Getal & Ruimte (13e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 5.2 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{7 x + 18}\)

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

a

(Kwadrateren)
\(x^{2} = 7 x + 18\)

1p

○

(Oplossen)
\(1 x^{2} + -7 x + -18 = 0\)
\((x + 2) (x + -9) = 0\)
\(x = -2 ∨ x = 9\)

1p

○

(Controleren)
\(x = -2\) voldoet niet, \(x = 9\) voldoet.

1p

3p

b

\(8 + 6 \sqrt{x} = 9\)

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

b

(Isoleren)
\(6 \sqrt{x} = 1\)

1p

○

(Kwadrateren)
\((6 \sqrt{x})^{2} = 1^{2}\)
\(36 x = 1\)
\(x = \frac{1}{36}\)

1p

○

(Controleren)
\(x = \frac{1}{36}\) voldoet.

1p

4p

c

\(9 x - 3 \sqrt{x} = 6\)

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

c

(Isoleren)
\(9 x - 6 = 3 \sqrt{x}\)

1p

○

(Kwadrateren)
\((9 x - 6)^{2} = (3 \sqrt{x})^{2}\)
\(81 x^{2} - 108 x + 36 = 9 x\)

1p

○

(Oplossen)
\(81 x^{2} + -117 x + 36 = 0\)
\(9 x^{2} + -13 x + 4 = 0\)
\(D = -13^{2} - 4 ⋅ 9 ⋅ 4 = 25\)
\(x = {13 - \sqrt{25} \over 2 ⋅ 9} ∨ x = {13 + \sqrt{25} \over 2 ⋅ 9}\)
\(x = {4 \over 9} ∨ x = 1\)

1p

○

(Controleren)
\(x = \frac{4}{9}\) voldoet niet, \(x = 1\) voldoet.

1p

4p

d

\(x = \sqrt{9 x + 19} - 3\)

Wortel (3)
008q - Wortelvergelijkingen - basis - 1ms - dynamic variables

d

(Isoleren)
\(x + 3 = \sqrt{9 x + 19}\)

1p

○

(Kwadrateren)
\((x + 3)^{2} = (\sqrt{9 x + 19})^{2}\)
\(x^{2} + 6 x + 9 = 9 x + 19\)

1p

○

(Oplossen)
\(1 x^{2} + -3 x + -10 = 0\)
\((x + 2) (x + -5) = 0\)
\(x = -2 ∨ x = 5\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

\(8 x - 3 \sqrt{4 x - 2} = 3\)

Wortel (5)
008r - Wortelvergelijkingen - basis - 511ms - dynamic variables

○

(Isoleren)
\(8 x - 3 = 3 \sqrt{4 x - 2}\)

1p

○

(Kwadrateren)
\((8 x - 3)^{2} = (3 \sqrt{4 x - 2})^{2}\)
\(64 x^{2} - 48 x + 9 = 9 ⋅ (4 x - 2)\)
\(64 x^{2} - 48 x + 9 = 36 x - 18\)

1p

○

(Oplossen)
\(64 x^{2} + -84 x + 27 = 0\)
\(D = -84^{2} - 4 ⋅ 64 ⋅ 27 = 144\)
\(x = {84 - \sqrt{144} \over 2 ⋅ 64} ∨ x = {84 + \sqrt{144} \over 2 ⋅ 64}\)
\(x = {9 \over 16} ∨ x = {3 \over 4}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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