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

'Wortelvergelijkingen'.

3 vwo 5.6 Wortelvergelijkingen

Wortelvergelijkingen (1)

opgave 1

Los exact op.

3p

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

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

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

1p

(Kwadrateren)
\((-9 \sqrt{x})^{2} = (-3)^{2}\)
\(81 x = 9\)
\(x = \frac{1}{9}\)

1p

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

1p

vwo wiskunde B 4.3 Regels voor het oplossen van vergelijkingen

Wortelvergelijkingen (4)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{-2 x + 35}\)

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

a

(Kwadrateren)
\(x^{2} = -2 x + 35\)

1p

(Oplossen)
\(1 x^{2} + 2 x + -35 = 0\)
\((x + 7) (x + -5) = 0\)
\(x = -7 ∨ x = 5\)

1p

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

1p

4p

b

\(9 x + 3 \sqrt{x} = 6\)

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

b

(Isoleren)
\(9 x - 6 = -3 \sqrt{x}\)

1p

(Kwadrateren)
\((9 x - 6)^{2} = (-3 \sqrt{x})^{2}\)
\(81 x^{2} - 108 x + 36 = 9 x\)

1p

(Oplossen)
\(81 x^{2} + -117 x + 36 = 0\)
\(9 x^{2} + -13 x + 4 = 0\)
\(D = -13^{2} - 4 ⋅ 9 ⋅ 4 = 25\)
\(x = {13 - \sqrt{25} \over 2 ⋅ 9} ∨ x = {13 + \sqrt{25} \over 2 ⋅ 9}\)
\(x = {4 \over 9} ∨ x = 1\)

1p

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

1p

4p

c

\(x = \sqrt{7 x + 35} - 5\)

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

c

(Isoleren)
\(x + 5 = \sqrt{7 x + 35}\)

1p

(Kwadrateren)
\((x + 5)^{2} = (\sqrt{7 x + 35})^{2}\)
\(x^{2} + 10 x + 25 = 7 x + 35\)

1p

(Oplossen)
\(1 x^{2} + 3 x + -10 = 0\)
\((x + 5) (x + -2) = 0\)
\(x = -5 ∨ x = 2\)

1p

(Controleren)
Beide oplossingen voldoen.

1p

4p

d

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

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

d

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

1p

(Kwadrateren)
\((3 x - 4)^{2} = (2 \sqrt{6 x - 3})^{2}\)
\(9 x^{2} - 24 x + 16 = 4 ⋅ (6 x - 3)\)
\(9 x^{2} - 24 x + 16 = 24 x - 12\)

1p

(Oplossen)
\(9 x^{2} + -48 x + 28 = 0\)
\(D = -48^{2} - 4 ⋅ 9 ⋅ 28 = 1296\)
\(x = {48 - \sqrt{1296} \over 2 ⋅ 9} ∨ x = {48 + \sqrt{1296} \over 2 ⋅ 9}\)
\(x = {2 \over 3} ∨ x = {14 \over 3}\)

1p

(Controleren)
\(x = \frac{2}{3}\) voldoet niet, \(x = 4\frac{2}{3}\) voldoet.

1p

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