Moderne Wiskunde (12.1e editie) - havo wiskunde B

'Wortelvergelijkingen'.

havo wiskunde B 1.3 Wortelvergelijkingen

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-x + 72}\)

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

a

(Kwadrateren)
\(x^{2} = -x + 72\)

1p

○

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

1p

○

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

1p

3p

b

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

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

b

(Isoleren)
\(-5 \sqrt{x} = -1\)

1p

○

(Kwadrateren)
\((-5 \sqrt{x})^{2} = (-1)^{2}\)
\(25 x = 1\)
\(x = \frac{1}{25}\)

1p

○

(Controleren)
\(x = \frac{1}{25}\) voldoet.

1p

4p

c

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

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

c

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

1p

○

(Kwadrateren)
\((3 x - 2)^{2} = (-5 \sqrt{x})^{2}\)
\(9 x^{2} - 12 x + 4 = 25 x\)

1p

○

(Oplossen)
\(9 x^{2} + -37 x + 4 = 0\)
\(D = -37^{2} - 4 ⋅ 9 ⋅ 4 = 1225\)
\(x = {37 - \sqrt{1225} \over 2 ⋅ 9} ∨ x = {37 + \sqrt{1225} \over 2 ⋅ 9}\)
\(x = {1 \over 9} ∨ x = 4\)

1p

○

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

1p

4p

d

\(x = \sqrt{5 x + 10} - 2\)

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

d

(Isoleren)
\(x + 2 = \sqrt{5 x + 10}\)

1p

○

(Kwadrateren)
\((x + 2)^{2} = (\sqrt{5 x + 10})^{2}\)
\(x^{2} + 4 x + 4 = 5 x + 10\)

1p

○

(Oplossen)
\(1 x^{2} + -1 x + -6 = 0\)
\((x + 2) (x + -3) = 0\)
\(x = -2 ∨ x = 3\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p

opgave 2

Los exact op.

4p

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

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

○

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

1p

○

(Kwadrateren)
\((6 x - 6)^{2} = (3 \sqrt{2 x - 2})^{2}\)
\(36 x^{2} - 72 x + 36 = 9 ⋅ (2 x - 2)\)
\(36 x^{2} - 72 x + 36 = 18 x - 18\)

1p

○

(Oplossen)
\(36 x^{2} + -90 x + 54 = 0\)
\(2 x^{2} + -5 x + 3 = 0\)
\(D = -5^{2} - 4 ⋅ 2 ⋅ 3 = 1\)
\(x = {5 - \sqrt{1} \over 2 ⋅ 2} ∨ x = {5 + \sqrt{1} \over 2 ⋅ 2}\)
\(x = 1 ∨ x = {3 \over 2}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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