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${(x - 3)}^{2} - 9 + {(y - 3)}^{2} - 9 + 1 = 0$
${(x - 3)}^{2} + {(y - 3)}^{2} = 17$
Middelpunt $(3,3)$ en straal $\sqrt{17}$ .

Zie opgave b.

${(x - 3)}^{2} + {(3 x + 5 - 3)}^{2} = 17$
${(x - 3)}^{2} + {(3 x + 2)}^{2} = 17$
$x^{2} - 6 x + 9 + 9 x^{2} + 12 x + 4 = 17$
$10 x^{2} + 6 x + 13 = 17$
$10 x^{2} + 6 x - 4 = 0$
$x^{2} + \frac{6}{10} x - \frac{4}{10} = 0$
${(x + \frac{3}{10})}^{2} - \frac{9}{100} - \frac{4}{10} = 0$
${(x + \frac{3}{10})}^{2} = \frac{49}{100}$
$x + \frac{3}{10} = \sqrt{\frac{49}{100}} = \frac{7}{10}$ of $x + \frac{3}{10} = ‐ \frac{7}{10}$
$x = \frac{4}{10} = \frac{2}{5}$ of $x = ‐ 1$
Als $x = \frac{2}{5}$ , dan $y = 3 \cdot \frac{2}{5} + 5 = 6 \frac{1}{5}$ .
Als $x = ‐ 1$ , dan $y = 3 \cdot ‐ 1 + 5 = 2$ .
Snijpunten $(\frac{2}{5},6 \frac{1}{5})$ en $(‐ 1,2)$ .

${(60 - 2 x)}^{2} = 500$
$60 - 2 x = \sqrt{500} = 10 \sqrt{5}$ of $60 - 2 x = ‐ 10 \sqrt{5}$
$‐ 2 x = ‐ 60 + 10 \sqrt{5}$ of $‐ 2 x = ‐ 60 - 10 \sqrt{5}$
$x = 30 - 5 \sqrt{5}$ of $x = 30 + 5 \sqrt{5}$
Maar $0 < x < 30$ , dus $x = 30 - 5 \sqrt{5}$ .

${(60 - 2 x)}^{2} = 4 \cdot x \cdot (60 - 2 x)$
$3600 - 240 x + 4 x^{2} = 240 x - 8 x^{2}$
$12 x^{2} - 480 x + 3600 = 0$
$x^{2} - 40 x + 300 = 0$
$(x - 30) (x - 10) = 0$
$x = 30$ of $x = 10$
Maar $0 < x < 30$ , dus $x = 10$ .

$2 (x + 3) = x (x + 1)$
$2 x + 6 = x^{2} + x$
$x^{2} - x - 6 = 0$
$(x - 3) (x + 2) = 0$
$x = 3$ of $x = ‐ 2$
Geen van beide oplossingen maken noemers 0, dus de oplossingen zijn $x = 3$ en $x = ‐ 2$ .

$2 (x + 1) = 1 (x^{2} + x)$
$2 x + 2 = x^{2} + x$
$x^{2} - x - 2 = 0$
$(x - 2) (x + 1) = 0$
$x = 2$ of $x = ‐ 1$
$x = ‐ 1$ maakt noemers 0, dus de enige oplossing is $x = 2$ .

$C = (6 - 2 \cdot 1 \frac{1}{2}) \cdot 1 \frac{1}{2} \cdot 10 = 3 \cdot 1 \frac{1}{2} \cdot 10 = 45$

$C = (6 - 2 x) \cdot x \cdot 10 = 60 x - 20 x^{2}$

$60 x - 20 x^{2} = 20$
$0 = 20 x^{2} - 60 x + 20$
$x^{2} - 3 x + 1 = 0$
${(x - 1 \frac{1}{2})}^{2} - 2 \frac{1}{4} + 1 = 0$
${(x - 1 \frac{1}{2})}^{2} = 1 \frac{1}{4} = \frac{5}{4}$
$x - 1 \frac{1}{2} = \sqrt{\frac{5}{4}} = \frac{1}{2} \sqrt{5}$ of $x - 1 \frac{1}{2} = ‐ \frac{1}{2} \sqrt{5}$
$x = 1 \frac{1}{2} + \frac{1}{2} \sqrt{5}$ of $x = 1 \frac{1}{2} - \frac{1}{2} \sqrt{5}$
Allebei de oplossingen voldoen.

Zie opgave a.

$W A = \sqrt{9^{2} + 3^{2}} = \sqrt{90} = 3 \sqrt{10}$ en $W B = \sqrt{1^{2} + 3^{2}} = \sqrt{10}$

Geldt $W A^{2} + W B^{2} = A B^{2}$ ?
Ja, want ${\sqrt{90}}^{2} + {\sqrt{10}}^{2} = 90 + 10 = 100 = 10^{2}$ , dus hoek $W$ is recht.

Zie opgave a.

$\sqrt{{(x - 5)}^{2} + y^{2}}$

$P A^{2} + P B^{2} = A B^{2}$ , dus ${(x + 5)}^{2} + y^{2} + {(x - 5)}^{2} + y^{2} = 100$

$x^{2} + 10 x + 25 + y^{2} + x^{2} - 10 x + 25 + y^{2} = 100$
$2 x^{2} + 2 y^{2} + 50 = 100$
$x^{2} + y^{2} = 25$
Middelpunt $(0,0)$ en straal 5.

In de kleine: $\frac{x - 3}{x} = \tan (α)$

In de grote: $\frac{8}{x + 7} = \tan (α)$

$\frac{x - 3}{x} = \frac{8}{x + 7}$
$8 x = (x - 3) (x + 7)$
$8 x = x^{2} + 4 x - 21$
$x^{2} - 4 x - 21 = 0$
$(x - 7) (x + 3) = 0$
$x = 7$ of $x = ‐ 3$
Maar $x > 0$ , dus $x = 7$ .

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

$12 x - 2 x^{2} = 2$
$0 = 2 x^{2} - 12 x + 2$
$x^{2} - 6 x + 1 = 0$
${(x - 3)}^{2} - 9 + 1 = 0$
${(x - 3)}^{2} = 8$
$x - 3 = \sqrt{8} = 2 \sqrt{2}$ of $x - 3 = ‐ 2 \sqrt{2}$
$x = 3 + 2 \sqrt{2}$ of $x = 3 - 2 \sqrt{2}$
Maar $0 < x < 3$ , dus $x = 3 - 2 \sqrt{2}$ .

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 $x = ‐ 3 + \sqrt{17}$ .

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