Moderne Wiskunde (12.1e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 1.3 Wortelvergelijkingen

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-4 x + 21}\)

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

a

(Kwadrateren)
\(x^{2} = -4 x + 21\)

1p

○

(Oplossen)
\(1 x^{2} + 4 x + -21 = 0\)
\((x + 7) (x + -3) = 0\)
\(x = -7 ∨ x = 3\)

1p

○

(Controleren)
\(x = -7\) voldoet niet, \(x = 3\) voldoet.

1p

3p

b

\(6 + 4 \sqrt{x} = 7\)

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

b

(Isoleren)
\(4 \sqrt{x} = 1\)

1p

○

(Kwadrateren)
\((4 \sqrt{x})^{2} = 1^{2}\)
\(16 x = 1\)
\(x = \frac{1}{16}\)

1p

○

(Controleren)
\(x = \frac{1}{16}\) voldoet.

1p

4p

c

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

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

c

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

1p

○

(Kwadrateren)
\((4 x - 8)^{2} = (4 \sqrt{x})^{2}\)
\(16 x^{2} - 64 x + 64 = 16 x\)

1p

○

(Oplossen)
\(16 x^{2} + -80 x + 64 = 0\)
\(1 x^{2} + -5 x + 4 = 0\)
\((x + -1) (x + -4) = 0\)
\(x = 1 ∨ x = 4\)

1p

○

(Controleren)
\(x = 1\) voldoet niet, \(x = 4\) voldoet.

1p

4p

d

\(x = \sqrt{3 x + 76} - 2\)

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

d

(Isoleren)
\(x + 2 = \sqrt{3 x + 76}\)

1p

○

(Kwadrateren)
\((x + 2)^{2} = (\sqrt{3 x + 76})^{2}\)
\(x^{2} + 4 x + 4 = 3 x + 76\)

1p

○

(Oplossen)
\(1 x^{2} + 1 x + -72 = 0\)
\((x + 9) (x + -8) = 0\)
\(x = -9 ∨ x = 8\)

1p

○

(Controleren)
\(x = 8\) voldoet, \(x = -9\) voldoet niet.

1p

opgave 2

Los exact op.

4p

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

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

○

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

1p

○

(Kwadrateren)
\((2 x - 7)^{2} = (4 \sqrt{2 x - 7})^{2}\)
\(4 x^{2} - 28 x + 49 = 16 ⋅ (2 x - 7)\)
\(4 x^{2} - 28 x + 49 = 32 x - 112\)

1p

○

(Oplossen)
\(4 x^{2} + -60 x + 161 = 0\)
\(D = -60^{2} - 4 ⋅ 4 ⋅ 161 = 1024\)
\(x = {60 - \sqrt{1024} \over 2 ⋅ 4} ∨ x = {60 + \sqrt{1024} \over 2 ⋅ 4}\)
\(x = {7 \over 2} ∨ x = {23 \over 2}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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