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

'Wortelvergelijkingen'.

havo wiskunde B 5.2 Wortelfuncties

Wortelvergelijkingen (5)

opgave 1

Los exact op.

3p

a

\(x = \sqrt{7 x + 18}\)

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

a

(Kwadrateren)
\(x^{2} = 7 x + 18\)

1p

○

(Oplossen)
\(1 x^{2} + -7 x + -18 = 0\)
\((x + 2) (x + -9) = 0\)
\(x = -2 ∨ x = 9\)

1p

○

(Controleren)
\(x = -2\) voldoet niet, \(x = 9\) voldoet.

1p

3p

b

\(3 + 8 \sqrt{x} = 7\)

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

b

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

1p

○

(Kwadrateren)
\((8 \sqrt{x})^{2} = 4^{2}\)
\(64 x = 16\)
\(x = \frac{1}{4}\)

1p

○

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

1p

4p

c

\(-2 x + 3 \sqrt{x} = -2\)

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

c

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

1p

○

(Kwadrateren)
\((-2 x + 2)^{2} = (-3 \sqrt{x})^{2}\)
\(4 x^{2} - 8 x + 4 = 9 x\)

1p

○

(Oplossen)
\(4 x^{2} + -17 x + 4 = 0\)
\(D = -17^{2} - 4 ⋅ 4 ⋅ 4 = 225\)
\(x = {17 - \sqrt{225} \over 2 ⋅ 4} ∨ x = {17 + \sqrt{225} \over 2 ⋅ 4}\)
\(x = {1 \over 4} ∨ x = 4\)

1p

○

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

1p

4p

d

\(x = \sqrt{3 x + 37} + 1\)

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

d

(Isoleren)
\(x - 1 = \sqrt{3 x + 37}\)

1p

○

(Kwadrateren)
\((x - 1)^{2} = (\sqrt{3 x + 37})^{2}\)
\(x^{2} - 2 x + 1 = 3 x + 37\)

1p

○

(Oplossen)
\(1 x^{2} + -5 x + -36 = 0\)
\((x + 4) (x + -9) = 0\)
\(x = -4 ∨ x = 9\)

1p

○

(Controleren)
\(x = 9\) voldoet, \(x = -4\) voldoet niet.

1p

opgave 2

Los exact op.

4p

\(5 x - 8 \sqrt{3 x - 8} = 4\)

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

○

(Isoleren)
\(5 x - 4 = 8 \sqrt{3 x - 8}\)

1p

○

(Kwadrateren)
\((5 x - 4)^{2} = (8 \sqrt{3 x - 8})^{2}\)
\(25 x^{2} - 40 x + 16 = 64 ⋅ (3 x - 8)\)
\(25 x^{2} - 40 x + 16 = 192 x - 512\)

1p

○

(Oplossen)
\(25 x^{2} + -232 x + 528 = 0\)
\(D = -232^{2} - 4 ⋅ 25 ⋅ 528 = 1024\)
\(x = {232 - \sqrt{1024} \over 2 ⋅ 25} ∨ x = {232 + \sqrt{1024} \over 2 ⋅ 25}\)
\(x = 4 ∨ x = {132 \over 25}\)

1p

○

(Controleren)
Beide oplossingen voldoen.

1p
"