1

a

${(x + 6)}^{2}$	$=$	$16$	haakjes weg
$x^{2} + 12 x + 36$	$=$	$16$	op nul herleiden
$x^{2} + 12 x + 20$	$=$	$0$	ontbinden
$(x + 2) (x + 10)$	$=$	$0$
$x = ‐ 2$	of	$x = ‐ 10$

b

$4$ of $‐ 4$

c

$x + 6 = 4$ of $x + 6 = ‐ 4$
dus $x = ‐ 2$ of $x = ‐ 10$

d

Linkerkolom

${(x - 3)}^{2}$	$=$	$25$
$x - 3 = ‐ 5$	of	$x - 3 = 5$
$x = ‐ 2$	of	$x = 8$

${(x - 2)}^{2}$	$=$	$0$
$x - 2$	$=$	$0$
$x$	$=$	$2$

${(x + 2)}^{2}$	$=$	$‐ 9$
geen oplossing

Rechterkolom

${(x + 5)}^{2}$	$=$	$100$
$x + 5 = ‐ 10$	of	$x + 5 = 10$
$x = ‐ 15$	of	$x = 5$

${(x - 1)}^{2}$	$=$	$\frac{9}{4}$
$x - 1 = ‐ \frac{3}{2}$	of	$x - 1 = \frac{3}{2}$
$x = ‐ \frac{1}{2}$	of	$x = 2 \frac{1}{2}$

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

2

$x = ‐ 5 - \sqrt{7}$ of $x = ‐ 5 + \sqrt{7}$	$x = 5 - \sqrt{3}$ of $x = 5 + \sqrt{3}$
$x = ‐ 2 - \sqrt{2}$ of $x = ‐ 2 + \sqrt{2}$	$x = 1 - \sqrt{7}$ of $x = 1 + \sqrt{7}$
$x = 1 - \sqrt{10}$ of $x = 1 + \sqrt{10}$	$x = 2 - \frac{1}{2} \sqrt{6}$ of $x = 2 + \frac{1}{2} \sqrt{6}$

3

a

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

b

$x {}^{2}+ 6 x = 7$
$x {}^{2}+ 6 x - 7 = 0$
$(x - 1) (x + 7) = 0$
$x = 1$ of $x = ‐ 7$
Dus $x = 1$ , want $x$ kan niet negatief zijn.

c

$x {}^{2}+ 6 x = 16$
$x {}^{2}+ 6 x - 16 = 0$
$(x - 2) (x + 8) = 0$
$x = 2$ of $x = ‐ 8$
Dus $x = 2$ , want $x$ kan niet negatief zijn.

d

${(‐ 3 + \sqrt{19})}^{2} + 6 (‐ 3 + \sqrt{19}) = 9 - 6 \sqrt{19} + 19 - 18 + 6 \sqrt{19} = 10$ , klopt.

e

9

f

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

g

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

h

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

4

a

$x + 5$

b

25

c

$x = ‐ 5 + \sqrt{5}$ of $x = ‐ 5 - \sqrt{5}$

d

${(x + 5)}^{2} = 37$

e

$x = ‐ 5 + \sqrt{37}$ of $x = ‐ 5 - \sqrt{37}$

5

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 $x = ‐ 6 - 4 \sqrt{2}$

6

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

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

7

a

$‐ x^{2} + 3 x = 4 x - 5$
$x^{2} - 3 x = ‐ 4 x + 5$
$x^{2} + x = 5$
${(x + \frac{1}{2})}^{2} - \frac{1}{4} = 5$
${(x + \frac{1}{2})}^{2} = 5 \frac{1}{4} = \frac{21}{4}$
$x + \frac{1}{2} = \sqrt{\frac{21}{4}} = \frac{1}{2} \sqrt{21}$ of $x + \frac{1}{2} = ‐ \frac{1}{2} \sqrt{21}$
$x = ‐ \frac{1}{2} + \frac{1}{2} \sqrt{21}$ of $x = ‐ \frac{1}{2} - \frac{1}{2} \sqrt{21}$

b

$2 x^{2} = 4 x + 6$
$x^{2} = 2 x + 3$
$x^{2} - 2 x = 3$
${(x - 1)}^{2} - 1 = 3$
${(x - 1)}^{2} = 4$
$x - 1 = 2$ of $x - 1 = ‐ 2$
$x = 3$ of $x = ‐ 1$

