x - 3)}^{2} - 6 \frac{1}{2}$

9

Als $a = 0$ , dan is $b x + c = 0$ , dus $x = ‐ \frac{c}{b}$ .

10

linker kolom:

$2 x^{2} - 3 x - 35 = 0$ $\to$ $2 x^{2} - 3 x - 35 = 0$ $\to$ $\begin{array}{l} a = 2 \\ b = ‐ 3 \\ c = ‐ 35 \end{array}\} \begin{matrix} D = 9 - 4 \cdot 2 \cdot ‐ 35 = 289 \\ \sqrt{D} = \sqrt{289} = 17 \end{matrix}$
$\to$ $x = \frac{3 + 17}{4} = 5$ of $x = \frac{3 - 17}{4} = ‐ 3 \frac{1}{2}$
$2 x^{2} + 4 x - 1 = 0$ $\to$ $\begin{array}{l} a = 2 \\ b = 4 \\ c = ‐ 1 \end{array}\} \begin{matrix} D = 16 - 4 \cdot 2 \cdot ‐ 1 = 24 \\ \sqrt{D} = \sqrt{24} = 2 \sqrt{6} \end{matrix}$
$\to$ $x = \frac{‐ 4 + 2 \sqrt{6}}{4} = ‐ 1 + \frac{1}{2} \sqrt{6}$ of $x = \frac{‐ 4 - 2 \sqrt{6}}{4} = ‐ 1 - \frac{1}{2} \sqrt{6}$
$7 x^{2} - 6 x + 2 = 0$ $\to$ $\begin{array}{l} a = 7 \\ b = ‐ 6 \\ c = 2 \end{array}\} D = 36 - 4 \cdot 7 \cdot 2 = ‐ 20$
$\to$ $D < 0$ , dus géén oplossingen
$\frac{1}{2} x^{2} - 3 x - 4 \frac{1}{2} = 0$ $\to$ $\begin{array}{l} a = \frac{1}{2} \\ b = ‐ 3 \\ c = ‐ 4 \frac{1}{2} \end{array}\} \begin{matrix} D = 9 - 4 \cdot \frac{1}{2} \cdot ‐ 4 \frac{1}{2} = 18 \\ \sqrt{D} = \sqrt{18} = 3 \sqrt{2} \end{matrix}$
$\to$ $x = \frac{3 + 3 \sqrt{2}}{1} = 3 + 3 \sqrt{2}$ of $x = \frac{3 - 3 \sqrt{2}}{1} = 3 - 3 \sqrt{2}$

Rechter kolom:

$4 x = 1 + 4 x^{2}$ $\to$ $4 x^{2} - 4 x + 1 = 0$ $\to$ $\begin{array}{l} a = 4 \\ b = ‐ 4 \\ c = 1 \end{array}\} D = 16 - 4 \cdot 4 \cdot 1 = 0$
$\to$ $x = \frac{‐ (‐ 4)}{8} = \frac{1}{2}$
${(x - 3)}^{2} = 5 - 3 x$ $\to$ $x^{2} - 3 x + 4 = 0$ $\to$ $\begin{array}{l} a = 1 \\ b = ‐ 3 \\ c = 4 \end{array}\} D = 9 - 4 \cdot 1 \cdot 4 = ‐ 7$
$\to$ $D < 0$ , dus géén oplossingen
$5 x - 3 x^{2} = 0$ $\to$ $\begin{array}{l} a = ‐ 3 \\ b = 5 \\ c = 0 \end{array}\} \begin{matrix} D = 25 - 4 \cdot ‐ 3 \cdot 0 = 25 \\ \sqrt{D} = \sqrt{25} = 5 \end{matrix}$
$\to$ $x = \frac{‐ 5 + 5}{‐ 6} = 0$ of $x = \frac{‐ 5 - 5}{‐ 6} = \frac{10}{6} = 1 \frac{2}{3}$