1

a

$x (x - 5) = 2 \cdot 7$
$x^{2} - 5 x = 14$
$x^{2} - 5 x - 14 = 0$
$(x - 7) (x + 2) = 0$
$x = 7$ of $x = ‐ 2$
Beide oplossingen voldoen.

b

$2 x^{2} - 4 x - 7 = 0$
$x^{2} - 2 x - 3 \frac{1}{2} = 0$
${(x - 1)}^{2} - 1 - 3 \frac{1}{2} = 0$
${(x - 1)}^{2} = 4 \frac{1}{2} = \frac{18}{4}$
$x - 1 = \sqrt{\frac{18}{4}} = \frac{1}{2} \sqrt{18} = 1 \frac{1}{2} \sqrt{2}$ of $x - 1 = ‐ 1 \frac{1}{2} \sqrt{2}$
$x = 1 + 1 \frac{1}{2} \sqrt{2}$ of $x = 1 - 1 \frac{1}{2} \sqrt{2}$

c

$x^{2} - 5 x - 5 = 0$
${(x - 2 \frac{1}{2})}^{2} - 6 \frac{1}{4} - 5 = 0$
${(x - 2 \frac{1}{2})}^{2} = 11 \frac{1}{4} = \frac{45}{4}$
$x - 2 \frac{1}{2} = \sqrt{\frac{45}{4}} = \frac{1}{2} \sqrt{45} = 1 \frac{1}{2} \sqrt{5}$ of $x - 2 \frac{1}{2} = ‐ 1 \frac{1}{2} \sqrt{5}$
$x = 2 \frac{1}{2} + 1 \frac{1}{2} \sqrt{5}$ of $x = 2 \frac{1}{2} - 1 \frac{1}{2} \sqrt{5}$

d

$4 {(x + 1)}^{2} = 25 \cdot 1$
${(x + 1)}^{2} = \frac{25}{4}$
$x + 1 = \sqrt{\frac{25}{4}} = \frac{5}{2} = 2 \frac{1}{2}$ of $x + 1 = ‐ 2 \frac{1}{2}$
$x = 1 \frac{1}{2}$ of $x = ‐ 3 \frac{1}{2}$
Beide oplossingen voldoen.

e

$2 x^{2} + 8 x + 10 = 4 x^{2}$
$2 x^{2} - 8 x - 10 = 0$
$x^{2} - 4 x - 5 = 0$
$(x - 5) (x + 1) = 0$
$x = 5$ of $x = ‐ 1$

f

$x^{2} - 2 x + 4 = 0$
${(x - 1)}^{2} - 1 + 4 = 0$
${(x - 1)}^{2} = ‐ 3$
Er zijn geen oplossingen.

g

$x^{2} = x + 2$
$x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0$
$x = 2$ of $x = ‐ 1$ , maar $x = ‐ 1$ mag niet.

h

${(x + \frac{1}{2})}^{2} - \frac{1}{4} = 35 \frac{3}{4}$
${(x + \frac{1}{2})}^{2} = 36$
$x + \frac{1}{2} = 6$ of $x + \frac{1}{2} = ‐ 6$
$x = 5 \frac{1}{2}$ of $x = ‐ 6 \frac{1}{2}$

i

$x + 1 = 49$
$x = 48$

j

$2 \cdot x^{2} = 1 \cdot (x^{2} + 3 x)$
$x^{2} - 3 x = 0$
$x (x - 3) = 0$
$x = 0$ of $x = 3$
$x = 0$ maakt de noemer 0, dus de enige oplossing is $x = 3$ .

2

a

Parabool A:	Parabool B:
Top $(0,2)$ en punt $(1,1)$	Top $(‐ 2,3)$ en punt $(0,1)$
$1 = c {(1 - 0)}^{2} + 2$	$1 = c {(0 + 2)}^{2} + 3$
$1 = c + 2$	$1 = 4 c + 3$
$‐ 1 = c$	$‐ 2 = 4 c$
$y = ‐ x^{2} + 2$	$‐ \frac{1}{2} = c$
	$y = ‐ \frac{1}{2} {(x + 2)}^{2} + 3$

