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

'Wortelvergelijkingen'.

havo wiskunde B 5.3 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{3 x + 54}\)

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

a

(Kwadrateren)
\(x^{2} = 3 x + 54\)

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

3p

b

\(7 - 8 \sqrt{x} = 6\)

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

b

(Isoleren)
\(-8 \sqrt{x} = -1\)

1p

○

(Kwadrateren)
\((-8 \sqrt{x})^{2} = (-1)^{2}\)
\(64 x = 1\)
\(x = \frac{1}{64}\)

1p

○

(Controleren)
\(x = \frac{1}{64}\) voldoet.

1p

4p

c

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

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

c

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

1p

○

(Kwadrateren)
\((6 x - 8)^{2} = (2 \sqrt{x})^{2}\)
\(36 x^{2} - 96 x + 64 = 4 x\)

1p

○

(Oplossen)
\(36 x^{2} + -100 x + 64 = 0\)
\(9 x^{2} + -25 x + 16 = 0\)
\(D = -25^{2} - 4 ⋅ 9 ⋅ 16 = 49\)
\(x = {25 - \sqrt{49} \over 2 ⋅ 9} ∨ x = {25 + \sqrt{49} \over 2 ⋅ 9}\)
\(x = 1 ∨ x = {16 \over 9}\)

1p

○

(Controleren)
\(x = 1\) voldoet niet, \(x = 1\frac{7}{9}\) voldoet.

1p

4p

d

\(x = \sqrt{2 x + 22} - 7\)

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

d

(Isoleren)
\(x + 7 = \sqrt{2 x + 22}\)

1p

○

(Kwadrateren)
\((x + 7)^{2} = (\sqrt{2 x + 22})^{2}\)
\(x^{2} + 14 x + 49 = 2 x + 22\)

1p

○

(Oplossen)
\(1 x^{2} + 12 x + 27 = 0\)
\((x + 9) (x + 3) = 0\)
\(x = -9 ∨ x = -3\)

1p

○

(Controleren)
\(x = -9\) voldoet niet, \(x = -3\) voldoet.

1p

opgave 2

Los exact op.

4p

\(3 x - 2 \sqrt{5 x - 4} = 5\)

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

○

(Isoleren)
\(3 x - 5 = 2 \sqrt{5 x - 4}\)

1p

○

(Kwadrateren)
\((3 x - 5)^{2} = (2 \sqrt{5 x - 4})^{2}\)
\(9 x^{2} - 30 x + 25 = 4 ⋅ (5 x - 4)\)
\(9 x^{2} - 30 x + 25 = 20 x - 16\)

1p

○

(Oplossen)
\(9 x^{2} + -50 x + 41 = 0\)
\(D = -50^{2} - 4 ⋅ 9 ⋅ 41 = 1024\)
\(x = {50 - \sqrt{1024} \over 2 ⋅ 9} ∨ x = {50 + \sqrt{1024} \over 2 ⋅ 9}\)
\(x = 1 ∨ x = {41 \over 9}\)

1p

○

(Controleren)
\(x = 1\) voldoet niet, \(x = 4\frac{5}{9}\) voldoet.

1p
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