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

'Wortelvergelijkingen'.

havo wiskunde B 5.3 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-6 x + 16}\)

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

a

(Kwadrateren)
\(x^{2} = -6 x + 16\)

1p

○

(Oplossen)
\(x^{2} + 6 x - 16 = 0\)
\((x + 8) (x - 2) = 0\)
\(x = -8 ∨ x = 2\)

1p

○

(Controleren)
\(x = -8\) voldoet niet, \(x = 2\) voldoet.

1p

3p

b

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

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

b

(Isoleren)
\(-7 \sqrt{x} = -5\)

1p

○

(Kwadrateren)
\((-7 \sqrt{x})^{2} = (-5)^{2}\)
\(49 x = 25\)
\(x = \frac{25}{49}\)

1p

○

(Controleren)
\(x = \frac{25}{49}\) voldoet.

1p

4p

c

\(4 x + 3 \sqrt{x} = 7\)

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

c

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

1p

○

(Kwadrateren)
\((4 x - 7)^{2} = (-3 \sqrt{x})^{2}\)
\(16 x^{2} - 56 x + 49 = 9 x\)

1p

○

(Oplossen)
\(16 x^{2} - 65 x + 49 = 0\)
\(D = (-65)^{2} - 4 ⋅ 16 ⋅ 49 = 1\,089\)
\(x = {65 - \sqrt{1\,089} \over 2 ⋅ 16} ∨ x = {65 + \sqrt{1\,089} \over 2 ⋅ 16}\)
\(x = 1 ∨ x = 3\frac{1}{16}\)

1p

○

(Controleren)
\(x = 1\) voldoet, \(x = 3\frac{1}{16}\) voldoet niet.

1p

4p

d

\(x = \sqrt{8 x + 44} - 3\)

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

d

(Isoleren)
\(x + 3 = \sqrt{8 x + 44}\)

1p

○

(Kwadrateren)
\((x + 3)^{2} = (\sqrt{8 x + 44})^{2}\)
\(x^{2} + 6 x + 9 = 8 x + 44\)

1p

○

(Oplossen)
\(x^{2} - 2 x - 35 = 0\)
\((x + 5) (x - 7) = 0\)
\(x = -5 ∨ x = 7\)

1p

○

(Controleren)
\(x = 7\) voldoet, \(x = -5\) voldoet niet.

1p

opgave 2

Los exact op.

4p

\(3 x - 2 \sqrt{6 x - 8} = 4\)

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

○

(Isoleren)
\(3 x - 4 = 2 \sqrt{6 x - 8}\)

1p

○

(Kwadrateren)
\((3 x - 4)^{2} = (2 \sqrt{6 x - 8})^{2}\)
\(9 x^{2} - 24 x + 16 = 4 ⋅ (6 x - 8)\)
\(9 x^{2} - 24 x + 16 = 24 x - 32\)

1p

○

(Oplossen)
\(9 x^{2} - 48 x + 48 = 0\)
\(3 x^{2} - 16 x + 16 = 0\)
\(D = (-16)^{2} - 4 ⋅ 3 ⋅ 16 = 64\)
\(x = {16 - \sqrt{64} \over 2 ⋅ 3} ∨ x = {16 + \sqrt{64} \over 2 ⋅ 3}\)
\(x = 1\frac{1}{3} ∨ x = 4\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
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