Parabool C:	Parabool D:
Top $(‐ 1, ‐ 4)$ en punt $(0, ‐ 2)$	Top $(‐ 1,0)$ en punt $(0,1)$
$‐ 2 = c {(0 + 1)}^{2} - 4$	$1 = c {(0 + 1)}^{2} + 0$
$‐ 2 = c - 4$	$1 = c$
$2 = c$	$y = {(x + 1)}^{2}$
$y = 2 {(x + 1)}^{2} - 4$

b

A en C:
$‐ x^{2} + 2 = 2 {(x + 1)}^{2} - 4$
$‐ x^{2} + 2 = 2 x^{2} + 4 x + 2 - 4$
$3 x^{2} + 4 x - 4 = 0$
$\begin{array}{l} a = 3 \\ b = 4 \\ c = ‐ 4 \end{array}\} \begin{matrix} D = 16 - 4 \cdot 3 \cdot ‐ 4 = 64 \\ \sqrt{D} = \sqrt{64} = 8 \end{matrix}$

$x = \frac{‐ 4 + 8}{6} = \frac{2}{3}$	of	$x = \frac{‐ 4 - 8}{6} = ‐ 2$
$y = ‐ {(\frac{2}{3})}^{2} + 2 = 1 \frac{5}{9}$		$y = ‐ {(‐ 2)}^{2} + 2 = ‐ 2$
$(\frac{2}{3},1 \frac{5}{9})$		$(‐ 2, ‐ 2)$

B en D:
$‐ \frac{1}{2} {(x + 2)}^{2} + 3 = {(x + 1)}^{2}$
$‐ \frac{1}{2} x^{2} - 2 x - 2 + 3 = x^{2} + 2 x + 1$
$1 \frac{1}{2} x (x + \frac{8}{3}) = 0$

$x = 0$	of	$x = ‐ \frac{8}{3}$
$y = {(0 + 1)}^{2} = 1$		$y = {(‐ \frac{8}{3} + 1)}^{2} = \frac{25}{9} = 2 \frac{7}{9}$
$(0,1)$		$(‐ \frac{8}{3},2 \frac{7}{9})$

3

a

$y = \frac{1}{4} x^{2} + 3 x + 5$	$y = ‐ 2 x^{2} + 4 x + 6$
$4 y = x^{2} + 12 x + 20$	$y = ‐ 2 (x^{2} - 2 x - 3)$
$4 y = {(x + 6)}^{2} - 36 + 20$	$y = ‐ 2 ({(x - 1)}^{2} - 1 - 3)$
$y = \frac{1}{4} {(x + 6)}^{2} - 4$	$y = ‐ 2 {(x - 1)}^{2} + 8$
Top $(‐ 6, ‐ 4)$	Top $(1,8)$

b

$y = \frac{1}{4} x^{2} + 3 x + 5$	$y = ‐ 2 x^{2} + 4 x + 6$
$4 y = x^{2} + 12 x + 20$	$‐ \frac{1}{2} y = x^{2} - 2 x - 3$
$4 y = (x + 10) (x + 2)$	$‐ \frac{1}{2} y = (x - 3) (x + 1)$
$y = \frac{1}{4} (x + 10) (x + 2)$	$y = ‐ 2 (x - 3) (x + 1)$

4

a

$0 = c {(2 - 1)}^{2} - 2$
$2 = c$
$y = 2 {(x - 1)}^{2} - 2$

b

c

Zie vraag b.

d

$3 x = 2 {(x - 1)}^{2} - 2$
$2 x^{2} - 7 x = 0$
$2 x (x - 3 \frac{1}{2}) = 0$

$x = 0$	of	$x = 3 \frac{1}{2}$
$y = 3 \cdot 0 = 0$		$y = 3 \cdot 3 \frac{1}{2} = 10 \frac{1}{2}$
$(0,0)$		$(3 \frac{1}{2},10 \frac{1}{2})$

e

$3 x + p = 2 {(x - 1)}^{2} - 2$
$2 x {}^{2}- 7 x - p = 0$
$\begin{array}{l} a = 2 \\ b = ‐ 7 \\ c = ‐ p \end{array}\} D = 49 - 4 \cdot 2 \cdot ‐ p = 49 + 8 p$
Eén oplossing als $D = 0$ .
$49 + 8 p = 0$
$8 p = ‐ 49$
$p = ‐ \frac{49}{8} = ‐ 6 \frac{1}{8}$

