Getal & Ruimte (13e editie) - vwo wiskunde B

'Wortelvergelijkingen'.

3 vwo 5.6 Wortelvergelijkingen

Wortelvergelijkingen (1)

opgave 1

Los exact op.

3p

\(3 + 5 \sqrt{x} = 9\)

Wortel (1)
008o - Wortelvergelijkingen - basis - 1ms - dynamic variables

○

(Isoleren)
\(5 \sqrt{x} = 6\)

1p

○

(Kwadrateren)
\((5 \sqrt{x})^{2} = 6^{2}\)
\(25 x = 36\)
\(x = 1\frac{11}{25}\)

1p

○

(Controleren)
\(x = 1\frac{11}{25}\) voldoet.

1p

vwo wiskunde B 4.3 Regels voor het oplossen van vergelijkingen

Wortelvergelijkingen (4)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{3 x + 18}\)

Wortel (2)
008n - Wortelvergelijkingen - basis - 0ms - dynamic variables

a

(Kwadrateren)
\(x^{2} = 3 x + 18\)

1p

○

(Oplossen)
\(1 x^{2} + -3 x + -18 = 0\)
\((x + 3) (x + -6) = 0\)
\(x = -3 ∨ x = 6\)

1p

○

(Controleren)
\(x = -3\) voldoet niet, \(x = 6\) voldoet.

1p

4p

b

\(8 x + 2 \sqrt{x} = 3\)

Wortel (4)
008p - Wortelvergelijkingen - basis - 5ms - dynamic variables

b

(Isoleren)
\(8 x - 3 = -2 \sqrt{x}\)

1p

○

(Kwadrateren)
\((8 x - 3)^{2} = (-2 \sqrt{x})^{2}\)
\(64 x^{2} - 48 x + 9 = 4 x\)

1p

○

(Oplossen)
\(64 x^{2} + -52 x + 9 = 0\)
\(D = -52^{2} - 4 ⋅ 64 ⋅ 9 = 400\)
\(x = {52 - \sqrt{400} \over 2 ⋅ 64} ∨ x = {52 + \sqrt{400} \over 2 ⋅ 64}\)
\(x = {1 \over 4} ∨ x = {9 \over 16}\)

1p

○

(Controleren)
\(x = \frac{1}{4}\) voldoet, \(x = \frac{9}{16}\) voldoet niet.

1p

4p

c

\(x = \sqrt{2 x + 16} - 8\)

Wortel (3)
008q - Wortelvergelijkingen - basis - 0ms - dynamic variables

c

(Isoleren)
\(x + 8 = \sqrt{2 x + 16}\)

1p

○

(Kwadrateren)
\((x + 8)^{2} = (\sqrt{2 x + 16})^{2}\)
\(x^{2} + 16 x + 64 = 2 x + 16\)

1p

○

(Oplossen)
\(1 x^{2} + 14 x + 48 = 0\)
\((x + 8) (x + 6) = 0\)
\(x = -8 ∨ x = -6\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

4p

d

\(4 x - 4 \sqrt{3 x - 8} = 9\)

Wortel (5)
008r - Wortelvergelijkingen - basis - 489ms - dynamic variables

d

(Isoleren)
\(4 x - 9 = 4 \sqrt{3 x - 8}\)

1p

○

(Kwadrateren)
\((4 x - 9)^{2} = (4 \sqrt{3 x - 8})^{2}\)
\(16 x^{2} - 72 x + 81 = 16 ⋅ (3 x - 8)\)
\(16 x^{2} - 72 x + 81 = 48 x - 128\)

1p

○

(Oplossen)
\(16 x^{2} + -120 x + 209 = 0\)
\(D = -120^{2} - 4 ⋅ 16 ⋅ 209 = 1024\)
\(x = {120 - \sqrt{1024} \over 2 ⋅ 16} ∨ x = {120 + \sqrt{1024} \over 2 ⋅ 16}\)
\(x = {11 \over 4} ∨ x = {19 \over 4}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

"