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

'Wortelvergelijkingen'.

havo wiskunde B 5.2 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-4 x + 45}\)

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

a

(Kwadrateren)
\(x^{2} = -4 x + 45\)

1p

○

(Oplossen)
\(x^{2} + 4 x - 45 = 0\)
\((x + 9) (x - 5) = 0\)
\(x = -9 ∨ x = 5\)

1p

○

(Controleren)
\(x = -9\) voldoet niet, \(x = 5\) 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

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

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

1p

○

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

1p

4p

d

\(x = \sqrt{4 x + 61} - 4\)

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

d

(Isoleren)
\(x + 4 = \sqrt{4 x + 61}\)

1p

○

(Kwadrateren)
\((x + 4)^{2} = (\sqrt{4 x + 61})^{2}\)
\(x^{2} + 8 x + 16 = 4 x + 61\)

1p

○

(Oplossen)
\(x^{2} + 4 x - 45 = 0\)
\((x + 9) (x - 5) = 0\)
\(x = -9 ∨ x = 5\)

1p

○

(Controleren)
\(x = -9\) voldoet niet, \(x = 5\) voldoet.

1p

opgave 2

Los exact op.

4p

\(6 x - 5 \sqrt{2 x - 2} = 6\)

Wortel (5)
008r - Wortelvergelijkingen - basis - 594ms - dynamic variables

○

(Isoleren)
\(6 x - 6 = 5 \sqrt{2 x - 2}\)

1p

○

(Kwadrateren)
\((6 x - 6)^{2} = (5 \sqrt{2 x - 2})^{2}\)
\(36 x^{2} - 72 x + 36 = 25 ⋅ (2 x - 2)\)
\(36 x^{2} - 72 x + 36 = 50 x - 50\)

1p

○

(Oplossen)
\(36 x^{2} - 122 x + 86 = 0\)
\(18 x^{2} - 61 x + 43 = 0\)
\(D = (-61)^{2} - 4 ⋅ 18 ⋅ 43 = 625\)
\(x = {61 - \sqrt{625} \over 2 ⋅ 18} ∨ x = {61 + \sqrt{625} \over 2 ⋅ 18}\)
\(x = 1 ∨ x = 2\frac{7}{18}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
"