Getal & Ruimte (12e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 5.3 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{x + 42}\)

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

a

(Kwadrateren)
\(x^{2} = x + 42\)

1p

○

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

1p

○

(Controleren)
\(x = -6\) voldoet niet, \(x = 7\) voldoet.

1p

3p

b

\(2 + 7 \sqrt{x} = 6\)

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

b

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

1p

○

(Kwadrateren)
\((7 \sqrt{x})^{2} = 4^{2}\)
\(49 x = 16\)
\(x = \frac{16}{49}\)

1p

○

(Controleren)
\(x = \frac{16}{49}\) voldoet.

1p

4p

c

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

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

c

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

1p

○

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

1p

○

(Oplossen)
\(9 x^{2} + -34 x + 25 = 0\)
\(D = -34^{2} - 4 ⋅ 9 ⋅ 25 = 256\)
\(x = {34 - \sqrt{256} \over 2 ⋅ 9} ∨ x = {34 + \sqrt{256} \over 2 ⋅ 9}\)
\(x = 1 ∨ x = {25 \over 9}\)

1p

○

(Controleren)
\(x = 1\) voldoet, \(x = 2\frac{7}{9}\) voldoet niet.

1p

4p

d

\(x = \sqrt{8 x - 23} + 2\)

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

d

(Isoleren)
\(x - 2 = \sqrt{8 x - 23}\)

1p

○

(Kwadrateren)
\((x - 2)^{2} = (\sqrt{8 x - 23})^{2}\)
\(x^{2} - 4 x + 4 = 8 x - 23\)

1p

○

(Oplossen)
\(1 x^{2} + -12 x + 27 = 0\)
\((x + -3) (x + -9) = 0\)
\(x = 3 ∨ x = 9\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

\(2 x - 2 \sqrt{5 x + 4} = 8\)

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

○

(Isoleren)
\(2 x - 8 = 2 \sqrt{5 x + 4}\)

1p

○

(Kwadrateren)
\((2 x - 8)^{2} = (2 \sqrt{5 x + 4})^{2}\)
\(4 x^{2} - 32 x + 64 = 4 ⋅ (5 x + 4)\)
\(4 x^{2} - 32 x + 64 = 20 x + 16\)

1p

○

(Oplossen)
\(4 x^{2} + -52 x + 48 = 0\)
\(1 x^{2} + -13 x + 12 = 0\)
\((x + -1) (x + -12) = 0\)
\(x = 1 ∨ x = 12\)

1p

○

(Controleren)
\(x = 1\) voldoet niet, \(x = 12\) voldoet.

1p
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