Moderne Wiskunde (12.1e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 1.3 Wortelvergelijkingen

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{x + 20}\)

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

a

(Kwadrateren)
\(x^{2} = x + 20\)

1p

○

(Oplossen)
\(1 x^{2} + -1 x + -20 = 0\)
\((x + 4) (x + -5) = 0\)
\(x = -4 ∨ x = 5\)

1p

○

(Controleren)
\(x = -4\) voldoet niet, \(x = 5\) voldoet.

1p

3p

b

\(6 - 2 \sqrt{x} = 5\)

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

b

(Isoleren)
\(-2 \sqrt{x} = -1\)

1p

○

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

1p

○

(Controleren)
\(x = \frac{1}{4}\) voldoet.

1p

4p

c

\(4 x - 6 \sqrt{x} = -2\)

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

c

(Isoleren)
\(4 x + 2 = 6 \sqrt{x}\)

1p

○

(Kwadrateren)
\((4 x + 2)^{2} = (6 \sqrt{x})^{2}\)
\(16 x^{2} + 16 x + 4 = 36 x\)

1p

○

(Oplossen)
\(16 x^{2} + -20 x + 4 = 0\)
\(4 x^{2} + -5 x + 1 = 0\)
\(D = -5^{2} - 4 ⋅ 4 ⋅ 1 = 9\)
\(x = {5 - \sqrt{9} \over 2 ⋅ 4} ∨ x = {5 + \sqrt{9} \over 2 ⋅ 4}\)
\(x = {1 \over 4} ∨ x = 1\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

4p

d

\(x = \sqrt{7 x - 14} + 2\)

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

d

(Isoleren)
\(x - 2 = \sqrt{7 x - 14}\)

1p

○

(Kwadrateren)
\((x - 2)^{2} = (\sqrt{7 x - 14})^{2}\)
\(x^{2} - 4 x + 4 = 7 x - 14\)

1p

○

(Oplossen)
\(1 x^{2} + -11 x + 18 = 0\)
\((x + -2) (x + -9) = 0\)
\(x = 2 ∨ x = 9\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

\(3 x - 3 \sqrt{3 x - 3} = 3\)

Wortel (5)
008r - Wortelvergelijkingen - basis - 560ms - dynamic variables

○

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

1p

○

(Kwadrateren)
\((3 x - 3)^{2} = (3 \sqrt{3 x - 3})^{2}\)
\(9 x^{2} - 18 x + 9 = 9 ⋅ (3 x - 3)\)
\(9 x^{2} - 18 x + 9 = 27 x - 27\)

1p

○

(Oplossen)
\(9 x^{2} + -45 x + 36 = 0\)
\(1 x^{2} + -5 x + 4 = 0\)
\((x + -1) (x + -4) = 0\)
\(x = 1 ∨ x = 4\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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