Moderne Wiskunde (12.1e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 1.3 Wortelvergelijkingen

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-4 x + 12}\)

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

a

(Kwadrateren)
\(x^{2} = -4 x + 12\)

1p

○

(Oplossen)
\(x^{2} + 4 x - 12 = 0\)
\((x + 6) (x - 2) = 0\)
\(x = -6 ∨ x = 2\)

1p

○

(Controleren)
\(x = -6\) voldoet niet, \(x = 2\) voldoet.

1p

3p

b

\(7 - 5 \sqrt{x} = 6\)

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

b

(Isoleren)
\(-5 \sqrt{x} = -1\)

1p

○

(Kwadrateren)
\((-5 \sqrt{x})^{2} = (-1)^{2}\)
\(25 x = 1\)
\(x = \frac{1}{25}\)

1p

○

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

1p

4p

c

\(-6 x + 2 \sqrt{x} = -8\)

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

c

(Isoleren)
\(-6 x + 8 = -2 \sqrt{x}\)

1p

○

(Kwadrateren)
\((-6 x + 8)^{2} = (-2 \sqrt{x})^{2}\)
\(36 x^{2} - 96 x + 64 = 4 x\)

1p

○

(Oplossen)
\(36 x^{2} - 100 x + 64 = 0\)
\(9 x^{2} - 25 x + 16 = 0\)
\(D = (-25)^{2} - 4 ⋅ 9 ⋅ 16 = 49\)
\(x = {25 - \sqrt{49} \over 2 ⋅ 9} ∨ x = {25 + \sqrt{49} \over 2 ⋅ 9}\)
\(x = 1 ∨ x = 1\frac{7}{9}\)

1p

○

(Controleren)
\(x = 1\) voldoet niet, \(x = 1\frac{7}{9}\) voldoet.

1p

4p

d

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

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

d

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

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

\(4 x - 7 \sqrt{2 x - 7} = 8\)

Wortel (5)
008r - Wortelvergelijkingen - basis - 594ms - dynamic variables

○

(Isoleren)
\(4 x - 8 = 7 \sqrt{2 x - 7}\)

1p

○

(Kwadrateren)
\((4 x - 8)^{2} = (7 \sqrt{2 x - 7})^{2}\)
\(16 x^{2} - 64 x + 64 = 49 ⋅ (2 x - 7)\)
\(16 x^{2} - 64 x + 64 = 98 x - 343\)

1p

○

(Oplossen)
\(16 x^{2} - 162 x + 407 = 0\)
\(D = (-162)^{2} - 4 ⋅ 16 ⋅ 407 = 196\)
\(x = {162 - \sqrt{196} \over 2 ⋅ 16} ∨ x = {162 + \sqrt{196} \over 2 ⋅ 16}\)
\(x = 4\frac{5}{8} ∨ x = 5\frac{1}{2}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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