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

'Wortelvergelijkingen'.

havo wiskunde B 5.3 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-x + 20}\)

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

a

(Kwadrateren)
\(x^{2} = -x + 20\)

1p

○

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

1p

○

(Controleren)
\(x = -5\) voldoet niet, \(x = 4\) voldoet.

1p

3p

b

\(8 - 3 \sqrt{x} = 5\)

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

b

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

1p

○

(Kwadrateren)
\((-3 \sqrt{x})^{2} = (-3)^{2}\)
\(9 x = 9\)
\(x = 1\)

1p

○

(Controleren)
\(x = 1\) voldoet.

1p

4p

c

\(3 x - 8 \sqrt{x} = -5\)

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

c

(Isoleren)
\(3 x + 5 = 8 \sqrt{x}\)

1p

○

(Kwadrateren)
\((3 x + 5)^{2} = (8 \sqrt{x})^{2}\)
\(9 x^{2} + 30 x + 25 = 64 x\)

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)
Beide oplossingen voldoen.

1p

4p

d

\(x = \sqrt{5 x + 55} - 1\)

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

d

(Isoleren)
\(x + 1 = \sqrt{5 x + 55}\)

1p

○

(Kwadrateren)
\((x + 1)^{2} = (\sqrt{5 x + 55})^{2}\)
\(x^{2} + 2 x + 1 = 5 x + 55\)

1p

○

(Oplossen)
\(1 x^{2} + -3 x + -54 = 0\)
\((x + 6) (x + -9) = 0\)
\(x = -6 ∨ x = 9\)

1p

○

(Controleren)
\(x = -6\) voldoet niet, \(x = 9\) voldoet.

1p

opgave 2

Los exact op.

4p

\(4 x - 7 \sqrt{3 x - 9} = 3\)

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

○

(Isoleren)
\(4 x - 3 = 7 \sqrt{3 x - 9}\)

1p

○

(Kwadrateren)
\((4 x - 3)^{2} = (7 \sqrt{3 x - 9})^{2}\)
\(16 x^{2} - 24 x + 9 = 49 ⋅ (3 x - 9)\)
\(16 x^{2} - 24 x + 9 = 147 x - 441\)

1p

○

(Oplossen)
\(16 x^{2} + -171 x + 450 = 0\)
\(D = -171^{2} - 4 ⋅ 16 ⋅ 450 = 441\)
\(x = {171 - \sqrt{441} \over 2 ⋅ 16} ∨ x = {171 + \sqrt{441} \over 2 ⋅ 16}\)
\(x = {75 \over 16} ∨ x = 6\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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