Handiger:
$x^{2} - 2 x - 3 = 0$
$(x - 3) (x + 1) = 0$
$x = 3$ of $x = ‐ 1$

c

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

d

$x^{2} + 5 x + 3 = 0$
${(x + 2 \frac{1}{2})}^{2} - 6 \frac{1}{4} + 3 = 0$
${(x + 2 \frac{1}{2})}^{2} = 3 \frac{1}{4} = \frac{13}{4}$
$x + 2 \frac{1}{2} = \sqrt{\frac{13}{4}} = \frac{1}{2} \sqrt{13}$ of $x + 2 \frac{1}{2} = ‐ \frac{1}{2} \sqrt{13}$
$x = ‐ 2 \frac{1}{2} + \frac{1}{2} \sqrt{13}$ of $x = ‐ 2 \frac{1}{2} - \frac{1}{2} \sqrt{13}$

e

$3 x^{2} + 6 x + 9 = 0$
$x^{2} + 2 x + 3 = 0$
${(x + 1)}^{2} - 1 + 3 = 0$
${(x + 1)}^{2} = ‐ 2$
Er zijn geen oplossingen.

f

$‐ 2 x^{2} - 4 x = 20$
$‐ 2 x^{2} - 4 x - 20 = 0$
$x^{2} + 2 x + 10 = 0$
${(x + 1)}^{2} - 1 + 10 = 0$
${(x + 1)}^{2} = ‐ 9$
Er zijn geen oplossingen.

g

${(2 x)}^{2} = 4 x - 1$
$4 x^{2} - 4 x + 1 = 0$
$x^{2} - x + \frac{1}{4} = 0$
${(x - \frac{1}{2})}^{2} - \frac{1}{4} + \frac{1}{4} = 0$
${(x - \frac{1}{2})}^{2} = 0$
$x = \frac{1}{2}$

8

a

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

b

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

c

$a = 3$ , $b = 5$ en $c = ‐ 2$
$x = \frac{‐ 5 + \sqrt{5^{2} - 4 \cdot 3 \cdot ‐ 2}}{2 \cdot 3} = \frac{‐ 5 + \sqrt{49}}{6} = \frac{‐ 5 + 7}{6} = \frac{2}{6} = \frac{1}{3}$
$x = \frac{‐ 5 - \sqrt{5^{2} - 4 \cdot 3 \cdot ‐ 2}}{2 \cdot 3} = \frac{‐ 5 - \sqrt{49}}{6} = \frac{‐ 5 - 7}{6} = \frac{‐ 12}{6} = ‐ 2$

d

$a = \frac{1}{3}$ , $b = ‐ 2$ en $c = 3$
$x = \frac{‐ (‐ 2) + \sqrt{{(‐ 2)}^{2} - 4 \cdot \frac{1}{3} \cdot 3}}{2 \cdot \frac{1}{3}} = \frac{2 + \sqrt{0}}{\frac{2}{3}} = 3$
$x = \frac{‐ (‐ 2) - \sqrt{{(‐ 2)}^{2} - 4 \cdot \frac{1}{3} \cdot 3}}{2 \cdot \frac{1}{3}} = \frac{2 - \sqrt{0}}{\frac{2}{3}} = 3$
Je krijgt twee keer dezelfde oplossing, dus er is maar één oplossing.

e

$a = ‐ 2$ , $b = 3$ en $c = ‐ 5$
$x = \frac{‐ 3 + \sqrt{3^{2} - 4 \cdot ‐ 2 \cdot ‐ 5}}{2 \cdot ‐ 2} = \frac{‐ 3 + \sqrt{‐ 31}}{‐ 4}$ kan niet!
$x = \frac{‐ 3 - \sqrt{3^{2} - 4 \cdot ‐ 2 \cdot ‐ 5}}{2 \cdot ‐ 2} = \frac{‐ 3 - \sqrt{‐ 31}}{‐ 4}$ kan niet!
$\sqrt{‐ 31}$ bestaat niet, dus er is géén oplossing.

f

Met kwadraatafsplitsen:

$3 x^{2} + 5 x - 1$	$=$	$0$	DELEN DOOR 3
$x^{2} + \frac{5}{3} x - \frac{1}{3}$	$=$	$0$	kwadraatafsplitsen
${(x + \frac{5}{6})}^{2} - \frac{25}{36} - \frac{1}{3}$	$=$	$0$	breuken samennemen
${(x + \frac{5}{6})}^{2} - \frac{37}{36}$	$=$	$0$	PLUS $\frac{37}{36}$
${(x + \frac{5}{6})}^{2}$	$=$	$\frac{37}{36}$
$x + \frac{5}{6} = ‐ \sqrt{\frac{37}{36}}$	of	$x + \frac{5}{6} = \sqrt{\frac{37}{36}}$
$x = ‐ \frac{5}{6} - \sqrt{\frac{37}{36}}$	of	$x = ‐ \frac{5}{6} + \sqrt{\frac{37}{36}}$
( $x = ‐ \frac{5}{6} - \frac{1}{6} \sqrt{37}$	of	$x = ‐ \frac{5}{6} + \frac{1}{6} \sqrt{37}$ )