5

a

Border: $2 x^{2} + 3 \cdot 4 x = 2 x^{2} + 12 x$
Dus $2 x^{2} + 12 x = 4 \cdot 4 = 16$
$2 x^{2} + 12 x - 16 = 0$
$x^{2} + 6 x - 8 = 0$
${(x + 3)}^{2} - 9 - 8 = 0$
${(x + 3)}^{2} = 17$
$x + 3 = \sqrt{17}$ of $x + 3 = ‐ \sqrt{17}$
$x = ‐ 3 + \sqrt{17}$ of $x = ‐ 3 - \sqrt{17}$
Maar $x > 0$ , dus $x = ‐ 3 + \sqrt{17}$ .

b

$2 (2 x^{2} + 12 x) = 16$
$2 x^{2} + 12 x = 8$
$2 x^{2} + 12 x - 8 = 0$
$x^{2} + 6 x - 4 = 0$
${(x + 3)}^{2} - 9 - 4 = 0$
${(x + 3)}^{2} = 13$
$x + 3 = \sqrt{13}$ of $x + 3 = ‐ \sqrt{13}$
$x = ‐ 3 + \sqrt{13}$ of $x = ‐ 3 - \sqrt{13}$
Maar $x > 0$ , dus $x = ‐ 3 + \sqrt{13}$ .

6

linker kolom:

$x^{2} - 8 x + 22 = 0$ $\to$ $\begin{array}{l} a = 1 \\ b = ‐ 8 \\ c = 22 \end{array}\} D = 64 - 4 \cdot 1 \cdot 22 = ‐ 24$
$\to$ $D < 0$ , dus geen oplossingen
$2 x^{2} = 5 x - 3$ $\to$ $2 x^{2} - 5 x + 3 = 0$ $\to$ $\begin{array}{l} a = 2 \\ b = ‐ 5 \\ c = 3 \end{array}\} \begin{matrix} D = 25 - 4 \cdot 2 \cdot 3 = 1 \\ \sqrt{D} = \sqrt{1} = 1 \end{matrix}$
$\to$ $x = \frac{5 + 1}{4} = 1 \frac{1}{2}$ of $x = \frac{5 - 1}{4} = 1$

rechter kolom:

$3 {(x + 2)}^{2} - 2 x = 9$ $\to$ $3 x^{2} + 10 x + 3 = 0$ $\to$ $\begin{array}{l} a = 3 \\ b = 10 \\ c = 3 \end{array}\} \begin{matrix} D = 100 - 4 \cdot 3 \cdot 3 = 64 \\ \sqrt{D} = \sqrt{64} = 8 \end{matrix}$
$\to$ $x = \frac{‐ 10 + 8}{6} = ‐ \frac{1}{3}$ of $x = \frac{‐ 10 - 8}{6} = ‐ 3$
$‐ 5 x^{2} + 4 x - \frac{4}{5} = 0$ $\to$ $\begin{array}{l} a = ‐ 5 \\ b = 4 \\ c = ‐ \frac{4}{5} \end{array}\} D = 16 - 4 \cdot ‐ 5 \cdot ‐ \frac{4}{5} = 0$
$\to$ $x = ‐ \frac{4}{‐ 10} = \frac{2}{5}$

7

$x^{2} + 3 x + p = 0$
$\begin{array}{l} a = 1 \\ b = 3 \\ c = p \end{array}\} D = 9 - 4 \cdot 1 \cdot p = 9 - 4 p$
Twee oplossingen als $D > 0$ : $9 - 4 p > 0$ $\to$ $9 > 4 p$ $\to$ $2 \frac{1}{4} > p$ .
Dus twee oplossingen als $p < 2 \frac{1}{4}$ .

Eén oplossing als $D = 0$ : $p = 2 \frac{1}{4}$ .

Geen oplossingen als $D < 0$ : $p > 2 \frac{1}{4}$ .

