Moderne Wiskunde (12.1e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 1.3 Wortelvergelijkingen

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-2 x + 24}\)

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

a

(Kwadrateren)
\(x^{2} = -2 x + 24\)

1p

○

(Oplossen)
\(1 x^{2} + 2 x + -24 = 0\)
\((x + 6) (x + -4) = 0\)
\(x = -6 ∨ x = 4\)

1p

○

(Controleren)
\(x = -6\) voldoet niet, \(x = 4\) voldoet.

1p

3p

b

\(4 + 2 \sqrt{x} = 8\)

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

b

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

1p

○

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

1p

○

(Controleren)
\(x = 4\) voldoet.

1p

4p

c

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

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

c

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

1p

○

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

1p

○

(Oplossen)
\(4 x^{2} + -20 x + 16 = 0\)
\(1 x^{2} + -5 x + 4 = 0\)
\((x + -1) (x + -4) = 0\)
\(x = 1 ∨ x = 4\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

4p

d

\(x = \sqrt{5 x + 39} - 9\)

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

d

(Isoleren)
\(x + 9 = \sqrt{5 x + 39}\)

1p

○

(Kwadrateren)
\((x + 9)^{2} = (\sqrt{5 x + 39})^{2}\)
\(x^{2} + 18 x + 81 = 5 x + 39\)

1p

○

(Oplossen)
\(1 x^{2} + 13 x + 42 = 0\)
\((x + 7) (x + 6) = 0\)
\(x = -7 ∨ x = -6\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

\(5 x - 2 \sqrt{6 x - 5} = 3\)

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

○

(Isoleren)
\(5 x - 3 = 2 \sqrt{6 x - 5}\)

1p

○

(Kwadrateren)
\((5 x - 3)^{2} = (2 \sqrt{6 x - 5})^{2}\)
\(25 x^{2} - 30 x + 9 = 4 ⋅ (6 x - 5)\)
\(25 x^{2} - 30 x + 9 = 24 x - 20\)

1p

○

(Oplossen)
\(25 x^{2} + -54 x + 29 = 0\)
\(D = -54^{2} - 4 ⋅ 25 ⋅ 29 = 16\)
\(x = {54 - \sqrt{16} \over 2 ⋅ 25} ∨ x = {54 + \sqrt{16} \over 2 ⋅ 25}\)
\(x = 1 ∨ x = {29 \over 25}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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