Met de abc-formule:

a = 3

,

b = 5

en

c = ‐ 1

x = \frac{‐ 5 + \sqrt{5^{2} - 4 \cdot 3 \cdot ‐ 1}}{2 \cdot 3} = \frac{‐ 5 + \sqrt{37}}{6} = ‐ \frac{5}{6} + \frac{1}{6} \sqrt{37}

x = \frac{‐ 5 - \sqrt{5^{2} - 4 \cdot 3 \cdot ‐ 1}}{2 \cdot 3} = \frac{‐ 5 - \sqrt{37}}{6} = ‐ \frac{5}{6} - \frac{1}{6} \sqrt{37}

9

Dan staat er een lineaire vergelijking.

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 $x = \frac{‐ 5 - 5}{‐ 6} = \frac{10}{6} = 1 \frac{2}{3}$

11

a

$0 = x^{2} - 3 x = x (x - 3)$ , dus $x = 0$ of $x = 3$ .
Dus de nulpunten zijn $0$ en $3$ .

b

Dat zijn de snijpunten van de parabool met de $x$ -as.

c

Die zit midden tussen nulpunten, dus de symmetrieas: $x = \frac{0 + 3}{2} = 1 \frac{1}{2}$

d

$y = {(1 \frac{1}{2})}^{2} - 3 \cdot 1 \frac{1}{2} = ‐ 2 \frac{1}{4}$ ; top $(1 \frac{1}{2}, ‐ 2 \frac{1}{4})$

e

$0 = ‐ x^{2} - 3 x + 40$ $\to$ $0 = ‐ (x - 5) (x + 8)$ $\to$ $x = 5$ of $x = ‐ 8$
symmetrieas: $x = \frac{5 + ‐ 8}{2} = ‐ 1 \frac{1}{2}$
$y = ‐ {(‐ 1 \frac{1}{2})}^{2} - 3 \cdot ‐ 1 \frac{1}{2} + 40 = 42 \frac{1}{4}$ dus top $(‐ 1 \frac{1}{2},42 \frac{1}{4})$

f

$0 = x^{2} - 4 x + 2$ $\to$ $x = \frac{4 - \sqrt{8}}{2} = 2 - \sqrt{2}$ of $x = \frac{4 + \sqrt{8}}{2} = 2 + \sqrt{2}$
symmetrieas: $x = \frac{2 - \sqrt{2} + 2 + \sqrt{2}}{2} = 2$
$y = 2^{2} - 4 \cdot 2 + 2 = ‐ 2$ dus top $(2, ‐ 2)$

12

a

$0 = x^{2} - 2 x + 4$ $\to$ $D = {(‐ 2)}^{2} - 4 \cdot 1 \cdot 4 = ‐ 12$
$D < 0$ dus géén oplossingen.

b

$x = 0$ geeft $y = 0^{2} - 2 \cdot 0 + 4 = 4$ , dus snijpunt $(0,4)$

c

$x^{2} - 2 x + 4 = 4$ $\to$ $x^{2} - 2 x = 0$ $\to$ $x (x - 2) = 0$ $\to$ $x = 0$ of $x = 2$

d

symmetrieas: $x = \frac{0 + 2}{2} = 1$
$y = 1^{2} - 2 \cdot 1 + 4 = 3$ dus top $(1,3)$

e

Snijpunt $y$ -as: $(0, ‐ 5)$
$‐ x^{2} + 4 x - 5 = ‐ 5$ $\to$ $0 = x^{2} - 4 x = x (x - 4)$ $\to$ $x = 0$ of $x = 4$
symmetrieas: $x = \frac{0 + 4}{2} = 2$
$y = ‐ 1$ dus top $(2, ‐ 1)$

13

a

$3 x^{2} - 3 x + 3 = 3$ $\to$ $x^{2} - x = 0$ $\to$ $x (x - 1) = 0$ $\to$ $x = 0$ of $x = 1$ $\to$ $x_{t o p} = \frac{1}{2}$ $\to$ top: $(\frac{1}{2},2 \frac{1}{4})$

b

$D = {(‐ p)}^{2} - 4 \cdot p \cdot 3 = p^{2} - 12 p$
Één oplossing, dus $D = 0$ $\to$ $p^{2} - 12 p = 0$ $\to$ $p (p - 12) = 0$ $\to$ $p = 0$ of $p = 12$
Maar $p = 0$ voldoet niet, dus $p = 12$

c

Op de $x$ -as