8

a

$0 = \frac{1}{2} x^{2} + 2 x + \frac{1}{2}$
$0 = x^{2} + 4 x + 1$
$x = \frac{‐ 4 - \sqrt{12}}{2}$ of $x = \frac{‐ 4 + \sqrt{12}}{2}$
$x = ‐ 2 - \sqrt{3}$ of $x = ‐ 2 + \sqrt{3}$ (want $\sqrt{12} = 2 \sqrt{3}$ )
Dus: $(‐ 2 - \sqrt{3},0)$ en $(‐ 2 + \sqrt{3},0)$

b

$y = \frac{1}{2} x^{2} + 2 x + \frac{1}{2}$
$2 y = x^{2} + 4 x + 1$
$2 y = {(x + 2)}^{2} - 4 + 1$
$2 y = {(x + 2)}^{2} - 3$
$y = \frac{1}{2} {(x + 2)}^{2} - 1 \frac{1}{2}$

c

De top is $(‐ 2, ‐ 1 \frac{1}{2})$ , dus moet de parabool $1 \frac{1}{2}$ omhoog worden geschoven.
De formule is dan $y = \frac{1}{2} {(x + 2)}^{2}$ ( $= \frac{1}{2} x + 2 x + 2$ )

d

$\frac{1}{2} x^{2} + 2 x + \frac{1}{2} = ‐ 2 x - 3$
$x^{2} + 4 x + 1 = ‐ 4 x - 6$
$x^{2} + 8 x + 7 = 0$
$(x + 1) (x + 7) = 0$
$x = ‐ 1$ of $x = ‐ 7$
Snijpunten: $(‐ 1, ‐ 1)$ en $(‐ 7,11)$

9

a

$‐ 2 x^{2} + 12 x - 13 = 0$
$x = \frac{‐ 12 - \sqrt{40}}{‐ 4}$ of $x = \frac{‐ 12 + \sqrt{40}}{‐ 4}$
$x = 3 + \frac{1}{2} \sqrt{10}$ of $x = 3 - \frac{1}{2} \sqrt{10}$ (want $\sqrt{40} = 2 \sqrt{10}$ )
Dus: $(3 - \frac{1}{2} \sqrt{10},0)$ en $(3 + \frac{1}{2} \sqrt{10},0)$

b

$y = ‐ 2 x^{2} + 12 x - 13$ $\to$ $‐ \frac{1}{2} y = x^{2} - 6 x + 6 \frac{1}{2}$
$‐ \frac{1}{2} y = {(x - 3)}^{2} - 9 + 6 \frac{1}{2}$
$‐ \frac{1}{2} y = {(x - 3)}^{2} - 2 \frac{1}{2}$
$y = ‐ 2 {(x - 3)}^{2} + 5$ ;
top $(3,5)$

c

Vergelijking lijn: $y = 2 x + b$
$‐ 2 x^{2} + 12 x - 13 = 2 x + b$ $\to$ $2 x^{2} - 10 x + 13 + b = 0$
Raken, dus de discriminant = 0: ${(‐ 10)}^{2} - 4 \cdot 2 \cdot (13 + b) = 0$ $\to$ $100 - 104 - 8 b = 0$ ,
dus $b = ‐ \frac{1}{2}$
Raakpunt: $2 x^{2} - 10 x + 12 \frac{1}{2} = 0$ $\to$ $x = \frac{10 \pm \sqrt{0}}{4} = 2 \frac{1}{2}$ , dus raakpunt $(2 \frac{1}{2},4 \frac{1}{2})$

10

$‐ 2 x^{2} + 2 x = ‐ 2 x + p$ $\to$ $0 = 2 x^{2} - 4 x + p$
Twee snijpunten, dus $D = 16 - 8 p > 0$ $\to$ $16 > 8 p$ $\to$ $2 > p$
Antwoord: $p < 2$

11

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

$\sqrt{x^{2} + 3 x} = 3 \sqrt{2}$
$x^{2} + 3 x = 18$
$x^{2} + 3 x - 18 = 0$
$(x - 3) (x + 6) = 0$
$x = 3$ of $x = ‐ 6$

12

a

$‐ 0,012 a^{2} + 1,152 a = 0$
$a (‐ 0,012 a + 1,152) = 0$
$a = 0$ of $‐ 0,012 a + 1,152 = 0$
$a = 0$ of $a = \frac{1,152}{0,012} = 96$
De bal landt na $96$ meter.

b

Top zit halverwege, dus bij $a = 48$ .
Invullen: $h = 27,648 = 27,6$ m (of $276$ dm)

c

Top $(50,32)$ , dus $h = c {(a - 50)}^{2} + 32$
Punt $(100,0)$ invullen: $0 = c {(100 - 50)}^{2} + 32 = 2500 c + 32$ $\to$ $c = \frac{‐ 32}{2500} = ‐ 0,0128$
$h = ‐ 0,0128 {(a - 50)}^{2} + 32$